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Magnetic Effect Of Current And Magnetism

Magnetic Properties Of Materials Introduction

Electric dipole end electric dipole moment:

From the chapter ‘Electric Field’ we know that, two equal but opposite charges +q and -q kept close to each other form an electric dipole.

Electric dipole moment \(\vec{p}\) of this di[o;e is defined as:

⇒ \(\vec{p}=q \vec{r}\)….(1)

= magnitude of any charge x position vector of charge +q with respect to charge -q

So, the magnitude of \(\vec{p} \text { is } p=|\vec{p}|=q r\) and the direction is from -q to +q.

Electric field on the axis of an electric dipole: Let p be a point on the axis of an electric dipole. If the distance

X of the point P from the mid-point of the dipole be much greater than the length r of the dipole, the glottic field at the point P due to that dipole,

⇒ \(\vec{E}=\frac{1}{4 \pi \epsilon_0} \cdot \frac{2 \vec{p}}{x^3}\)

Here, ∈₀ = permittivity of air or Vacuum

Read and Learn More Class 12 Physics Notes

= 8.854 x 10^-12 C².N^-1.m^-2

Torque on an electric dipole in a uniform electric field: \(\vec{E}\) is a uniform electric field (-q, +q) is an electric dipole whose dipole moment = \(\vec{p}\).

Now, the torque acting on the dipole due to the electric field \(\vec{E}\) is,

⇒ \(\vec{\tau}=\vec{p} \times \vec{E}\)

In the direction of \(\vec{\tau}\) is particularly downward concerning the page of the tired book, which Is denoted by the symbol\(\otimes\).

Magnetic field on the axis of a current loop:

r = radius of a circle conductor of a single turn, i.e., the radius of the current loop,

I = current in that loop.

P is any point on the axis of the current loop which is at a distance x from the center of the loop.

In the chapter ‘Electromagnetism’ we know that the magnetic field produced at the point P due to the current loop is,

⇒ \(B=\frac{\mu_0 I}{2} \cdot \frac{r^2}{\left(r^2+x^2\right)^{3 / 2}}\) [μ₀ = magnetic permeability of air or vacuum = 4π x 10^-7 Hm^-1]

⇒ \(\text { If } r \ll x \text {, then } B \approx \frac{\mu_0 I}{2} \cdot \frac{r^2}{x^3}=\frac{\mu_0 I}{2 \pi} \cdot \frac{\pi r^2}{x^3}=\frac{\mu_0 I}{2 \pi} \cdot \frac{A}{x^3}\)

where, A = πr² = area of the current loop.

∴ \(B=\frac{\mu_0}{4 \pi} \cdot \frac{2 I A}{x^3}\)

From the corkscrew rule we get, that the direction of \(\vec{B}\) is along the axis of the loop; the direction of \(\vec{B}\) at the point P is outward along the axis. Again, taking that direction as the direction of the area A, it can be expressed as \(\vec{A}\).

Therefore, \(\vec{B}=\frac{\mu_0}{4 \pi} \cdot \frac{2(\overrightarrow{I A})}{x^3}\)….(4)

Torque on a current loop in a uniform magnetic field:

In the chapter ‘Electromagnetism’ we know that, if a current loop of single turn is kept in a uniform magnetic field B, the torque acting on the loop is,

⇒ \(\vec{\tau}=\overrightarrow{I A} \times \vec{B}\)….(5)

The direction of this torque is perpendicularly downward concerning the page of the book, which is denoted by the symbol.

Magnetic Effect Of Current And Magnetism

Magnetic Properties Of Materials Magnetic Dipole And Magnetic Dipole Moment Or Magnetic Moment

Comparing equations (2) and (4) and at the same time equations (3) and (5) as described, we get,

1. Electric field \(\vec{E}\) in electrostatics plays the same role as that of magnetic field \(\vec{B}\) in magnetism.

2. The role of the quantity \(\frac{1}{4 \pi \epsilon_0}\) in electrostatics is the same as that of the quantity \(\frac{\mu_0}{4 \pi}\) in magnetism.

3. The role of electric dipole moment \(\vec{p}\) in electrostatics is the same as the quantity I\(\vec{A}\) related to a current loop in magnetism.

Inference: Any current loop behaves as a magnetic dipole. The dipole moment of this dipole is,

⇒ \(\vec{p}_m=I \vec{A}\)

where, I = current through the loop, \(\vec{A}\) = area vector of the loop.

The magnitude of \(\vec{A}\) is the same as the magnitude of the area of the loop; the direction of \(\vec{A}\) is in the direction of advancement of the screw-head when a right-handed screw is rotated in the direction of current I through the loop.

Naturally, if the current loop contains N turns, it’s the magnetic moment, becoming \(\vec{p}_m=N I \vec{A}\).

Unit of magnetic moment:

Unit of pm = unit of I x unit of A

= ampere.metrer² (A.m²)

In the CGS or Gaussian system:

The magnetic dipole moment of a current loop, \(\vec{p}_m \equiv I \vec{A}\); if the current loop contains N turns instead of a single turn, \(\vec{p}_m=N I \vec{A}\). Here, the units of

A, I and pm are cm², emu of current, and emu.m², respectively.

∴ 1 emu.cm² = 10A x 10^-4 m²

= 10^-3 A.m²

Significance: Any current loop behaves as a magnetic dipole-it means that a current loop and a magnet having a north and a south pole are qualitatively equivalent. This similarity is discussed with the help of the following two examples.

Similarities between a circular conductor and a magnet:

Magnetic lines of force near the center of a circular conductor are almost parallel to each other and they remain perpendicular to the plane of the circle.

So, almost a uniform magnetic field is generated at that region, which acts normally to the plane of the circular conductor.

From the properties of magnetic lines of force of a permanent magnet, we know that, if the circular conductor is replaced by a small permanent magnet in that region, similar lines of force will be obtained.

So, we can conclude that a circular current-carrying coil behaves as a permanent magnet.

We can also get the rule for the determination of the polarity of the circular conductor.

1. The face of a circular conductor on which the current appears to flow clockwise, develops a magnetic south pole.

2. The face of a circular conductor, on which the current appears to flow anticlockwise, develops a magnetic north pole.

With the help of the following experiment, the magnetic property of a circular conductor can be shown.

De la Rives floating battery: In a wide test tube some mercury is taken so that it can float upright in water

Some dilute sulphuric acid (H₂SO₄) is poured into the test tube above mercury, and zinc and copper plates are dipped in the acid so that a voltaic cell is formed. This is known as a floating battery.

Two conducting wires from the two plates are brought outside the test tube through a cork fitted at the top.

These two wires act as two poles of the battery. Now the two ends of a circular coil are joined with those two poles. Usually, the coil contains several number of turns instead of a single turn.

When the coil is connected to the battery, the whole system begins to oscillate about a vertical axis.

At last, when the battery comes to rest, the axis of the circular coil sets itself in the north-south direction.

From this directive property, it is understood that the current-carrying coil behaves as a magnet.

With the help of a bar magnet, it can also be observed that like poles repel and unlike poles attract each other.

Similarities between a current-carrying solenoid and a magnet:

The arrangement of the lines of force in the chapter ‘Electromagnetism’ is similar to the arrangement of lines of force of a bar magnet.

On the face, on which the direction of current flow appears clockwise, a magnetic south pole develops, and on the other face of the solenoid, a magnetic north pole develops.

So, we can conclude that a current-carrying solenoid behaves as a permanent bar magnet.

Experimental demonstration:

The magnetic properties of a solenoid can be shown by replacing the circular coil attached to the de la Rive’s floating battery with a solenoid.

In this case, also, the directive and the attractive or repulsive properties of the solenoid with another magnet are observed.

Actually, from the similarities of a magnet and a current loop, it can be concluded that magnetism is not a separate branch of physics, rather it is a part of electricity and this specific subject is known as electromagnetism.

Pule-sought of a Magnet:

In modem theory of magnetism, the concept of the pole-strength of a magnet is not essential; but for the comparison with electrostatics and also to get an idea about the old theory of magnetism, we may discuss here about the pole-strength of a magnet.

Let a bar magnet NS of effective length r be kept at an angle θ with the direction of a uniform magnetic f field \(\vec{B}\).

The effect on the N and S poles are equal and opposite; hence due to the magnetic field \(\vec{B}\), two equal but opposite forces will act on the two poles.

The distance between the line of action of the two forces, NC = rsinθ. So, the torque acting on the bar magnet due to these two forces,

\(\tau\) = magnitude of any one force x perpendicular distance between the two forces

= Frsinθ….(1)

substituting \(\vec{p}_m=I \vec{A}\) we get,

⇒ \(\vec{\tau}=\vec{p}_m \times \vec{B}\)

We get the value of the torque \(\vec{\tau}\),

⇒ \(\tau=p_m B \sin \theta\)

Since a current loop and a magnet are identical, from equations (1) and (2) we get,

⇒ \(F r \sin \theta=p_m B \sin \theta \quad \text { or, } F=\frac{p_m}{r} B=q_m B\)….(3)

The term qm is known as the pole strength of a magnet. The value of the pole-strength of each of the N and S poles is qm; conventionally the pole-strength of the N pole is taken as + q_m and that of the S pole is taken as -q_m.

Definition:

The ratio of the magnetic moment of a magnet to its effective length is called the pole strength of that magnet.

The strength of the two poles of the magnet is equal but opposite; the strength of the north pole is taken as positive and that of the south pole as negative.

Using the vector symbol, equation (3) can be written as

⇒ \(\vec{p}_m=q_m \vec{r} \text { and } \vec{F}=q_m \vec{B}\)

These two equations are identical to the equations \(\vec{p}=q \vec{r} \text { and } \vec{F}=q \vec{E}\) in electrostatics.

Due to the similarities between the electric field \(\vec{E}\) and magnetic field \(\vec{B}\), we can say that the role of positive arid negative charges (±q) in electrostatics is the same as that of the north and south poles (±qm) in electromagnetism.

Unit of pole-strength:

In SI: \(\text { Unit of } q_m=\frac{\text { unit of } p_m}{\text { unit of } r}=\frac{\mathrm{A} \cdot \mathrm{m}^2}{\mathrm{~m}}=\mathrm{A} \cdot \mathrm{m}\)

In CGS: Unit of qm is emu of current cm;

⇒ \(1 \mathrm{emu} \text { of current } \cdot \mathrm{cm}=10 \mathrm{~A} \times 10^{-2} \mathrm{~m}=\frac{1}{10} \mathrm{~A} \cdot \mathrm{m}\)

The magnetic moment of a magnet: The magnetic moment of a magnet can also be defined from the concept of pole strength.

Definition: The product of the pole-strength of any pole of a magnet and its effective, length is called the magnetic moment of that magnet.

If the distance vector from the south pole to the north pole of a bar magnet is 2\(\vec{l}\) and its pole strength is qm, then the magnetic moment,

⇒ \(\vec{p}_m=2 q_m \vec{l}\)

Mutual force between two magnetic poles: We know that, for a charge q, the electric field at any point at a distance r from the charge = force acting on unit positive charge placed at that point,

⇒ \(\text { i.e., } E=\frac{1}{4 \pi \epsilon_0} \cdot \frac{q}{r^2}\) [∈₀ = electrical permittivity of vacuum]

From analogy, we can say that the magnetic field at any point at a distance r from a magnetic pole of pole-strength q_m = force acting on a unit north pole placed at that point,

⇒ \(\text { i.e., } B=\frac{\mu_0}{4 \pi} \cdot \frac{q_m}{r^2}\) [μ₀ = magnetic permeability of vacuum]

If another magnetic pole of pole-strength ‘m is placed at a distance r from q_m, from the equation F = q_mB we can say that, force acting on that pole

⇒ \(F=\frac{\mu_0}{4 \pi} \cdot \frac{q_m \cdot q_m^{\prime}}{r^2}\)…(4)

According to Newton’s third law of motion, it is the mutual force acting between the poles having pole strengths qm and I’m. Equation (4) is called Coulomb’s law in magnetism.

If both the poles are north poles or both the poles are south poles, the product qmq’m becomes positive, and hence F is also positive. It means that the direction of F is the same as r, i.e., the force F is repulsive.

If the two poles are unlike, the product qmq am becomes negative. It means that F is in the direction opposite to r, i.e., the force F is attractive. From equation (4), we can say that the force acting between two magnetic poles is,

1. Directly proportional to the product of the pole strengths of the two poles, and

2. Inversely proportional to the square of the distance between the two poles.

In electrostatics, the similar law is the Coulomb’s law:

⇒ \(F=\frac{1}{4 \pi \epsilon_0} \cdot \frac{q q^{\prime}}{r^2}\)

In CGS or Gaussian system:

The corresponding relations of electric field E, magnetic intelligence H, and magnetic force Fin vacuum or in air are respectively,

⇒ \(E=\frac{q}{r^2}, H=\frac{q_m}{r^2} \text { and } F=\frac{q_m \dot{q}_m^{\prime}}{r^2}\)

Magnetic Field due to a Bar Magnet:

End-on or axial position:

A bar magnet SN is kept in the air, Its pole strength is q_m, and its magnetic length = SN = 21. A point P is taken on the axis of the magnet at a distance d from its mid-point O’.

The position of point P is called the end-on or axial position: We have to determine the magnetic field at point F, due to the magnet

Calculation: Magnetic field at the point P due to N-pole,

⇒ \(B_1=\frac{\mu_0}{4 \pi} \cdot \frac{q_m}{(d-l)^2} ; \text { the direction of } \vec{B}_1 \text { is along } \overrightarrow{O P}\)

Again, the magnetic field at the point P due to S-the pole,

⇒ \(B_2=\frac{\mu_0}{4 \pi} \cdot \frac{q_m}{(d+l)^2} ; \text { the direction of } \vec{B}_2 \text { is along } \overrightarrow{P O}\)

∵ The N-pole is nearer to P than the S-pole, and the value of B₁ is greater than that of B₂.

Therefore, the resultant magnetic field at the point P,

⇒ \(B=B_1-B_2=\frac{\mu_0}{4 \pi}\left[\frac{q_m}{(d-l)^2}-\frac{q_m}{(d+l)^2}\right]\)

⇒ \(\frac{\mu_0}{4 \pi} \cdot \frac{q_m \cdot 4 d l}{\left(d^2-l^2\right)^2}\)

= \(\frac{\mu_0}{4 \pi} \cdot \frac{2 p_m d}{\left(d^2-l^2\right)^2}\) [p_m = q_m.21 = magnetic moment of the bar magnet]

The direction of \(\vec{B}\) is along \(\vec{OP}\).

The vector representation of the equation is,

⇒ \(\vec{B}=\frac{\mu_0}{4 \pi} \cdot \frac{2 \vec{p}_m d}{\left(d^2-l^2\right)^2}\)

The resultant magnetic intensity in the CGS system

⇒ \(H=\frac{2 p_m d}{\left(d^2-l^2\right)^2}\)

If the length of the bar magnet is very small and the point p is taken at a very- large distance from the bar magnet, l.e., d>>l, then (d²- l²)² ≈ d⁴.

∴ \(B=\frac{\mu_0}{4 \pi} \cdot \frac{2 p_m}{d^3}\)

Broadside on or equatorial position:

A bar magnet SN is kept in the air, its pole strength is q_m, and magnetic length = SN = 2l.

A point P is taken on the perpendicular bisector of the magnetic length and at a distance d from the mid-point O of the magnet.

The position of the point p is called the broadside-oil or equatorial position. We have to determine the magnetic field at the point P due to the magnet.

Calculation: Magnetic field at the point P due to N-pole,

⇒ \(B_1=\frac{\mu_0}{4 \pi} \cdot \frac{q_m}{N P^2} \text {; the direction of } \vec{B}_1 \text { is along } \overrightarrow{P Q}\)

Again, the magnetic field at the point P due to the S-pole,

⇒ \(B_2=\frac{\mu_0}{4 \pi} \cdot \frac{q_m}{S P^2} ; \text { the direction of } \vec{B}_2 \text { is along } \overrightarrow{P S}\)

The components of B₁ and B₂, i.e., B₁sinθ and B₂sinθ, along with PT and PO respectively, cancel each other.

On the other hand, the components B₁cosθ and B₂cosθ along PR are added together.

So, the resultant magnetic field at the point P,

B = 2B₁cosθ [∵ magnitudes of B₁ , B₂ are equal]

or, \(B=\frac{2 \mu_0}{4 \pi} \cdot \frac{q_m}{N P^2} \cos \theta=\frac{2 \mu_0}{4 \pi} \cdot \frac{q_m}{\left(d^2+l^2\right)} \cdot \frac{l}{\sqrt{d^2+l^2}}\)

⇒ \(\frac{\mu_0}{4 \pi} \cdot \frac{p_m}{\left(d^2+l^2\right)^{3 / 2}} \text {; the direction of } \vec{B} \text { is along } \overrightarrow{P R}\)

The vector representation of the equation is,

⇒ \(\vec{B}=-\frac{\mu_0}{4 \pi} \cdot \frac{\vec{p}_m}{\left(d^2+l^2\right)^{3 / 2}}\)

The resultant magnetic intensity in the CGS system,

⇒ \(H=\frac{p_m}{\left(d^2+l^2\right)^{3 / 2}}\)

If the length of the bar magnet is very small and the point P is
taken at a very large distance from the bar magnet, i.e., l<<d

then, \(B=\frac{\mu_0}{4 \pi} \cdot \frac{p_m}{d^3}\)

So, magnetic field at the end-on or axial position = 2 x magnetic field at the broadside-on or equatorial position

Any position:

Concerning the bar magnet of magnetic length 2l, the point P under consideration is at the position (r, θ), i.e., the point P is at a distance r from the mid-point of the magnet and lies at an angle θ with the magnetic axis.

The component of pm along r is p_mcosθ; concerning this component, the point P is at the end of the position, and hence the magnetic field at the point P,

⇒ \(B_r=\frac{\mu_0}{4 \pi} \cdot \frac{2 p_m \cos \theta}{r^3}(∵ r \gg l)\)

Again, the component of pm normal to r is p_msinθ; concerning this component point P is at the equatorial position, and hence the magnetic field at point P,

⇒ \(B_\theta=\frac{\mu_0}{4 \pi} \cdot \frac{p_m \sin \theta}{r^3}\)

So, according to the resultant magnetic field at the point P,

⇒ \(B=\sqrt{B_r^2+B_\theta^2}=\sqrt{\left(\frac{\mu_0}{4 \pi} \cdot \frac{p_m}{r^3}\right)^2\left(4 \cos ^2 \theta+\sin ^2 \theta\right)}\)

⇒ \(\frac{\mu_0}{4 \pi} \cdot \frac{p_m}{r^3} \sqrt{3 \cos ^2 \theta+1}\)

The direction of \(\vec{B} \text { is along } \overrightarrow{P R} \text {; where } \angle T P R=\phi \text { (say) }\)

∴ \(\tan \phi=\frac{B_\theta}{B_r}=\frac{1}{2} \tan \theta\)

The resultant magnetic intensity in the CGS system,

⇒ \(H=\frac{p_m}{r^3} \sqrt{3 \cos ^2 \theta+1}\)

Equivalence of a solenoid and a bar magnet: Let us assume a solenoid of radius = a, length = 2l, number of turns per unit length = n. Current through it = I.

So, the total number of turns of the solenoid = 2ln, and hence the effective net current around its axis =2lnI. As the cross-sectional area of the solenoid = πa², so, the magnetic dipole moment of the current-carrying solenoid,

⇒ \(p_m=(2 \ln I) \cdot\left(\pi a^2\right)=2 \pi \ln a^2 I\)

Now the center (O) of the solenoid is taken as the origin and its axis as the x-axis. Considering a small length dx at a distance x from 0, we get a circular current-carrying coil with several turns dx. For this coil, the magnetic field is thus produced at a point P on the axis (where, OP = r and r>>a, r>>x),

⇒ \(d B=\frac{\mu_0(n d x) I}{2} \cdot \frac{a^2}{\left\{a^2+(r-x)^2\right\}^{3 / 2}}=\frac{\mu_0 n I a^2}{2 r^3} d x\) [As a and x are negligible with respect to r, \(\left\{a^2+(r-x)^2\right\}^{3 / 2} \approx r^3\)]

So, the magnetic field at P for the entire solenoid,

⇒ \(B=\frac{\mu_0 n I a^2}{2 r^3} \int_{-l}^l d x=\frac{\mu_0 n I a^2}{2 r^3} \cdot 2 l=\frac{\mu_0}{4 \pi} \cdot \frac{2\left(2 \pi \ln a^2 I\right)}{r^3}\)

i.e., \(B=\frac{\mu_0}{4 \pi} \frac{2 p_m}{r^3}\)

In this very section, we have already found the same expression due to a bar magnet. Thus we conclude that a bar magnet and a current-carrying solenoid are equivalent to each other.

Comparing their magnetic moments, we get,

⇒ \(q_m \cdot 2 l=2 \pi \ln a^2 I \quad \text { or, } q_m=\pi n a^2 I\)

Magnetic Moment of Charged Particle Moving in a Circle:

We know that a revolving charged particle is equivalent to an electric current. So, the orbit of that particle is equivalent to a current loop and as a result, it behaves as a magnetic dipole which must have a magnetic moment.

Suppose a particle having a charge q is revolving in a plane circular path; r = radius of the orbit of that particle and v = velocity of the particle.

The period of revolution of the particle, \(T=\frac{2 \pi r}{v}\)

∴ Equivalent current, \(I=\frac{q}{T}=\frac{q v}{2 \pi r}\)

So, the magnetic moment of the particle,

⇒ \(p_m=\frac{q \nu}{2 \pi r} \cdot A=\frac{q \nu}{2 \pi r} \cdot \pi r^2=\frac{q v r}{2}\)

The direction of this magnetic moment pm can be obtained by applying a corkscrew rule. If the mass of the particle is m then along the axis of the circular path, Le., along the direction of p_m, the angular momentum of the particle,

L = mv [L = moment of momentum = momentum x radius of circular path]

∴ \(p_m=\frac{q v r}{2}=\frac{q}{2 m} \cdot m v r=\frac{q}{2 m} L\)

Since the directions of pm and L are the same, using vector notation we can write,

⇒ \(\vec{p}_m=\frac{q}{2 m} \vec{L}\)…(1)

Magnetic dipole moment of an electron:

Electrons inside an atom revolve around the nucleus continuously. As a revolving charged particle, each electron behaves as a magnetic dipole. Substituting the charge of an electron -e in place of q in equation (1), we can write,

⇒ \(\vec{p}_m=-\frac{e}{2 m} \cdot \vec{L}\)…(2)

The negative sign on the right-hand side indicates that the charge of an electron is negative, its angular momentum and magnetic moment are oppositely directed.

From Bohr’s theory related to atomic structure: Atoms, it is known that electrons revolving in different orbits have discrete angular momenta, defined as

⇒ \(L=n \frac{h}{2 \pi}=n \hbar \quad[n=1,2,3, \cdots]\)

Here, \(h=\text { Planck’s constant }=6.626 \times 10^{-34} \mathrm{~J} \cdot \mathrm{s} \text { and } \dot{\hbar}=\frac{h}{2 \pi}\)

= 1.05 x 10^-34 J.s = reduced Planck’s constant

So, the magnitude of the magnetic moment of an electron [from equation (2)],

⇒ \(p_m=\frac{e}{2 m} n \hbar=n \frac{e \hbar}{2 m}\)….(3)

In the case of the innermost orbit of an atom (K-orbit), n = 1; then \(p_m=\frac{e \hbar}{2 m}\) – this quantity is called Bohr magneton. It is denoted by μ_B and it is the atomic unit of the magnetic moment

⇒ \(I \mu_B=\frac{e h}{2 m}=\frac{\left.\left(1.6 \times 10^{-19} \mathrm{C}\right) \times\left(1.05 \times 10^{-34}\right) \cdot s\right)}{2 \times\left(9.1 \times 10^{-31} \mathrm{~kg}\right)}\)

= 9.23 x 10^-24 A m²

= 9.23 x 10^-24 J. T^-1

The spin of an electron: Besides the revolution of each electron in its orbit, it also rotates about its axis (like the diurnal motion of the earth). This is known as the spin of an electron. An electron may have any of two oppositely directed spins.

One kind of spin (say clockwise) is called up spin and is denoted by ( symbol) The opposite kind of spin is then known as down spin and is denoted by the symbol.

Due to the spin of any electron. a magnetic moment generates. By theoretical analysis, its Value is found to be l Bohr magneton. It is allied tire intrinsic spin magnetic moment.

The concept of rotation of an electron about its axis is oversimplified and requires quantum physics for a proper explanation.

The magnetism of an atom: If the resultant moment i.e., the vector sum of tire magnetic moments of the electrons rotating, around the nucleus inside an atom is zero, we can say that the atom as a whole does not show the airy magnetic property.

On the other hand, if the resultant is not zero, the tire atom behaves like a magnetic dipole. Due to this property, each atom or molecule of iron, nickel eta behaves like a small magnet and is called an atomic magnet.

Magnetic Effect Of Current And Magnetism

Magnetic Properties Of Materials Numerical Examples

Example 1. A torque of 8 units is applied on a magnet when it is kept at 30° with the direction of a uniform magnetic field of intensity 0.32 units. Determine the magnetic moment of the magnet.
Solution:

If a magnet having magnetic moment pm is placed inclined at an angle θ with a uniform magnetic field B, the torque acting on the magnet,

⇒ \(\tau=p_m B \sin \theta\)

Here, B = 0.32 units, θ = 30° and \(\tau\) = 8 units.

∴ \(p_m=\frac{\tau}{B \sin \theta}=\frac{8}{0.32 \times \sin 30^{\circ}}\)

⇒ \(\frac{8}{0.32 \times \frac{1}{2}}=50 \text { units }\)

Example 2. lf the distance between two north poles of equal strength is 2 cm, the mutual force of repulsion between them becomes 2.5 dyn. What should be the distance of separation between them for which the repulsive force becomes 3.6 dyn?
Solution:

Magnetic force between two magnetic poles,

⇒ \(F=\frac{\mu_0}{4 \pi} \cdot \frac{q_m \cdot q_m^{\prime}}{r^2}\)

So, if the pole strength remains unchanged then, \(F \propto \frac{1}{r^2}\)

Hence, for two separate distances,

⇒ \(\frac{F_1}{F_2}=\left(\frac{r_2}{r_1}\right)^2 \quad \text { or, } \frac{r_2}{r_1}=\sqrt{\frac{F_1}{F_2}}\)

∴ \(r_2=r_1 \sqrt{\frac{F_1}{F_2}}=2 \times \sqrt{\frac{2.5}{3.6}}=2 \times \frac{5}{6}\)

= 1.67cm

Example 3. The length of a bar magnet is 20 cm and its magnetic moment is 0.6 Am². Determine the magnetic field at a point 30 cm away from either end.
Solution:

PN = PS

= 30 cm

= 0.3 m

= \(\sqrt{d^2+l^2}\)

The magnetic field at the point P due to the magnet,

⇒ \(B=\frac{\mu_0}{4 \pi} \cdot \frac{p_m}{\left(d^2+l^2\right)^{3 / 2}}\)

⇒ \(\frac{4 \pi \times 10^{-7}}{4 \pi} \times \frac{0.6}{(0.3)^3}\)

= 2.22 x 10^-6 Wb.m^-2

Example 4. The length of a bar magnet is 20 cm and its magnetic moment is 0.6 A.m². Determine the magnetic field at a point on the axis of the magnet and 30 cm away from the north pole.
Solution:

Length of the magnet. 2l = 20 cm

= 0.2 m

So, l = 0.1 m

Distance of the given point from the center of the magnet,

d = (0.3 + 0.1)m

= 0.4 m

∴ The magnetic field at the given point due to the magnet,

⇒ \(B=\frac{\mu_0}{4 \pi} \cdot \frac{2 p_m d}{\left(d^2-l^2\right)^2}=\frac{4 \pi \times 10^{-7}}{4 \pi} \times \frac{2 \times 0.6 \times 0.4}{\left\{(0.4)^2-(0.1)^2\right\}^2}\)

⇒ \(10^{-7} \times \frac{0.48}{0.0225}=2.13 \times 10^{-6} \mathrm{~Wb} \cdot \mathrm{m}^{-2}\)

Example 5. The radius of a circular conducting coil of 100 turns Is 10 cm. If 2A current passes through the coil, what will be the magnetic moment generated?
Solution:

Magnetic moment,

pm = NIA = NIπr²

= 100 x 2 x (π x 0.1 x 0.1) [∵ 10 cm = 0.1 m]

= 6.28 A.m²

Example 6. If the magnetic moment of a straight magnetized wire is pm, what will be Its magnetic moment when the wire is bent in the form of a semicircle?
Solution:

Let the length of the magnetized wire be l.

∴ Pole-strength of the wire, \(m=\frac{p_m}{l}\)

When the wire is bent in the form of a semicircle, its effective length = diameter of the semicircle = 2r.

According to the problem, \(\pi r=l \quad \text { or, } r=\frac{l}{\pi}\)

∴ The changed magnetic moment of the wire = effective length x pole-strength

⇒ \(2 r \times \frac{p_m}{l}=\frac{2 l}{\pi} \cdot \frac{p_m}{l}=\frac{2 p_m}{\pi}\)

Example 7. The magnetic moments of two small magnets are pm and p’_m; they are kept on a table. What will be the magnitude and direction of the magnetic field produced by the gents at point P? [p_m = 2.7 A m², p_m = 3,2 A m²; d₁ = 30 cm, d₂ = 40 cm]

Solution:

According to the point, P lies on the perpendicular bisectors of both magnets.

The magnetic field at the point P due to the magnet having magnetic moment p_m,

⇒ \(B_1=\frac{\mu_0}{4 \pi} \cdot \frac{p_m}{d_1^3}=10^{-7} \times \frac{2.7}{(0.3)^3}=10^{-5} \mathrm{~T}\)

At point P, the magnetic field due to the magnet having magnetic moment I’m is,

⇒ \(B_2=\frac{\mu_0}{4 \pi} \cdot \frac{p_m^{\prime}}{d_2^3}=10^{-7} \times \frac{3.2}{(0.4)^3}=5 \times 10^{-6} \mathrm{~T}\)

Since these two magnetic fields are perpendicular to each other, the resultant magnetic field at the point P,

⇒ \(B=\sqrt{B_1^2+B_2^2}\)

⇒ \(\sqrt{\left(10^{-5}\right)^2+\left(5 \times 10^{-6}\right)^2}\)

= 1.12 x 10^-5T

If the resultant magnetic field makes an angle θ with B₁, then,

⇒ \(\tan \theta=\frac{B_2}{B_1}=\frac{5 \times 10^{-6}}{10^{-5}}=0.5\)

∴ \(\theta=\tan ^{-1}(0.5)=26.57^{\circ}\)

Example 8. Two bar magnets A and B, each having a magnetic length of 4 cm are placed along a straight line with their north poles 8 cm apart and facing each other. The neutral point lies on the axis, 2 cm from the north pole of the magnet A. Calculate the ratio of the magnetic moments of A and B.
Solution:

O is die neutral point.

Given, NO = 2 cm

N’O = (8-2)

= 6 cm

⇒ \(X N=X^{\prime} N^{\prime}=l=\frac{4}{2} \mathrm{~cm}=2 \times 10^{-2} \mathrm{~m}\)

d₁ = XO

= (2 + 2) cm

= 4 x 10^-3 m

d₂ = X’O

= (6 + 2) cm

= 8 x 10^-3 m

∴ The magnetic field at point O due to the magnet A,

⇒ \(B_1=\frac{\mu_0}{4 \pi} \frac{2 p_{m_1} d_1}{\left(d_1^2-l^2\right)^2}\)

or, \(B_1=\frac{\mu_0}{4 \pi} \frac{2 p_{m_1} \times 4 \times 10^{-2}}{\left[\left(4 \times 10^{-2}\right)^2-\left(2 \times 10^{-2}\right)^2\right]^2}\)

⇒ \(=\frac{\mu_0}{4 \pi} \cdot \frac{2 p_{m_1} \times 4 \times 10^{-2}}{(16-4)^2 \times 10^{-8}}\)

∴ The magnetic field at O due to the magnet B,

⇒ \(B_2=\frac{\mu_0}{4 \pi} \frac{2 p_{m_2} \times 8 \times 10^{-2}}{\left[\left(8 \times 10^{-2}\right)^2-\left(2 \times 10^{-2}\right)^2\right]^2}\)

⇒ \(\frac{\mu_0}{4 \pi} \cdot \frac{2 p_{m_2} \times 8 \times 10^{-2}}{(64-4)^2 \times 10^{-8}}\)

According to the question, B₁ = B₂

∴ \(\frac{\mu_0}{4 \pi} \cdot \frac{2 p_{m_1} \times 4 \times 10^{-2}}{144 \times 10^{-8}}=\frac{\mu_0}{4 \pi} \cdot \frac{2 p_{m_2} \times 8 \times 10^{-2}}{3600 \times 10^{-8}}\)

or, \(\frac{p_{m_1}}{p_{m_2}}=\frac{2 \times 144}{3600}=\frac{2}{25}\)

∴ Ratio of the magnetic moments \(p_{m_1}: p_{m_2}=2: 25\)

Magnetic Effect Of Current And Magnetism

Magnetic Properties Of Materials Magnetic Lines Of Induction

The magnetic lines of force in a uniform magnetic field are equidistant parallel straight lines.

If a piece of any magnetic material (i.e., iron, steel, nickel, etc.) is placed in this kind of external magnetic field, a magnetic field is induced in the specimen which is converted into a temporary magnet. The external magnetic field is referred to as the inducing magnetic field.

Let an iron bar PQ be placed parallel to a uniform magnetic field. At the end of PQ through which the magnetic lines of force enter, an S-pole is developed and at the other end, an AT-pole is developed.

So, due to induction, the magnetic lines of force are rearranged. This rearrangement inside and outside the magnetic material is discussed below.

The lines inside the magnetic material are generated due to the superposition of two kinds of lines of force:

1. The lines of force are due to the inducing magnetic field.
2. The lines of magnetization (the magnetic material is temporarily converted into a magnet due to induction and the lines of the magnetic field of this temporary magnet are called lines of magnetization).

The lines of force outside the magnetic material are generated due to the superposition of two types of lines of force:

1. The lines of force due to inducing magnetic field
2. The lines of force are due to the temporary magnetic field of the magnetized specimen.

Outside the iron bar, the lines of force suffer bending and inside the bar are densely crowded. In the region adjacent to points A and B outside the iron bar, the lines of force are less dense.

Note that, inside the bar PQ, the lines of force of the inducing field and the lines of magnetization are unidirectional. It means that inside the bar PQ, the resultant of the two magnetic fields increases, and hence in that region and the adjacent regions C and D, the lines of force are densely crowded.

In the regions adjacent to points A and B, the inducing magnetic field and the temporary magnetic field are oppositely directed and hence the resultant magnetic field decreases there. So the lines of force become less dense in that region

Definition:

The lines of force generated inside a magnetic material due to the superposition of the lines of force of the inducing magnetic field and the lines of magnetization are called the lines of induction.

Outside a magnetic material lines of magnetic induction follow magnetic lines of force. In some materials, called diamagnetic materials, the scenario would be the reverse of the above picture

Magnetic Effect Of Current And Magnetism

Magnetic Properties Of Materials Some Magnetic Quantities And Their Relations

Magnetic permeability of a material:

Let the current through a long straight solenoid = I; the number of turns per unit length of the solenoid = n.

If air or vacuum is taken as the core of the solenoid, the magnetic field produced along the axis of the solenoid,

B0 = μ₀nI…(1)

where, μ₀ = magnetic permeability of vacuum (or air)

= 4π x 10^-7 H.m^-1

If a rod of any material is introduced inside the solenoid co-axially, the magnetic field along the axis of the solenoid will change; this magnetic field can be expressed as,

B = μnI….(2)

This quantity ju is called the magnetic permeability of the material used.

Relative magnetic permeability: If air or vacuum is replaced by a material, the fractional change of magnetic field in that material is called the relative magnetic permeability of that material.

So, relative magnetic permeability,

⇒ \(\mu_r=\frac{B}{B_0}=\frac{\mu n I}{\mu_o n I}=\frac{\mu}{\mu_0}\)….(3)

Unit: Equations (1) and (2) show that the units of μ₀ are the same, i.e., H m^-1.

Again, equation (3) shows that being the ratio of two identical quantities μ and μ₀, μ_r has no unit.

Classification of materials: Depending on the value of relative magnetic permeability μ_r, different materials can be divided into three groups.

1. \(\mu_r<1 \text {, i.e., } \mu<\mu_0\): Diamagnetic material;

2. \(\mu_r>1 \text {, i.e., } \mu>\mu_0\): Paramagnetic material;

3. \(\mu_r \gg 1 \text {, i.e., } \mu \gg \mu_0\): Ferromagnetic material.

For Example, in the case of aluminum, μ_r = 1.00002, hence it is a paramagnetic material; in the case of copper, μ_r = 0.9999904, hence it is a diamagnetic material; in the case of iron, μ_r = 1000 to 5000 (approx.), hence it is a ferromagnetic material. These materials are discussed elaborately later

Magnetization: If the magnetizing field at any point in vacuum or air is \(\vec{B}\), the magnetic field at that point, \(\vec{B}_0=\mu_0 \vec{H}\).

If vacuum or air at a point is replaced by any other material, magnetism is induced in that material and hence the magnetic field acting will also change from \(\vec{B}_0 \text { to } \vec{B}\) (say).

Naturally, \(\vec{B} \neq \mu_0 \vec{H}\). If the material is paramagnetic or ferromagnetic, \(\vec{B}>\mu_0 \vec{H}\). Here, it is assumed that,

⇒ \(\vec{B}=\mu_0 \vec{H}+\mu_0 \vec{M}=\mu_0(\vec{H}+\vec{M})\)….(4)

The first term on the right-hand side \(\left(\mu_0 \vec{H}\right)\) comes from the magnetizing field \(\vec{H}\) produced because of electric current or some other external cause, and tire second term \(\left(\mu_0 \vec{M}\right)\) comes due to induced magnetism inside the material.

So, the term \(\mu_0 \vec{M}\) indicates the additional magnetic field produced due to magnetic induction.

The quantity \(\vec{M}\) is called the intensity of magnetization of the material or simply the magnetization vector, the magnitude of which at a point is given as the net dipole moment per unit volume around that point.

By calculation, we can show that (this calculation is omitted here), the magnetic moment per unit volume of a material is \(\vec{M}\).

Definition: If the unit volume is considered around any point in a material, the magnetic moment of that volume is called the intensity of magnetization at that point. So, relative magnetic permeability

Unit: From equation (4) it is clear that the unit of \(\vec{M}\) is the same as the unit of \(\vec{H}\).

This unit is A.m^-1. \(\vec{B}\), \(\vec{H}\) and \(\vec{M}\) are called the three magnetic vectors,

Magnetic susceptibility of material: In most materials, the intensity of magnetization at a point is directly proportional to the magnetic intensity at that point, i.e.,

⇒ \(M \propto H \quad \text { or, } M=k H\)

This constant fc is called the magnetic susceptibility of the material. The property by which a material can be magnetized is its magnetic susceptibility.

In vector form, \(\vec{M}\) = k\(\vec{H}\)….(5)

Now, if H = 1, k = M; from this we can define k.

Definition: The magnetic moment induced per unit volume of a material due to unit magnetic intensity is called the magnetic susceptibility of that material.

Unit: Since the units of M and H are the same, k has no unit.

Mass (magnetic) susceptibility: The ratio of the magnetic susceptibility of a material and its density is called the mass susceptibility of that material. It Is denoted by the symbol \(\chi\).

∴ Mass susceptibility,

⇒ \(x=\frac{k}{\rho}\) [where, ρ = density of the material]

Rotation between magnetic susceptibility and relative magnetic permeability:

We know that,

⇒ \(\vec{B}=\mu \vec{H}\)

⇒ \(\vec{B}=\mu_0(\vec{H}+\vec{M}) \text { and } \vec{M}=k \vec{H}\) [from equations (4) and (5)]

Combining them we can write,

⇒ \(\mu \vec{H}=\mu_0(\vec{H}+\vec{M})=\mu_0(\vec{H}+k \vec{H})=\mu_0(1+k) \vec{H}\)

∴ \(\frac{\mu}{\mu_0}=1+k \quad \text { or, } \mu_r=1+k\) [\(\mu_r=\frac{\mu}{\mu_0}\) = relative magnetic permeability]

or, k = μ_r – 1….(6)

For vacuum or air, μ_r = 1 , hence, k = 0

The values of μ_r and k of some materials:

Hence, for paramagnetic material, k > 0 (positive); for diamagnetic material, k<0 (negative); for ferromagnetic material, k >> 0 (large positive number).

Significance:

1. In the case of paramagnetic and diamagnetic materials magnetic susceptibilities are very low and are respectively positive and negative numbers.

This means that magnetic induction in these materials is negligible. Thus, these are called non-magnetic materials.

2. On the other hand, in the case of ferromagnetic materials magnetic susceptibilities are large positive numbers.

Hence, in this kind of material, strong magnetic induction takes place. Thus, these are known as magnetic materials (e.g., iron, nickel, cobalt).

Since the value of k is very large, from the relation, \(\vec{B}=\mu_0(\vec{H}+\vec{M})=\mu_0(\vec{H}+k \vec{H})\), we come to know that the term pgM is much greater than the term p0H.

So, the magnitude of the magnetic field \(\vec{B}\) is mainly determined by induced magnetism.

In CGS or Gaussian system:

Equations (l) and (2) respectively, can be replaced by the relations, H₀ = 4π ni [magnetic permeability of vacuum or air =1 ] and H = 4πμni [magnetic permeability of the material = μ] So, relative magnetic permeability, \(\mu_r=\frac{H}{H_0}=\mu\); so in this system, magnetic permeability and relative magnetic permeability of a material are the same.

Again, in this system equation (4) is written as,

⇒ \(\vec{B}=\vec{H}+4 \pi \vec{M}\)

But we know that, \(\vec{B}=\mu \vec{H}\)

So, \(\mu \vec{H}=\vec{H}+4 \pi k \vec{H} \quad[∵ \vec{M}=k \vec{H}]\)

⇒ \(\mu=1+4 \pi k \quad \text { or, } k=\frac{\mu-1}{4 \pi}\)

Magnetic retentivity and coercivity:

When a magnetic material is placed in a magnetic field, the material acquires magnetism due to induction. This magnetism does not vanish completely even after the withdrawal of the magnetic field; some amount of magnetism is left behind in the material.

Magnetic retentivity:

The property by which a magnetic material retains some magnetism in it even after the withdrawal of the magnetizing field is called the retentivity of that material.

The magnetism, retained in a magnetic material even after the removal of the magnetic field applied to it, is called residual magnetism.

There are some magnetic materials for which the residual magnetism is almost zero, i.e., they are almost completely demagnetized.

Magnetic coercivity: The property by which magnetic material can retain induced magnetism even if used roughly, i.e., subjected to demagnetizing forces, is called the coercivity of that material.

Differences between wrought iron and steel on magnetic properties:

If two identical rods one of soft iron and the other of steel-are placed in the same magnetic field, both of them acquire approximately equal amounts of magnetism.

If the rods are then removed from the magnetic field, both soft iron and steel retain almost the entire magnetism.

Soft iron can hold slightly more magnetism than steel. But if this soft iron magnet is handled roughly, its magnetism dies out easily compared to that of steel.

So, it can be said that the magnetic retentivity of soft iron is slightly greater, but the coercivity of soft iron is much less than that of steel.

A ferromagnetic material can be converted into a strong magnet easily. In the case of a ferromagnetic material like soft iron or steel, if a graph of intensity of magnetization (M) vs magnetic field intensity (H) is drawn, we will get a closed loop.

It is known as the magnetization cycle. In the case of steel, the loop is OABCDA and in the case of soft iron, the loop is OA’B’C’D’A’. From the graph, it is evident that the intensity of magnetization (M) is not zero ( OB or OB’) even if the magnetic field intensity (H) is reduced to zero from its maximum value, i.e., M lags behind H.

This lagging of the intensity of magnetization is called hysteresis. However, M falls to zero if H is given a certain value (OC or OC’) in the opposite direction. According to the diagram, OB is the retentivity of steel, OB’ is the retentivity of soft iron, OC is the coercivity of steel, and OC’ is the coercivity of soft iron.

Selection of material to construct a permanent magnet:

The magnetic material chosen to construct a permanent magnet should have the following properties m The material should have high retentivity so that it can retain sufficient magnetism even after the withdrawal of the magnetizing field.

If The saturation magnetization of the material should be high enough it can make a strong polarity.

The material should also have high coercivity so that it can retain induced magnetization, even if used roughly. The magnetic susceptibility of the material should be of high magnitude.

Though all of the above properties do not match properly, steel, rather than soft iron is used to construct permanent magnets.

Besides steel, there are some metallic alloys like alnico (Fe 51%, Cu 3%, A1 8%, Ni 14%, Co 24% ); national (Fe 47%, Cu 3%, A1 8%, Ti 2%, Ni 15%, Co 25% ), etc. possess the above-mentioned qualities and can be used to construct permanent magnets.

Selection of material to construct an electromagnet: To construct an electromagnet, a material that possesses the following properties should be chosen.

1. The material should have low retentivity so that it can lose almost all of its magnetism as soon as the applied magnetizing field is withdrawn.

2. The saturation magnetisation of the material should be high enough which can make a strong polarity.

3. The material must have low coercivity so that it can be easily demagnetized.

4. Hysteresis loss for the material should be low so that during magnetization and demagnetization the temperature of the material should remain more or less constant.

Soft iron or alloy (an alloy of 5% silicon and 95% iron) possesses these qualities and hence can be used as the core of an electromagnet.

Selection of material as the core of a transformer or dynamo:

To prepare the core of a transformer or a dynamo, a material of high magnetic permeability should be chosen.

Soft iron possesses such qualities, and hence it is used as the core. Moreover, metallic alloys, like permalloy (50% iron and 50% nickel) and transformer steel (96% iron and 4% silicon), are used nowadays for the same purpose.

Magnetic Effect Of Current And Magnetism

Magnetic Properties Of Materials Numerical Examples

Example 1. The number of turns of a solenoid of length 10 cm is 1000. If the air inside it is replaced by a magnetic material and 1 A current is passed through the coil, the magnitude of the magnetic field at any point on its axis becomes 20 T. Determine the magnetic Intensity at that point and relative magnetic permeability of the magnetic material.
Solution:

Number of turns per unit length of the coil,

⇒ \(n=\frac{1000}{10}=100 \mathrm{~cm}^{-1}=10000 \mathrm{~m}^{-1}\)

∴ Magnetic intensity on the axis

H = nI = 10000 x 1

= 10000 A.m^-1

If the interior of the solenoid is a vacuum or contains air then the magnetic field on the axis,

B0 = μ₀nI = (4π x 10^-7) x 10000 = 12.56 x 10^-3T

Due to the presence of the magnetic material,

B = μnl = 20 T

∴ Relative magnetic permeability,

⇒ \(\mu_r=\frac{\mu}{\mu_0}=\frac{B}{B_0}=\frac{20}{12.56 \times 10^{-3}}=1592\)

Example 2. The relative magnetic permeability of a magnetic medium is 1000. If the magnetic field at any point in the medium is 0.1 Wb.m^-2, what will be the values of magnetic intensity and intensity of magnetization at that point?
Solution:

Here, B = 0.1 Wb.m^-2 and μ_r – 1000.

So, μ = μ₀ x μ_r

= 4π x 10^-7 x 1000

= 4π x 10^-4 H.m^-1

Now, magnetic intensity, \(H=\frac{B}{\mu}=\frac{0.1}{4 \pi \times 10^{-4}}=79.6 \mathrm{~A} \cdot \mathrm{m}^{-1}\)

Again, B = μ₀(H + M)

∴ The intensity of magnetization,

⇒ \(M=\frac{B}{\mu_0}-H=\frac{0.1}{4 \pi \times 10^{-7}}-79.6\)

= 79577.5 – 79.6

= 79497.9 A.m^-1

Example 3. An iron-cored toroid has a ring radius of 7 cm and several turns of 500. If 2 A current is passed through the wire, what will be the value of a magnetic field on the axis of the toroid? Given, the relative magnetic permeability of iron = 1500.
Solution:

Length of the circular axis of the ring

⇒ \(2 \pi r=2 \times \frac{22}{7} \times 7=44 \mathrm{~cm}=0.44 \mathrm{~m}\)

∴ The number of turns per unit length of the toroid,

⇒ \(n=\frac{500}{0.44} \mathrm{~m}^{-1}\)

∴ Magnetic field on the axis of the toroid,

B = μ₀nI

= μ₀μ_rnI

⇒ \(\left(4 \pi \times 10^{-7}\right) \times 1500 \times \frac{500}{0.44} \times 2=4.28 \mathrm{~Wb} \cdot \mathrm{m}^{-2}\)

Magnetic Effect Of Current And Magnetism

Magnetic Properties Of Materials Paramagnetic Diamagnetic And Ferromagnetic Materials

Depending on the behavior of a material placed in an external magnetic field.

Michael Faraday first classified all materials into three groups:

1. Paramagnetic material,
2. Diamagnetic material
3. Ferromagnetic material.

Paramagnetic Material:

The materials attracted feebly by a strong magnet are known as paramagnetic materials. Paramagnetic materials may be solid, liquid, or gaseous.

Examples: Solid elements like aluminum, potassium, platinum, sodium, tin, manganese, etc., copper sulfate, ferric chloride salts and their aqueous solutions, and gaseous materials like oxygen, air, etc.

The relative magnetic permeability of a paramagnetic material varies from 1 to 1.001 and its magnetic susceptibility has a very small positive value (10^-4 or less).

Properties:

1. When a paramagnetic material is placed in a non-uniform magnetic field, it moves gradually from the weaker part to the stronger part of that magnetic field, i.e., the material is attracted feebly by the magnet

2. If a paramagnetic material is placed in a uniform magnetic field, the lines of force close up a little and pass through the material. As a result, the lines inside that material get crowded slightly.

So, the magnetic field (B), increases slightly inside a paramagnetic material.

3. A small number of molecules of a paramagnetic material are magnetic dipoles. When heated, the random thermal motion of these dipoles increases, and hence their magnetic alignment is disturbed. As a result, both magnetic susceptibility and magnetic permeability decrease.

Magnetic susceptibility per unit mass of a paramagnetic material, \(\chi=\frac{k}{\rho}(\rho=\text { density })\) is inversely proportional to the absolute temperature; i.e., \(\chi=\frac{C}{T}\) (for a particular material C is constant).

This is known as Curie law. Experimental observations show that Curie’s law applies only to gases. In the case of paramagnetic solids, Curie- Weiss law is applicable.

According to this law,

Magnetic susceptibility \(\chi=\frac{C}{T-\theta}\)

where θ is a specific temperature known as the Curie temperature of that paramagnetic material. The values of Curie temperature are very small for almost all paramagnetic materials.

4. If a liquid is poured through one of the limbs of a vertical U-tube then the liquid rises to the same level in both limbs.

If the liquid is paramagnetic and if one of the limbs of the U-tube is placed between the two poles of a strong electromagnet, it is observed that the liquid rises in that limb. This is due to the attraction of paramagnetic liquid by a magnet.

Diamagnetic Material:

The material repelled feebly by strong magnets are known as diamagnetic materials. Diamagnetic materials may be solid, liquid, or gaseous.

Examples: Antimony, bismuth, zinc, copper, silver, gold, lead, mercury, water, hydrogen, etc.

The relative permeability of a diamagnetic material is slightly less than 1 and its magnetic susceptibility is slightly negative (≥ -10^-4).

Properties:

1. If a diamagnetic material is placed in a non-uniform magnetic field, it tries to move from the stronger region to the weaker region of the magnetic field. So, the material is fully repelled by a magnet.

2. If a diamagnetic, material is kept in a uniform magnetic field, the lines of force move away a little. As a result, the lines are more sparse inside the material than outside. For a sphere.

So, the magnetic field (B) decreases slightly inside a diamagnetic material.

3. In a diamagnetic material, almost none of the molecules are magnetic dipoles. So the random molecular motion due to increasing temperature has a negligible effect on the magnetic properties.

An increase in the applied magnetic field also does not affect the magnetic alignment of molecules.

As a result, the magnetic susceptibility of a diamagnetic material does not depend on the applied magnetic field and temperature.

4. Theoretically it is proved that diamagnetism is present in every material. The diamagnetic property of a material is very much weaker than its paramagnetic and ferromagnetic properties.

As a result, despite the magnetic properties present in paramagnetic and ferromagnetic materials, diamagnetism remains dormant in them. Hence, it can be said that diamagnetism is the most fundamental magnetic property.

5. If some diamagnetic liquid is poured into a U-tube and if one of the limbs is now placed between the pole pieces of a strong magnet, the level of the liquid in the limb falls. This is due to the repulsion of diamagnetic liquid by the magnet.

Ferromagnetic Material:

Five metals iron (Fe), nickel (Ni), cobalt (Co), gadolinium (Gd), dysprosium (Dy), and alloys like steel, etc. are attracted strongly by any magnet. Even if the inducing magnetic field is removed, these materials can retain some induced magnetism. These are known as ferromagnetic materials.

The relative magnetic permeability of a ferromagnetic material is usually very high (10² to 10⁶ ) and its magnetic susceptibility is also very high and positive (200 to 2000 approximately).

Though the magnetic properties of ferromagnetic materials are just like that of paramagnetic materials they show stronger paramagnetism and hence they are placed in a separate group.

Heusler alloy is a special kind of ferromagnetic material because none of its components (Cu, 64%, Mn 24%, A1 12% ) are ferromagnetic.

Properties:

1. Generally, ferromagnetic materials are solid and crystalline.

2. If a ferromagnetic material is placed in a non-uniform field, it moves very fast from the weaker to the stronger region of the field and sets itself parallel to the direction of the magnetic field. So, ferromagnetic materials are very strongly attracted by a magnet.

If a ferromagnetic material is placed in a uniform magnetic field, the lines of force crowd too much within the material than in the air.

3. So, the magnetic field (B) is very high inside a ferromagnetic material. The value of μ of a ferromagnetic material is also very high.

For Example, the relative magnetic permeability of iron is approximately 2000, for nickel it is 300, and for cobalt, it is 250. Magnetic susceptibility (k) of ferromagnetic material has a high positive value.

4. Magnetic permeability and magnetic susceptibility of a ferromagnetic material are not constants. They change with the change in the magnitude of the applied magnetic field.

5. Magnetic susceptibility of ferromagnetic materials changes with the temperature change.

If the magnetic field is weak, magnetic susceptibility increases with an increase in temperature. In a stronger field, the magnetic susceptibility decreases with an increase in temperature.

6. If a ferromagnetic material is heated gradually, at a particular temperature the material loses its ferromagnetic property completely and is converted into a paramagnetic material.

The transition temperature at which a ferromagnetic material is converted into a paramagnetic one is called the Curie point or Curie temperature for that material.

Above Curie point, magnetic susceptibility obeys Curie-Weiss law. Curie points of some materials are mentioned in Table.

7. Even if the inducing magnetic field is removed, a ferromagnetic material can retain some induced magnetism, i.e., a ferromagnetic material possesses magnetic retentivity.

It should be mentioned here that, though a ferromagnetic material can be converted into a paramagnetic one by heating above Curie temperature, ferromagnetism, and paramagnetism are two separate phenomena.

A paramagnetic material cannot be converted into a ferromagnetic material by cooling. The change is irreversible.

Magnetic screen:

A magnetic screen is an arrangement that works on the property of ferromagnetism and with this, a region can be kept free from the influence of any external magnetic field.

If a magnetic needle is kept in front of any pole of a bar magnet, the needle gets deflected. But if a plate of soft iron is placed in between them, the magnetic needle suffers no deflection.

This is because the magnetic permeability of the plate of soft iron is much greater than that of air.

Hence, the magnetic lines of force emerging from the bar magnet try to traverse the maximum distance through the plate of soft iron, and hence the lines of force cannot cross over to the other side of the plate.

Therefore, the region beyond the plate remains free from the influence of the magnetic field.

Similarly, if a soft-iron ring is placed in front of any pole of a bar magnet, most of the lines of force pass through the iron ring just avoiding the air gap inside the ring.

Since no line of force enters this region A inside the ring, it remains free from the influence of all external magnetic fields. If a magnetic needle is placed in that region, it sets itself at rest in any direction.

So, a magnetic material used to make a particular region free from the influence of any external magnetic field is called a magnetic screen.

Magnetic screens made of soft iron are used to protect delicate measuring instruments like galvanometers, ammeters, etc., from external magnetic fields. To make a costly wristwatch antimagnetic, a soft iron ring is fitted around it.

Comparison of Ferromagnetic, Paramagnetic and Diamagnetic Materials:

Magnetic Effect Of Current And Magnetism

Magnetic Properties Of Materials Geomagnetism

A freely suspended magnet always sets itself at rest along the north-south direction. If disturbed, it will again come to the previous position after a few oscillations.

This phenomenon is observed anywhere on the earth. Since only a magnetic field can influence a magnet, we can infer that a magnetic field exists throughout the earth, i.e., the earth behaves as a magnet.

The following two phenomena can be mentioned in support of this concept:

1. If a soft iron rod is kept at a place on the surface of the earth facing north-south direction, after a long period (say, six months), a feeble magnetism is found to have developed in the rod.

2. If a closed conductor is moved in a magnetic field an electric current is induced in that conductor. Similarly, if a closed conductor is moved at any place on the earth’s surface, a feeble electric current is seen to be induced in that conductor.

The Earth is a huge magnet: To explain the cause of the above-mentioned phenomena, scientists concluded that the Earth behaves as a huge magnet In 1600, physicist William Gilbert suggested this theory for the first time.

Later on, Gilbert performed a simple experiment. He shaped a lodestone into a sphere and demonstrated that small magnets placed at different positions on the sphere behave exactly as they do on Earth.

Locations of the Earth’s magnetic poles: Like an ordinary magnet, the Earth’s magnet also has two poles. The north pole of Earth’s magnet is situated at Ellef Ringnes Island in Canada.

The latitude and longitude of this region are 78.3°N and 104° W. This region is at an approximate distance of 1300 km from the geographical north pole.

The south pole of the earth’s magnet is located in the sea near the sea-shore of Wilkes Land belonging to Antarctica.

The latitude and longitude of this region are 65°S and 139°E. This region is at a distance of about 2550 km from the geographical south pole.

These poles are not stationary. The north pole shifts towards the northwest by 15 km every year; the shifting of the south pole is nearly the same. At present times, the magnetic axis of the earth is inclined at an angle of 11.5° with its geographical axis and the distance of the earth’s magnetic axis from the center of the earth is about 1120 km.

Nature of Earth’s magnetic poles:

The pole of a magnetic needle that points towards the north is called the north-seeking pole or simply the north pole, and the pole that points towards the south is called the south-seeking pole or simply the south pole.

The magnetic poles of Earth at the north and the south are called the magnetic north pole and magnetic south pole. Now, the north and south poles of a magnetic needle direct themselves towards the magnetic north and south poles of the earth.

It means that the north pole of the earth behaves as the south pole of an ordinary magnet and the south pole of the earth as the north pole of an ordinary magnet Sometimes the north pole of the earth’s magnet is called the blue pole and its south pole is called a red pole.

Geomagnetic Field:

The magnetic field of the earth extends up to a great height above its surface. Practically, the influence of this field is found up to a height of 105 kilometers (approx.) above the earth’s surface.

There was no clear concept about the source of the geomagnetic field for a long time. If we imagine a bar magnet kept inclined at an angle of 11.5° with the geographical axis at the center of the earth, it is possible to describe the alignment of the geomagnetic field more or less correctly. For a hypothetical bar magnet, the nature of the lines of force.

Sources of geomagnetic field: The geomagnetic field has three different sources. From these three sources, the following three magnetic fields are produced.

1. Main field: The source of this magnetic field is the outer core of Earth which is made of liquid iron. The electric current in this part produces the main field. The main field develops 97% to 99% of the geomagnetic field.

2. Crustal field: The source of this magnetic field is the earth’s crust Some parts of the earth’s crust made of hard rock become magnetized due to the presence of the main field. This magnetized part creates the crustal field. 1% to 2% of the geomagnetic field is due to this crustal field.

3. External field: The source of this magnetic field is the ionosphere present in the atmosphere of Earth. This part made up of ions, produces the external field under the influence of solar wind. 1% to 2% of the geomagnetic field is due to this external field.

Direction of the geomagnetic field: Magnetic lines of force of the geomagnetic field are naturally from the N-pole to the S-pole. But the geomagnetic N-pole is in the south and the S-pole is in the north, As a result, to show the direction of the geomagnetic field at a place, the arrow sign on each line of force should be from the south to the north direction.

Elements of Earth’s Magnetism:

To know the magnitude and direction of the geomagnetic field at any place on Earth, the quantities required are called the elements of Earth’s magnetism. There are three elements of earth’s magnetism.

These are:

1. Angle of dip
2. Angle of declination
3. Horizontal component of geomagnetic intensity.

Dip or angle Of dip: The angle made by the intensity of the earth’s magnetic field with the horizontal at any place on the earth is called dip or angle of dip at that place. If a bar magnet is suspended from its center of gravity by a thread, the magnet at rest, does not remain horizontal but inclines a little.

This means that the magnetic axis sets’ itself along the direction of the intensity of the geomagnetic field at that place. The angle (θ) made by the magnetic axis with the horizontal straight line drawn on the magnetic meridian is called dip or the angle of dip at that, place.

So, if the angle of a place is known,’ We can determine the direction of the intensity of the magnetic field at that place.

Positive and negative dips: The inclination of a magnet is different at different places on Earth, i.e., the values of the dip are different at different places. If the north pole of a magnet leans downwards at a place, the value of the dip is taken as positive; but if the south pole of the magnet leans downwards, the dip at that place is taken as negative.

Positive and negative dips are denoted by the symbols N and S, respectively. For any place in the northern hemisphere of the earth, the angle of dip becomes positive, but for places in the southern hemisphere, this angle of dip is negative.

‘The angle of dip at Kolkata is 31° N’, which means that if a magnet is suspended from its center of gravity at Kolkata, the north pole of the magnet leans downwards and the magnetic axis of the magnet inclines at an angle of 31° with the horizontal plane.

At the two magnetic poles of the earth, angles of dip are 90° each, i.e., at these two places, a freely suspended magnet remains vertical. At the magnetic equator, the angle of dip is 0°, i.e., a freely suspended magnet remains horizontal at the magnetic equator.

Declination or angle of declination: The geographical poles and the magnetic poles of the earth are not located in the same places. So if a magnetic needle, capable of rotating in the horizontal plane freely, is kept at a place, we will observe that, at rest, the magnetic axis of that needle makes a definite angle with the geographical north-south horizontal line at that place.

At different places on the earth’s surface, the value of this angle is different The vertical plane passing through the magnetic axis of a freely suspended magnetic die at any place is called the magnetic meridian at that place. Again, the vertical plane at a place containing the geographical north and south poles of Earth is called the geographical meridian at that place.

Definition: The angle between the magnetic meridian and the geographical meridian at a place is called the angle of declination at that place.

It is usually denoted by the symbol δ. If the north pole of the magnetic needle turns towards the east or the west concerning the geographical meridian, the angle of declination is called \(\delta^{\circ} E \text { or- } \delta^{\circ} W\), respectively.

For Example, the angle of declination of Delhi is 2°E means that at Delhi, the north pole of a magnetic needle capable of rotating freely on a horizontal plane moves away from the geographic meridian towards east through 2°.

Naturally, the angles of declination at different places on the earth are different. At a place, where the magnetic meridian and geographical meridian coincide, the angle of declination becomes zero.

Horizontal component of geomagnetic intensity: The direction of the geomagnetic intensity at a place does not lie on the horizontal plane usually, but it inclines at a definite angle with the horizontal plane.

This definite angle is called the angle of dip. Since magnetic intensity is a vector quantity, the geomagnetic intensity can be resolved into a horizontal component and a vertical component It should be clear that these two components lie on the magnetic meridian.

ABCD is the geographical meridian and GBCJ is the magnetic meridian. At point B, the magnitude and direction of the geomagnetic intensity I can be represented by BR. The magnitudes and directions of the vertical component V and the horizontal component H of I can be expressed by BM and BN, respectively.

Let the angle of dip be 8 and the angle of declination be δ.

∴ V = Isinθ and H = Icosθ

or, \(\frac{V}{H}=\frac{I \sin \theta}{I \cos \theta}=\tan \theta\)

∴ V = Htanθ…(1)

Again, V² + H² = I²sin²θ + I²cos²θ = I²

∴ \(I=\sqrt{V^2+H^2}\)….(2)

The value of the horizontal component H of geomagnetic intensity is not the same throughout the earth’s surface face. At the magnetic equator, θ = .0° and hence H = I; this is the maximum value of H. Again, at the magnetic poles, θ = 90° and hence H = 0; this is the minimum v value of H. Note that, at a place where the angle of dip is 45°, the values of horizontal and vertical components are equal.

The elements of earth’s magnetism for the northern hemisphere. In the case of the southern hemisphere, the elements.

‘Horizontal component of earth’s magnetic field at Kolkata is 0.37 Oe’ means that the magnitude of the horizontal component of the geomagnetic intensity, i.e., the force acting on a unit pole along the magnetic meridian at that place is 0.37 Dyn.

In most cases, we require the horizontal component of the geomagnetic Intensity, the vertical component Is of less importance. This is because the magnetic needles we use in the laboratory, can rotate in the horizontal plane but not In the vertical plane.

As a result,’ only the horizontal component acts on the needle to create deflection in it, but the vertical component does not affect it,

Values of the geomagnetic elements at some places:

Mariner’s Compass:

Mariner’s compass, as the name suggests, is used to ascertain the direction in the sea when the sun, the pole star, or other stars are not visible.

The construction of a compass is based on the directive property of a magnet.

Working principle: Noting the position of the crown mark on the compass disc, navigators determine the direction. The crown mark on the compass disc indicates the magnetic north.

To determine the geographical north at a place, the angle of declination at the place should be collected from the magnetic maps. If the value of that declination is δ°W, it should be understood that the crown mark lies at an angle of δ°, west of the geographical north.

In this way from the position of the crown mark on the compass disc, navigators can determine directions.

Magnetic Maps: F

Values of the elements of Earth’s magnetism are different at different places on Earth. But the value of any one of the elements at different places on earth’s surface may be the same.

The places on the earth’s surface possessing the same value of a particular element are along a line. Thus different lines are drawn for different values of that element.

The geographical map of Earth containing such lines is called a magnetic map. For three different elements of Earth’s magnetism, three different magnetic maps are obtained.

The values of Earth’s magnetic elements at a particular place change gradually with time. Hence new magnetic maps should be drawn at different times. Maps of this kind are essential to navigators and also for searching minerals under the earth’s crust Three different kinds of magnetic maps.

Isogonic lines:

Places on earth’s surface, having equal declination are joined by lines, called isogonic lines. The lines shown are isogonic lines drawn in 2000 AD. The lines of zero declination are called agonic lines.

Isoclinic lines:

The lines obtained by joining the places on the earth’s surface having equal dips are called isoclinic lines. The lines shown are isoclinic lines drawn in -2000 AD. The line of zero angle of dip corresponds to the magnetic equator, and it is called an aclinic line.

Isodynamic lines:

Places on the earth’s surface having equal horizontal components of geomagnetic field intensity are joined by lines, called isodynamic lines. The lines are isodynamic lines drawn in 2000 AD.

Variations of the Elements of Geomagnetism:

The values of the elements of geomagnetism are different at different places on the earth. Again, at a particular place, the values of these elements do not remain the same always; they vary with time. This change is periodic, i.e., after a definite period, the elements return to their initial values. This kind of variation is known as periodic or regular variation.

Besides this, the elements of geomagnetism suffer another kind of variation. This kind of variation is known as a magnetic storm.

Regular variation: Regular variations are of three kinds

1. Daily variation,
2. Annual variation
3. Secular variation.

Daily variation: The elements of geomagnetism undergo a kind of slow variation daily. At two different specific times on a day, any element of geomagnetism attains maximum and minimum values.

Annual variation: The elements of geomagnetism also undergo a kind of very slow variation. On two specific months in a year, the value of any element becomes maximum and minimum.

For Example, in London, the declination is found to have a maximum value in February and a minimum value in August every year. Variations of declination in the northern and southern hemispheres are just the opposite.

Secular variation: A kind of secular variation is observed in the elements of geomagnetism. The rate of this kind of variation is very slow, and its period is approximately 960 years. Perhaps this variation is due to the rotation of the earth’s magnetic north and south poles around its geographical south and north poles, respectively.

Magnetic storm: Sometimes sudden huge changes in the elements of geomagnetism are seen throughout a large region on the earth’s surface. This phenomenon is known as a magnetic storm. This kind of variation cannot be predicted.

However, after a while, the values of the elements return to their normal state. Magnetic storms usually occur due to earthquakes, volcanic eruptions, aurora borealis, the appearance of large sunspots, etc. During a magnetic storm, radio communication, television, telephone systems, etc. are disturbed greatly.

Lines of Force of a Bar Magnet in Earth’s Magnetic Field Neutral Points:

In the discussions of the magnetic lines of force of a bar magnet, the influence of the earth’s magnetic field has not been taken into account so far. The lines of force for the geomagnetic field at a place remain parallel to the magnetic meridian at that place and the direction of the lines of force is from south to north. The pattern of the lines of force near a bar magnet gets distorted under the influence of the geomagnetic field. A few cases are shown below.

N-pola pointing north: Let a bar magnet be placed along the magnetic meridian in such a manner that its 4-pole points north. In this case, the pattern of the magnetic lines of force due to the combined effect of the geomagnetic field and the magnetic field due to the bar magnet is shown.

In the vicinity of the magnet, the lines of force are curved. In this region, the influence 6f the .bar magnet is more effective. The greater the distance from the magnet, the lesser its influence but the greater the influence of the geomagnetic field.

At a sufficient distance from the magnet,” its influence almost vanishes. In that region, ‘the lines of force due to geomagnetic field only. Hence, those lines of force are straight, parallel, and directed from south to north

At different points on the axis of the bar magnet, as the lines of force due to the bar magnet and geomagnetic field are; in the same direction, the value of the magnetic intensity increases.

But at different points on the perpendicular bisector of the magnetic axis, as the lines of force due to the geomagnetic field and the bar magnet are opposite in direction, the value of the magnetic intensity decreases.

On this perpendicular bisector, there are two points X and, X where the intensities due to the geomagnetic field and the magnetic- field of the magnet become equal and opposite.

As a result, these two intensities cancel each other; i.e., at these two points, the resultant magnetic intensity becomes zero. These two points are known as neutral points.

They lie at equal distances on either side of the magnet. A magnetic needle placed at any of these two natural points does not show any ‘directive property. Naturally, ho line of force passes through the neutral points.

Neutral point: A point in a magnetic field where the resultant magnetic intensity due to the superposition of two or more magnetic fields becomes zero is called a neutral point.

When the N-pole of the bar magnet points north, at the neutral point,

F = H

or, \(\frac{p_m}{\left(d^2+l^2\right)^{3 / 2}}=H\) [in CGS system]

[where, F-resultant intensity due to the bar magnet, H = horizontal component of geomagnetic intensity, pm = magnetic moment of the bar magnet, d distance, of the neutral point from the center of the bar magnet, and 2l = magnetic length of the bar magnet.]

For a tiny bar magnet, \(\frac{p_m}{d^3}=H\). Usually, if the value of d is more than ten times l, then l₂ can be neglected compared to d₂.

S-pola pointing north: Let a bar magnet be placed along the magnetic meridian in such a way that its S-pole points north. In this case, the pattern of the magnetic lines of force is due to the combined effect of the geomagnetic field and magnetic field due to the bar magnet.

Here the neutral points (X, X) lie on the axis of the bar magnet at equal distances from its two ends.

When the S-pole of the bar magnet points north, at neutral points,

⇒ \(F=H \quad \text { or, } \frac{2 p_m d}{\left(d^2-l^2\right)^2}=H \text { [in CGS system] }\)

[where, F = intensity of the bar magnet, H = horizontal component of the geomagnetic field, pm = magnetic moment of the bar magnet, d = distance of the neutral point from the center of the magnet, and 21 = magnetic length of the bar magnet]

For a tiny bar magnet, \(\frac{2 p_m}{d^3}=H\). If the value of d is more than

10 times of l, then l₂ can be neglected compared to d₂.

N – pole pointing east: The north pole of a bar magnet points east, the pattern of magnetic lines of force due to the combined effect of the geomagnetic field and the magnetic field of the bar magnet. The two points X, X at the northwest and southeast of the bar magnet are the neutral points. These are at equal distances from the center of the magnet.

N-pole pointing west: If the north pole of a bar magnet points west, the pattern of the magnetic lines of force is due to the combined effect of the geomagnetic field and magnetic field due to the bar magnet. The two points X, X at the northeast and southwest of the magnet are the neutral points. Their distances from the center of the magnet are equal.

Magnetic Effect Of Current And Magnetism

Magnetic Properties Of Materials Numerical Examples

Example 1. The angle of dip at a place is 30° and the horizontal component of the earth’s magnetic field at that place is 0.39 CGS units. Determine the vertical component of the earth’s magnetic field at that place.
Solution:

Here, θ = 30° and H = 0.39 CGS units.

If the intensity of the earth’s magnetic field is I, then

⇒ \(H=I \cos \theta \quad \text { or, } I=\frac{H}{\cos \theta}\)

Vertical component,

⇒ \(V=I \sin \theta=\frac{H}{\cos \theta} \cdot \sin \theta=H \tan \theta\)

⇒ \(0.39 \times \tan 30^{\circ}=0.39 \times \frac{1}{\sqrt{3}}=0.225 \text { CGS units }\)

Example 2. At two places, the angles of dip are 30° 1V and 30° S and the intensity of the earth’s magnetic field is 0.42 Oe. Determine the horizontal and the vertical components of the earth’s magnetic field at these two places and also indicate their directions with the help of a diagram.
Solution:

At the first place, θ = 30° N, and at the second place, θ = 30°S.

The intensity of the earth’s magnetic field at both places, I = 0.42 Oe

Horizontal component,

⇒ \(H=I \cos 30^{\circ}=0.42 \times \frac{\sqrt{3}}{2}=0.364 \mathrm{Oe}\)

and vertical component,

⇒ \(V=I \sin 30^{\circ}=0.42 \times \frac{1}{2}=0.210 \mathrm{e}\)

Example 3. At a place, the horizontal and vertical components of the earth’s magnetic field are 0.3 Oe and 0.2 Oe, respectively. Determine the resultant intensity and angle of dip there.
Solution:

Here, H = 0.3 Oe, V = 0.2 Oe.

If the resultant intensity is I then

⇒ \(I=\sqrt{H^2+V^2}=\sqrt{(0.3)^2+(0.2)^2}=\sqrt{0.13}=0.3605 \mathrm{Oe}\)

If the angle of dip is θ then,

⇒ \(\tan \theta=\frac{V}{H}=\frac{0.2}{0.3}=0.6667\)

∴ \(\theta=\tan ^{-1}(0.6667)=33.69^{\circ}\)

Example 4. The angle of dip and the horizontal component of the earth’s magnetic field at a place is 30° S and 0.36 Oe. Determine the magnitude and direction of the vertical component of the earth’s magnetic field at that place.
Solution:

Here, θ = 30° S. So, in equilibrium, the south pole of the magnetic needle leans downwards.

We know that, tanθ = \(\frac{V}{H}\)

∴ V = vertical component of the earth’s magnetic field,

= Htanθ

= 0.36tan30° [∵ H = 0.36 Oe]

⇒ \(0.36 \times \frac{1}{\sqrt{3}}\)

= 0.208 Oe

Since the north pole of the magnetic needle lies above the horizontal line, the direction of the vertical component (V) will be vertically upwards.

Example 5. At a place, the angle of declination is 30°E and the angle of dip is 45°N. Determine the horizontal and vertical components of the geomagnetic intensity In the geographical meridian at that place. Given, the horizontal component of earth’s magnetic held at that place = 0.3 Oe.
Solution:

Here, 8 = 30°E, 0 = 45°A’ and H = 0.3 Oe.

If the horizontal component of the earth’s magnetic field in the geographical meridian is H’

⇒ \(H^{\prime}=H \cos \delta=0.3 \cos 30^{\circ}=0.3 \times \frac{\sqrt{3}}{2}=0.2598 \mathrm{Oe}\)

The vertical component remains the same in the geographical and the magnetic meridian planes, hence

V = Htanθ

= 0.3 tan45°

= 0.3 x 1

= 0.3 Oe

Example 6. The mass of a magnetic needle is -7.5 g and its magnetic moment is 98 units. To keep the magnetic needle horizontal in the northern hemisphere, what should be the position of its fulcrum concerning its center of gravity? The vertical component of the earth’s magnetic field = 0.25 Oe.
Solution:

Let the fulcrum be kept at a distance x (towards the north pole) from the center of gravity to keep the magnetic needle horizontal.

If the length of the magnetic needle = 2l and the strength of each pole = m, magnetic moment, pm = m.21.

∴ In equilibrium,

mV.2l = W.x

or, V.pm = W.x

or, \(x=\frac{V^{:} \dot{p}_m}{W}=\frac{0.25 \times 98}{7.5 \times 980}\)

= 0.0033cm

Example 7. The magnetic moment of a magnetic needle of mass 3.2 g is 980 CGS units. From which point should the needle be hung so that it will remain horizontal in the magnetic meridian? The horizontal component of the earth’s magnetic field at that place is 0.32 Oe and the angle of dip = 45°N. [g = 980 cm.s^-2]
Solution:

Let the magnetic needle be hung from a point at a distance x from its center of gravity (towards the north N pole)

In equilibrium,

mV x 2l = W.x

or, V.pm = W.x [∵ Pm = m.2l]

or, \(x=\frac{V \cdot p_m}{W}=\frac{H \tan \theta \cdot p_m}{W}\) [∵ V = Htanθ]

⇒ \(=\frac{0.32 \times \tan 45^{\circ} \times 980}{3.2 \times 980}\)

= 0.1cm

Example 8. The angle of dip at a place = θ; if the angle of dip in a vertical plane making angle δ with the magnetic meridian be θ’, show that tanθ’: tanθ = Secδ: 1
Solution:

If the true dip angle in the magnetic meridian is θ, the vertical and horizontal components of earth’s magnetic field are V and H, respectively then,

⇒ \(\tan \theta=\frac{\dot{V}}{H}\)

In a vertical plane inclined at an angle δ with the magnetic meridian, the horizontal component of intensity,

H’ = Hcosδ, apparent dip at that plane = θ’ and vertical component = V.

∴ \(\tan \theta^{\prime}=\frac{V}{H^{\prime}}=\frac{V}{H \cos \delta}=\tan \theta \cdot \sec \delta\)

or, \(\frac{\tan \theta^{\prime}}{\tan \theta}=\sec \delta\)

∴ \(\tan \theta^{\prime}: \tan \theta=\sec \delta: 1\)

Example 9. At a place, the apparent geomagnetic dip in a vertical plane is 40°, and in another plane, perpendicular to it is 30°. What is the real dip at the place? Similar problem: If θ₁ is the angle of dip of the magnetic axis of a magnetic needle with horizontal at any vertical plane and θ₂ is that in another vertical plane at right angles to the former, prove that the real angle of dip, θ is given by \(\cot ^2 \theta=\cot ^2 \theta_1+\cot ^2 \theta_2\).
Solution:

If the horizontal and vertical components of the earth’s magnetic field are H and V, respectively and the true dip angle at that place is θ then,

⇒ \(\tan \theta=\frac{V}{H}\)…..(1)

M is the magnetic meridian and X, and Y are two vertical planes inclined at a right angle to each other.

The angle between the planes X and M is δ. If the apparent dip in the plane X is θ₁ then,

⇒ \(\tan \theta_1=\frac{V}{H \cos \delta}\)….(2)

If the apparent dip in plane Y is θ₂ then,

⇒ \(\tan \theta_2=\frac{V}{H \sin \delta}\)…..(3)

From equations (2) and (3) we get,

⇒ \(\cot ^2 \theta_1+\cot ^2 \theta_2=\frac{H^2}{V^2}\left(\cos ^2 \delta+\sin ^2 \delta\right)=\frac{H^2}{V^2}\)

= cot²θ [from equation (1)]

Here, θ₁ = 40° and θ₂ = 30°.

∴ \(\cot ^2 \theta=\cot ^2 40^{\circ}+\cot ^2 30^{\circ}\)

or, \(\cot \theta=\sqrt{1.42+3}=\sqrt{4.42}=2.1\)

or, θ = cot^-1(2.1)

= 25.46°

Example 10. A bar magnet of length 6 cm is kept vertically with its north pole on the ground. If the distance of the neutral point on the ground is 8 cm from the north pole, what will be the magnetic moment of that magnet? [If = 0.36 CGS units]
Solution:

Let O be the neutral point

Here, NS = 6 cm, NO = 8 cm

So, \(S O=\sqrt{6^2+8^2}=10 \mathrm{~cm}\)

If the pole strength of the magnet NS is qm, magnetic intensity at the point O due to the north pole,

⇒ \(H_1=\frac{q_m}{O N^2} \text {, in the direction of } O A \text { [in CGS] }\)

Again, due to the south pole magnetic intensity at the point O,

⇒ \(H_2=\frac{q_m}{O S^2} \text {, in the direction of } O S\)

∴ Component of H₂ in the direction ON,

⇒ \(H_2 \cos \theta=\frac{q_m}{O S^2} \cdot \frac{O N}{O S}\)

So, the horizontal magnetic intensity at the point O due to the entire magnet,

⇒ \(H_1-H_2 \cos \theta=\frac{q_m}{O N^2}-\frac{q_m}{O S^2}: \frac{O N}{O S}=q_m\left(\frac{1}{8^2}-\frac{1}{10^2} \times \frac{8}{10}\right)\)

= q_m x 0.007625 Oe

Since O is the neutral point, the magnetic intensity at that point due to the magnet will be equal but opposite to the horizontal component H of Earth’s magnetic field.

∴ \(q_m \times 0.007625=0.36 \quad \text { or, } q_m=\frac{0.36}{0.007625} \mathrm{emu} \cdot \mathrm{cm}\)

∴ Magnetic moment of the magnet

⇒ \(q_m \cdot N S=\frac{0.36}{0.007625} \times 6=283.3 \mathrm{emu} \cdot \mathrm{cm}^2\)

Example 11. A bar magnet of length 8 cm is placed on a horizontal plane in the magnetic meridian with its north pole pointing north. If the magnetic moment of the magnet is 90 CGS units and the horizontal, component of the earth’s magnetic field is 0.35 Oe, determine the positions of the neutral points.
Solution:

Let each of the two neutral points be at a distance d from the center of the magnet along its perpendicular bisector. If the magnetic moment of the magnet is pm and its magnetic length is 2l, the magnetic intensity at the neutral points due to the magnet (in the CGS system) is,

⇒ \(H_m=\frac{p_m}{\left(d^2+l^2\right)^{3 / 2}}\)

At the neutral points, the horizontal component H of the earth’s magnetic field will be equal and opposite.

∴ \(H=H_m=\frac{p_m}{\left(d^2+l^2\right)^{3 / 2}} \quad \text { or, }\left(d^2+l^2\right)^{3 / 2}=\frac{p_m}{H}\)

or, \(d^2+l^2=\left(\frac{p_m}{H}\right)^{2 / 3} \quad\)

or, \(d=\sqrt{\left(\frac{p_m}{H}\right)^{2 / 3}-l^2}\)

Here, pm = 90 CGS units, H = 0.35 Oe, 2l = 8 cm i.e.,
I = 4 cm.

∴ \(d=\sqrt{\left(\frac{90}{0.35}\right)^{2 / 3}-4^2}=4.94 \mathrm{~cm}\)

Example 12. At a place, the vertical component of the earth’s magnetic field is V3 times its horizontal component. What will be the angle of drop at that place?
Solution:

If the intensity of the earth’s magnetic field is I and the angle of dip is θ, then the horizontal component of the earth’s magnetic field, H = Icosθ, and a vertical component, V = Isinθ

According to the problem,

⇒ \(V=\sqrt{3} H \quad\)

or, \(\frac{V}{H}=\sqrt{3} \quad\)

or,\(\frac{I \sin \theta}{I \cos \theta}=\sqrt{3}\)

or, \(\tan \theta=\sqrt{3}=\tan 60^{\circ}\)

∴ θ = 60°

Magnetic Effect Of Current And Magnetism

Magnetic Properties Of Materials Very Short Questions and Answers

Question 1. The magnetic moment of a small bar magnet is 1 A.m². What will be the magnitude of the magnetic field at a point 1 m away from the center of the magnet along its length?
Answer: 2 x 10^-7T

Question 2. If the magnetic moment of a bar magnet of length 5 cm is 1 A.m2, what is the strength of each pole?
Answer: 20 A.m

Question 3. Pole-strengths of two magnetic poles are 1 A.m and 2 A.m. If they are kept 1 m apart in the air, then what is the magnitude of the force acting between them?
Answer: 2 x 10^-7N

Question 4. A short bar magnet, placed with its axis at an angle 6 with a uniform external magnetic field B, experiences a torque r. What is the moment of the magnet?
Answer: \(\frac{\tau}{B \sin \theta}\)

Question 5. What is the unit of magnetic susceptibility?
Answer: No unit

Question 6. What is the unit of intensity of magnetization \(/vec{M}\)?
Answer: A.m^-1

Question 7. If the relative magnetic permeability of a material is 1,00004, what will be its magnetic susceptibility?
Answer: 0.00004

Question 8. If the magnetic intensity at a point in a material is 100 A.m-1 and the magnetic susceptibility is 1000, what will be the value of the magnetic field at that point?
Answer: 4π x 10^-2T

Question 9. If the magnetic susceptibility of a material is -0.0002, what is its relative magnetic permeability?
Answer: 0.99998

Question 10. The area of the hysteresis loop of a magnetic material A is larger than that of another magnetic material B. Which of the two materials is better suited for use as the core of an electromagnet?
Answer: Material B

Question 11. The relative permeability of iron is 5500. What is its magnetic susceptibility?
Answer: 5499

Question 12. The Curie temperature of nickel is 360°C. In which group of magnetic materials will you place nickel at 500°C?
Answer: Paramagnetic

Question 13. Name a non-magnetic alloy of iron.
Answer: Stainless Steel

Question 14. Name a material that is used as the core of an electromagnet.
Answer: Soft Iron

Question 15. For which type of material the magnetic susceptibility is negative?
Answer: Diamagnetic

Question 16. For which type of material, the magnetic susceptibility is independent of temperature?
Answer: Diamagnetic

Question 17. How does the magnetic susceptibility per unit mass (\(\chi\)) of a paramagnetic gas depend on absolute temperature T?
Answer: \(\chi \propto \frac{1}{T}\)

Question 18. If the intensity of the geomagnetic field at a place is 60 A.m^-1 and the horizontal component of that intensity is 30 A.m^-1, determine the angle of dip.
Answer: 60°

Question 19. Where on the surface of the earth, the value of the angle of dip is zero?
Answer: Geomagnetic equator

Question 20. Where on the earth’s surface, the value of the angle of dip is 90°?
Answer: Geomagnetic poles

Question 21. The vertical component of the earth’s magnetic field at a place is √3 times the horizontal component. What is the angle of dip at this place?
Answer: 60°

Question 22. The horizontal component of the earth’s magnetic field at a place is V3 times the vertical component. What is the angle of dip at this place?
Answer: 30°

Magnetic Effect Of Current And Magnetism

Magnetic Properties Of Materials Fill in The Blanks

1. The ratio of the magnetic field \(\vec{B}\) at a point in a material to the magnetic permeability of that material is called the intensity of the magnetic field \(\vec{H}\) at that point.

2. The magnetic moment per unit volume of a material due to unit magnetic intensity is called the magnetic susceptibility of that material.

3. The horizontal component of the earth’s magnetic field exists in all regions except at the geomagnetic poles

Magnetic Effect Of Current And Magnetism

Magnetic Properties Of Materials Assertion Reason Type

Direction: These questions have Statement 1 and Statement 2. Of the four choices given below, choose the one that best describes the two statements.

1. Statement 1 is true, Statement 2 is true; Statement 2 is a correct explanation for Statement 1.
2. Statement 1 is true, Statement 2 is true; Statement 2 is not a correct explanation for Statement 1.
3. Statement 1 is true, Statement 2 is false.
4. Statement 1 is false, and Statement 2 is true.

Question 1.

Statement 1: The magnetic moment of an electron rotating in an atom is proportional to its angular momentum.

Statement 2: The electrons in an atom can rotate only in those orbits for which the angular momentum of the electron is an integral multiple of \(\frac{h}{2 \pi}\)(h= Planck’s constant).

Answer: 2. Statement 1 is true, Statement 2 is true; Statement 2 is not a correct explanation for Statement 1.

Question 2.

Statement 1: Soft iron is preferred for making electromagnets.

Statement 2: Both the permeability and retentivity of soft iron are high.

Answer: 3. Statement 1 is true, Statement 2 is false.

Question 3.

Statement 1: The magnetic lines of force inside a piece of copper placed in a uniform magnetic field move away from each other.

Statement 2: The permeability of diamagnetic material is less than one.

Answer: 1. Statement 1 is true, Statement 2 is true; Statement 2 is a correct explanation for Statement 1.

Question 4.

Statement 1: The intensity of Earth’s magnetic field may be different even if the horizontal component of Earth’s magnetic field is the same at two different places.

Statement 2: The horizontal component of the earth’s magnetic field is Isinθ where I am the intensity of the earth’s magnetic field and θ is the angle of dip at a place.

Answer: 3. Statement 1 is true, Statement 2 is false.

Question 5.

Statement 1: The B-H curve of a ferromagnetic material is not linear which means that these materials do not obey \(\vec{B}=\mu \vec{H}\) rule.

Statement 2: The permeability of a ferromagnetic material is not constant, it can even have a negative value

Answer: 4. Statement 1 is false, Statement 2 is true.

Magnetic Effect Of Current And Magnetism

Magnetic Properties Of Materials Conclusion

• A current loop behaves as a magnetic dipole.
• The product of the effective area of a current-carrying loop and the current flowing through that loop is called the magnetic moment of the loop.
• The behavior of a current-carrying circular coil is similar to that of a permanent magnet.
• The behavior of a current-carrying solenoid is similar to that of a permanent bar magnet.
• The ratio of the magnetic moment to the effective length of a magnet is the pole strength of that magnet. The strengths of the two poles of a magnet are equal but opposite.
• Inside a magnetic material, the lines of force produced as a result of the superposition of the lines of force due to ‘the induced magnetic field and the lines of magnetization are called magnetic lines of induction.
• If air or vacuum is replaced by any other material the fractional change in the magnetic field in that material is called the relative magnetic permeability (r) of that material.

• Tire intensity of magnetization (M) at a point, is defined as the magnetic moment of a unit volume surrounding the point.
• The amount of induced magnetic moment per unit volume of a material due to unit magnetic intensity is called the magnetic susceptibility of that material.
• The ratio of magnetic susceptibility and its density is called the mass (magnetic) susceptibility of the material.
• The property by which a magnetic material retains some magnetism even after the withdrawal of the external magnetic field is called the magnetic retentivity of the material.
• The property by which a magnetic material can retain induced magnetism even if it is handled roughly is called the coercivity of the material.
• Magnetic susceptibility per unit mass of a paramagnetic material, \(\dot{\chi}=\frac{k}{\rho} \cdot(\rho=\text { density })\) varies inversely with the absolute temperature. So, \(\chi=\frac{C}{T}\) (for a particular-material C is constant). This is known as Curie law.
• Diamagnetism is the most fundamental magnetic property of all materials.
• The critical temperature at which a ferromagnetic material is converted into a paramagnetic material is called the Curie point dr Curie temperature of that material.
• To make a particular place free from any external magnetic influence, a magnetic material having high magnetic permeability is used as a magnetic screen.
• The earth behaves as a huge magnet.
• Earth’s magnetic north pole is situated near its geographic south pole and the magnetic south pole is situated near its geographic north pole,

The elements of earthly magnetism are:

1. The angle of dip,
2. Angle of declination
3. Horizontal component of the earth’s magnetic intensity.

The angle made by the tire intensity of the earth’s magnetic field with the horizontal at a place on the earth is called, the angle of dip at that place.

• The angle between the magnetic meridian and geographical meridian at a place is called the angle of declination at that place.
• Due to the superposition of two or more magnetic fields at a point, if the resultant, magnetic field becomes zero, then that point is called a neutral point.
• The magnetic moment of a current loop (it is a magnetic dipole), \(\vec{p}_m=N I \vec{A}\)
• where, I = current through the loop, \(\vec{A}\) = vector indicating the area of the loop and N = number of turns.
• The torque acting on a bar magnet in a magnetic field,

⇒ \(\vec{\tau}=\vec{p}_m \times \vec{B}\)

• The magnetic moment of a bar magnet of effective length 21 is,

⇒ \(\vec{p}_m=2 q_m \vec{l} \text { [where } q_m=\text { pole-strength] }\)

• The mutual force between two magnetic poles of pole strengths strengths \(q_m \text { and } q_m^{\prime}\) is,

⇒ \(F=\frac{\mu_0}{4 \pi} \cdot \frac{q_m \cdot q_m^{\prime}}{r^2}\)

• This is known as Coulomb’s law of magnetism.

Magnitudes of the magnetic field at different points due to a bar magnet:

Here, d = distance of the point from the mid-point of the tire magnet, 2l = effective length of the magnet, and pm = magnetic content of the bar magnet.

The magnetic moment of a revolving charged particle (q),

⇒ \(p_m=\frac{q v r}{2}=\frac{q}{2 m} L\)

[where, r = radius of tire circle, v = speed of the particle, m = mass of the particle, L = angular momentum]

A few relations:

1. \(\vec{B}_0=\mu_0 \vec{H}\)
2. \(\vec{B}=\mu_0(\vec{H}+\vec{M})\)
3. \(\vec{M}=k \vec{H}\)
4. \(\mu_r=\frac{\mu}{\mu_0}=1+k\)
5. \(\chi=\frac{k}{\rho}\)
6. \(k=\mu_r-1\)

[where, B₀ = magnetic field in air (or vacuum),

• B = magnetic field in airy medium,
• H = magnetic intensity in air (or vacuum),
• M = intensity of magnetization,
• μ₀ = magnetic permeability of air (or vacuum),
• μ = permeability of a medium,
• μ_r = relative permeability of a medium,
• k = magnetic susceptibility,
• ρ = density of the medium,
• \(\chi\) = mass (magnetic) susceptibility]

If the angle of dip at a place is θ and the magnitude of tire Intensity of earth’s magnetic field is I, then the vertical component of earth’s magnetic field, V = Isinθ, and Its horizontal component H = Icosθ.

In this case, \(I=\sqrt{V^2+H^2}\)

Some physical quantities related to the magnetic property of material:

• The magnetic moment for a straight current-carrying wire is zero.
• The magnetic moment for a toroid is zero.
• A magnetized wire having length L and magnetic moment M is bent.
• Now the magnetic moment becomes \(M^{\prime}=\frac{M}{\sqrt{2}}\)

• A magnetized wire having length L and magnetic moment M is bent.
• Now the magnetic moment becomes \({M}^{\prime}=\frac{M}{2}\)

• If the magnetic latitude of a place is λ and the angle of dip is θ, then tanθ = 2tanλ.

Unit 3 Magnetic Effect Of Current And Magnetism Chapter 2 Magnetic Properties Of Materialsm Match The Columns

Question 1. Some quantities are given in column A and corresponding units are in column B.

Answer: 1-C, 2-B, 3-D, 4-A

Question 2. The magnetic moment of a magnetized steel wire is p. It is bent to form different shapes, given in column A. The magnetic moments are given in column B.

Answer: 1-D, 2-A, 3-B, 4-C

Magnetic Effect Of Current And Magnetism

Electromagnetism Magnetism

Magnetite, a black stone found in nature is called a natural magnet. It has two specific properties

1. Attractive property
2. Directive property.

1. Attractive property: Magnetite is found to attract pieces of iron.

2. Directive property:

• If a piece of magnetite is suspended freely with the help of a thread, it aligns itself in the north-south direction.
• Navigators used magnetite as a compass for guiding their ships and hence it was called leading stone or lodestone.
• The two above-mentioned properties are called magnetic properties and the phenomenon is known as magnetism. Bodies showing these properties are called magnets.
• Magnetism is a physical property of matter because when a body is magnetized, no chemical change occurs.

Artificial Magnet:

Magnets found in nature have no definite shape. So the directive property cannot be understood clearly. Moreover, the attractive power of this magnet is weak and hence is not so useful.

Read and Learn More Class 12 Physics Notes

Later artificial magnets were invented for practical use. Using some special processes magnetic properties can be built up in iron, steel, nickel, and some alloys.

This process is known as magnetization and the magnets thus made are known as artificial magnets.

There are different shapes and sizes of artificial magnets:

1. Bar magnet
2. Magnetic needle
3. Horseshoe magnet
4. Ball-ended magnet etc.

North and South Poles of a Magnet:

If a bar magnet is dipped into some iron filings and then withdrawn, a good amount of filings clings at the two ends of the magnet but almost none at its middle

So, attractive power is maximum at the two ends of the magnet, and these two regions are called the poles of a magnet. The middle portion, where no attraction is observed, is called the neutral region.

If a bar magnet is suspended freely by a thread, it sets itself at rest in the north-south direction. The pole of the magnet which always faces north is called the north pole (N-pole) or positive pole. Similarly, the pole facing south is called the south pole (S-pole) or negative pole of the magnet.

The line joining the two poles of a magnet is called the magnetic axis. If a magnet is suspended freely from its mid-point, it comes to rest after some time. The imaginary vertical plane through the magnetic axis of the magnet at this position is called the magnetic meridian of that place.

Usually, poles are considered as points and these two points lie very close to the ends of the magnet. The effective length of the magnet is about 80- 85% of its geometrical length.

If different magnets are dipped into iron filings and withdrawn, the amounts of iron filings collected are not the same in all cases. So, the attractive power of different magnets is different, but there is no difference between powerful and weak magnets with respect to the directive property.

Mutual Action Between Two Magnetic Poles:

If the N-pole of a bar magnet is brought near the Af-pole of a magnetic needle, repulsion occurs between them, i.e., the Npole of the magnetic needle moves away from the bar magnet.

Again, if the N-pole of the bar magnet is brought near the S-poIe of the magnetic needle, they attract each other, i.e., the S-pole of the magnetic needle comes closer to the bar magnet.

If the experiment is performed with the S-pole of the bar magnet, the opposite action is observed, i.e., the S-pole of the bar magnet attracts the AT-pole of the magnetic needle but repels its S-pole.

Inference: Like poles repel each other but unlike poles attract each other.

Repulsion Is the conclusive tost of magnetization:

If a body is repelled by a magnet, it is sure that the body is a magnet

If one end of a body Is brought near the Af-poles of a powerful magnet and if an attraction is observed between them there are two probabilities:

1. The body may be an ordinary piece of iron
2. The body is a magnet and the end under investigation is the S-pole of that magnet. So, attraction cannot Identify whether a body is a magnet or not.

But if repulsion instead of attraction Is observed In the above experiment, it is definite that the body is a magnet and the end under investigation is the iV-pole of that magnet.

Hence, repulsion is the conclusive test of magnetization.

Magnetic Effect Of Current And Magnetism

Electromagnetism Magnetic Field

A magnet attracts iron, nickel, cobalt, and some metallic alloys. This force of attraction even penetrates the wood, glass, paper, or other obstructions. It is always observed that the influence of a magnet is felt in its surrounding region. The greater the distance from the magnet, the lesser is its influence. Again if a weak magnet is replaced by a stronger one, the range of this influence increases.

Magnetic Field Definition: The region surrounding a magnet in which the Influence of that magnet is felt, is called the magnetic field of that magnet.

Theoretically, the magnetic field of a magnet extends up to infinity; but In practice, the field Is assumed to be extended up to a limited region due to,

Limitation of the experimental arrangement used for identification

Presence of other magnetic fields (like Earth’s magnetic field) in the environment.

Magnetic Lines of Force or Field lines:

Experiment: A small but powerful bar magnet is kept on a fairly large piece of cork and is let to float on water kept in a large vessel. In the floating condition, the magnet ultimately sets Itself at rest along the Nordic-south direction.

So, its N-pole faces the north and the S-pole faces the sound. With the help of a small cork, a long but comparatively weak magnetic needle is set to float vertically on the water in such a manner that the n-pole of the needle is just above the water’s surface but its s-pole remains deep inside the water.

In this situation, the effect of the bar magnet on the s-pole of the needle becomes negligible due to its depth. Hence, the n-pole of the needle can be treated as an isolated free n-pole with respect to the bar magnet.

The N-pole of the needle is brought in contact with the N-pole of the bar magnet at point A and then released. It is seen that this n-pole moves over the surface of water and follows a curved path ABCD to reach the S-pole of the bar magnet

Explanation of the experiment: The N-pole of the bar magnet exerts a force of repulsion on the isolated n-pole of the needle but the S-pole exerts a force of attraction on it.

The isolated free n-pole then starts moving along the resultant of the above two forces. At different points of the magnetic field of the bar magnet, the direction of this resultant force is different.

Then, the direction of motion of the n-pole will also change., So, when an isolated and free n-pole is allowed to travel in the magnetic field of a magnet, the pole describes a curved path and this path extends from the north pole to the south pole of the magnet.

A magnetic line of force is an imaginary curved line of a magnetic field; the direction of this field at any point is given by the tangent drawn at that point on the line of force passing through that point.

Properties of magnetic Hass of force:

1. A magnetic line of force emerges from the north pole of a magnet and terminates at the south pole.

2. For different initial points adjacent to the north pole of a magnet, different lines of force are obtained in the magnetic field. Different lines of force at one side of a bar magnet are shown. A number of such lines of force indicate a magnetic field.

3. Two lines of force never intersect each other. If they intersect, through that point of intersection two tangents can be drawn on the two lines of force and each tangent will be the direction of the magnetic field at the point of inter the section. But two directions of the magnetic field at a single point are meaningless.

4. The concept of lines of force is totally imaginary, no such line exists in a magnetic field.

5. At any point in a magnetic field, if a unit area normal to the direction of the lines of force is imagined, the number of lines of force passing through that area is called the number density of the lines of force or magnetic flux at that point.

6. The greater this number density, the greater will be the strength of the magnetic field at that point For Example, the number density at point B is less than that at point A. So the magnetic field at point B is weaker than that at point A.

7. In general, the strength of a magnetic field is different at different points in that field and their directions are also different. Hence, the magnetic lines of force are usually curved lines at different distances.

8. But if the magnetic field is uniform, i.e., its magnitude and direction are the same up to a certain region, then it can be represented by equispaced parallel straight lines. Earth’s magnetic field, very close to the surface of the earth, is such a uniform magnetic field.

Magnetic Effect Of Current And Magnetism

Electromagnetism Action Of Current On Magnets

Oersted’s Experiment:

In 1820, scientist Hans Christian Oersted discovered that a magnetic field is generated around a current-carrying conductor. He inferred this through the following experiment.

A magnetic needle is kept just below a conducting wire stretched along a north-south direction. The magnetic needle should be free to rotate about its vertical axis. Now if current is passed through the straight conductor with the help of an external source, the magnetic needle gets deflected, i.e., the needle undergoes an angular deflection θ

Observations related to deflection:

1. As soon as the current stops flowing through the wire, the magnetic needle rotates back to its initial position.

2. With the increase in current through the conducting wire, the angular deflection of the needle increases.

3. If the direction of the current in the conductor is reversed, the magnetic needle deflects in the opposite direction.

4. If the conducting wire is rotated slowly from its north-south direction while carrying current, the deflection of the magnetic needle gradually decreases. Ultimately when the direction of the current is along east-west, there will be no deflection of the magnetic needle.

5. So, when the conducting wire is placed normally to the axis of the magnetic needle and current is passed through the wire, no deflection of the magnetic needle is observed.

The direction of deflection of the magnetic needle due to the flow of electric current through the conductor can be determined with the help of any one of the following two rules:

1. Ampere’s swimming rule: If a man is imagined to be swimming along the direction of the current facing the magnetic needle with his arms outstretched, the north pole of the needle will be deflected towards his left hand

2. Right-hand thumb rule: The right hand, with its thumb sticking out, is held in. Such a way that the conducting wire is in between the palm and the magnetic needle. If the other fingers point the direction of the current then the thumb will indicate the direction of deflection of the north pole of the needle.

The direction of deflection of the magnetic needle is shown in the following table:

Discussions:

1. Dependence of magnetic field: At any adjacent point of a current-carrying conductor, the magnitude of the magnetic field depends on the magnitude of current and the direction of the magnetic field depends on the direction of current and on the position of the point with respect to the current-carrying conductor.

2. Presence of insulating material: The magnetic field is not affected if the current-carrying conductor is covered with an insulating material.

3. Current-carrying material: The current-carrying conductor itself is not magnetized. If some iron filings are brought in contact with the conductor, no attraction is observed.

4. Magnetic field due to a moving charged particle: The motion of charged particles is the cause of electric current. Hence, a moving charge can produce a magnetic field around it. Obviously, when a charged particle is at rest, it cannot produce a magnetic field.

Mapping of Magnetic Lines of Force due to an Electric Current:

Any magnetic field can be represented by magnetic lines of force. The direction of the magnetic field at any point is denoted by the direction of the magnetic line of force at that point. To determine the direction of the magnetic field at any point around a current-carrying conductor, either of the following two rules can be used.

1. Maxwell’s corkscrew rule: If we imagine a right-handed corkscrew to be driven along the direction of current in a conductor, then the direction in which it rotates, gives the direction of the magnetic field.

2. Right-hand grip rule: If a current-carrying conductor is imagined to be held within the grip of the right hand and if the direction of current through the conductor is indicated by the thumb, direction then the other fingers will curl in the direction of the magnetic field.

Long straight conductor: A long straight conducting wire carrying current is passed through the center of a cardboard and the cardboard is held normally to the length of the wire.

Some light iron filings are scattered over the cardboard. Now if the cardboard is slightly tapped, the iron filings arrange themselves in some concentric circles around the conducting wire. These concentric circles indicate the magnetic lines of force on a plane perpendicular to the current carrying a long straight conductor.

With the help of the corkscrew rule, the direction of the lines of force can also be determined. For an upward current, the directions of the magnetic lines of force. If the direction of current flow is reversed, i.e., for a downward current, the direction of the lines of force will also be reversed.

In the laboratory, generally, a magnetic needle is used instead of iron filings for plotting magnetic lines of force.

Circular conductor: A circular current-carrying conductor is shown in penetrating a cardboard plate kept perpendicular to the plane of the circular conductor.

With the help of iron filings or a magnetic needle, if the lines of force of the magnetic field are drawn on the cardboard.

The directions of the magnetic lines of force may be determined by the corkscrew rule. If the direction of current in the circular conductor is reversed, the directions of the lines of force will also get reversed.

It is to be noted here that, at the centre of the circular conductor the lines of force are almost parallel to the axis going through the center.

Magnetic Effect Of Current And Magnetism

Electromagnetism Biot-Savart Law Or Laplace’s Law

A diagrammatic representation of a magnetic field is its representation using magnetic lines of force:

1. The direction of the tangent drawn at any point on a magnetic line of force is the direction of the magnetic field at the point;

2. Comparing the number density of magnetic lines of forces at different points in a magnetic field, the field strengths at those points can be compared. But to define magnetic field precisely as a definite physical quantity, at every point its magnitude should be represented by a number and an associated unit. This is not possible from the concept of magnetic lines of force only.

As a physical quantity, the usual symbol of the magnetic field is \(\vec{B}\) (as it is a vector). This vector B is named magnetic field magnetic induction or magnetic flux density. In the region surrounding a current-carrying conductor

1. The direction of \(\vec{B}\) is determined by Maxwell’s corkscrew rule
2. The magnitude of \(\vec{B}\) (i.e., B) is determined by Biot-Savart law.

Statement of Biot-Savart law:

Let δl= a small elementary part of a conducting wire

I = current through the wire

r = distance of any external point from the element δl

θ = angle between the element \(\delta \vec{l}\) and the position vector -r of the external point

δB = magnitude of the magnetic field at that external point.

Then the Biot-Savart law states that,

1. \(\delta B \propto \delta l\)
2. \(\delta B \propto I\)
3. \(\delta B \propto \frac{1}{r^2}\)
4. \(\delta B \propto \sin \theta\)

This means, \(\delta B \propto \frac{I \delta l \sin \theta}{r^2} \text { or, } \delta B=k \frac{I \delta l \sin \theta}{r^2}\)….(1)

This law is also called Laplace’s law. The value of the constant k in equation (1) depends on

1. The nature of the medium between the conductor and the point under consideration and

2. The system of units used for different physical quantities. In this chapter, we shall consider only vacuum as the medium.

SI unit: SI units of different physical quantities used in equation (1) are:

δl and r: meter (m); I: ampere (A); δB: weber/metre2 (Wb.m-2) or tesla (T).

For the definition of the unit of magnetic field (Wb.m-2)

If these units are used in equation (1) then another constant is traditionally used instead of k. This constant is μ₀ = 4πk,

⇒ \(k=\frac{\mu_0}{4 \pi}\)

So, the usual form of Biot-Savart law in vacuum,

⇒ \(\delta B=\frac{\mu_0}{4 \pi} \cdot \frac{I \delta l \sin \theta}{r^2}\)….(2)

The constant μ₀ is called the permeability of free space.

Unit of μ₀: From equation (2),

⇒ \(\mu_0=\frac{4 \pi r^2 \delta B}{I \delta l \sin \theta}\)

∴ The unit of \(\mu_0=\frac{\mathrm{m}^2 \times \mathrm{Wb} \cdot \mathrm{m}^{-2}}{\mathrm{~A} \times \mathrm{m}}=\mathrm{Wb} \cdot \mathrm{A}^{-1} \cdot \mathrm{m}^{-1}\)

Here, Wb.A^-1 is also called Henry (H).

Hence, the unit of μ₀ is henry/metre (H.m^-1)

Value of μ₀: = 4π x 10^-7 H m^-1.

It is to be noted that,

⇒ \(\frac{\mu_0}{4 \pi}=10^{-7} \mathrm{H} \cdot \mathrm{m}^{-1}\)

Biot-Savart law for extended conductors: An extended conductor is assumed to be composed of a number of smaller parts \(\delta l_1, \delta l_2, \cdots\), etc. and the Biot-Savart law is applied for each part. At any point, the magnetic field due to the whole conductor will be,

⇒ \(B=\sum \delta B=\frac{\mu_0}{4 \pi} \sum \frac{I \delta l \sin \theta}{r^2}\)

The \(\) (summation) sign indicates the sum of a number of terms. \(\vec{B}\) is a vector; so to determine the resultant, the rule of vector addition is applied. Usually, this vector addition is very complicated. But in the case of symmetrical conductors like straight conducting wire, circular coil, etc., determination of the resultant magnetic field is not so troublesome.

Vector form of Wot-Savort low: If the unit vector towards the point P with respect to \(\delta l \text { be } \hat{r} \text {, then } \vec{r}=r \hat{r}\).
Hence, the magnitude of the vector product \((\delta \vec{l} \times \hat{r})\) is,

⇒ \(|\delta \vec{l} \times \hat{r}|=\delta l \cdot 1 \cdot \sin \theta=\delta l \sin \theta \quad[∵|\hat{r}|=1]\)

Again, the direction of \((\delta \vec{l} \times \hat{r})\) is downward with respect, which is actually the direction of the magnetic field as per the corkscrew rule. So, the vector form of the equation (2) is,

⇒ \(\delta \vec{B}=\frac{\mu_0}{4 \pi} \cdot \frac{I \delta \vec{l} \times \hat{r}}{r^2}\)….(3)

From equation (3) we get,

⇒ \(\vec{B}=\frac{\mu_0}{4 \pi} \sum \frac{I \delta \vec{l} \times \hat{r}}{r^2}=\frac{\mu_0}{4 \pi} \sum \frac{I \delta \vec{l} \times \vec{r}}{r^3}\)….(4)

Usually with respect to the plane of the paper, an upward vector is denoted by \(\odot\) sign to mean the tip of an arrow and a downward vector by \(\otimes\) sign to mean the tail of an arrow. Thus, the direction of the magnetic field at point P is denoted by the \(\otimes\) sign, and at point Q, \(\odot\) sign is used.

The magnetic intensity or magnetic field strength or magnetizing field:

In SI: The magnetic permeability of vacuum is μ₀ this value of μ₀ is also used for air because, in the presence of air, no remarkable change in the magnetic field is observed.

But for any other medium, the permeability of that medium is denoted by n. For different media the magnitude of μ is different. In the case of any medium, the general form of Biot-Savart law is,

⇒ \(\delta B=\frac{\mu}{4 \pi} \cdot \frac{I \delta l \sin \theta}{r^2}\)….(5)

From equations (2) and (5) we see that the quantity \(\frac{1}{4 \pi} \cdot \frac{I \delta l \sin \theta}{r^2}\) can be treated as the cause of the magnetic field in a medium. In the absence of the multiplier μ₀ or μ in this expression, this quantity does not depend on the medium. To determine the magnetic field in a medium, this quantity is just multiplied by the magnetic permeability of that medium. This quantity is called magnetic intensity or magnetizing field. It is expressed by H.

From the equation (5) we can write,

⇒ \(\delta B=\frac{\mu}{4 \pi} \cdot \frac{I \delta l \sin \theta}{r^2}=\mu \delta H\)

where, \(\delta H=\frac{1}{4 \pi} \cdot \frac{I \delta l \sin \theta}{r^2}\)…(6)

Generally, \(B=\mu H \quad \text { or, } \quad H=\frac{1}{\mu} B\)

Like \(\vec{B}\), H is also a vector quantity whose direction is the same
as that of \(\vec{B}\).

So, \(\vec{B}=\mu \vec{H} \quad \text { or, } \quad \vec{H}=\frac{1}{\mu} \vec{B}\)…(7)

With the help of this equation (7), \(\vec{H}\) can be defined.

Definition: At any point in a medium the ratio of the acting magnetic, field and the magnetic permeability of the medium is called the magnetic intensity or magnetizing field at that point.

Unit: Unit of \(H=\frac{\text { unit of, } B}{\text { unit of } \mu}=\frac{\mathrm{Wb} \cdot \mathrm{m}^{-2}}{\mathrm{~Wb} \cdot \mathrm{A}^{-1} \cdot \mathrm{m}^{-1}}\)

= A.m^-1 (ampere/metre)

CGS or Gaussian system:

In this system \(\vec{H}\), i.e., the magnetic intensification (instead of \(\vec{B}\)) is considered as the primary vector in magnetism. In this case, the unit of magnetic intensity H is d (Oe) arid the unit of magnetic field B is gauss (G). It may be of interest to note that 1 oersted is the same as 1 dyn per unit pole, and 1 gauss is the same as 1 maxwell/cm² according to the old definition of magnetic intensity.

Relation between SI and CGS units:

1 A.m^-1 = 4π x 10^-3 Oe, 1 Wb.m^-2 = 10^-4 G

In the CGS system, the electromagnetic, unit (abbreviated as emu) of current is used as the unit of current and is expressed by the symbol i. This electromagnetic unit is so defined that if the other quantities in equation (6) are expressed in CGS units, thÿf-,ÿ constant \(\frac{1}{4 \pi}\) can be replaced by 1. In that case, the CGS form of the equation will be,

⇒ \(\delta H=\frac{i \delta l \sin \theta}{r^2}\)…(8)

So, the different units used in this equation are

δl and r: cm; i: emu of current; δH: Oe.

This equation (δ) indicates the Biot-Savart law or Laplace’s law in the CGS or Gaussian system.

Emu current: The current which, when flowing through a conducting wire of length in the form of an arc of a circle of radius 1 cm, produces a magnetic Intensity of 1 Oe at the center of the arc, is called 1 electromagnetic unit (emu) of current

The two parts of the wire except the circular are kept along the radius of the circle; thus no magnetic field is produced at the center of the circle due to the current flowing through these two parts.

The definition of the electromagnetic unit of current can be explained from the discussion of the magnetic field produced due to current in a circular conductor

Relation between ampere and emu of current:

1 emu of current = 10 A

Rules of conversion from SI to CGS:

The following replacements convert SI expressions into the corresponding ones. CGS expressions;

1. Magnetic intensity \(\vec{H}\), in place of magnetic field \(\vec{B}\);

2. Electric current in the emu, in place of I in amperes;

3. The constant 1 in place of \(\frac{\mu_0}{4 \pi}\), i.e., 4π in place of μ₀

Application of Biot-Savart Law Long Straight Conductor:

Let I = strength of current in an electrical circuit. AB is a straight conductor and OP = r = perpendicular distance of the point P from the current carrying wire.

The two endpoints of the conducting wire make angles θ₁ and θ₂ at point P with respect to OP. In the direction of the current, if the angle θ₂ is taken as positive then in the opposite direction, θ₁ will be negative, i.e., it will be -θ₁.

The conducting wire and its adjacent region in a magnified form. From a very small part dl of the wire, the distance of the point P is x and the angle between the direction of current and x is a.

So, according to Iliot-Savnrt law, the magnitude of the magnetic field at die point P due to dl is,

⇒ \(d B=\frac{\mu_0}{4 \pi} \cdot \frac{I d l \sin \alpha}{x^2}\)…(1)

From the diagram,

⇒ \(\alpha=180^{\circ}-\beta=180^{\circ}-\left(90^{\circ}-\theta\right)=90^{\circ}+\theta\)

So, sinaaa = sin(90° + θ) = cosθ ….(2)

Again, l = rtanθ

Since, r = constant, differentiating the equation we get,

dl = rsec²θdθ….(3)

Again, cosθ = \(\frac{r}{x}\)

or, \(x^2=\frac{r^2}{\cos ^2 \theta}=r^2 \sec ^2 \theta\)…..(4)

Now putting the values of sinaaa, dl, and x2 from equations (2), (3), and (4) into equation (1) we get,

⇒ \(d B=\frac{\mu_0}{4 \pi} \cdot \frac{I \cdot r \sec ^2 \theta d \theta \cdot \cos \theta}{r^2 \sec ^2 \theta}=\frac{\mu_0}{4 \pi} \cdot \frac{I}{r} \cdot \cos \theta d \theta\)….(5)

So, for the entire wire AB, the magnetic field at the point P is,

⇒ \(B=\int d B=\frac{\mu_0}{4 \pi} \cdot \frac{I}{r} \int_{-\theta_1}^{\theta_2} \cos \theta d \theta\)

⇒ \(\frac{\mu_0}{4 \pi} \cdot \frac{I}{r}[\sin \theta]_{-\theta_1}^{\theta_2}=\frac{\mu_0}{4 \pi} \cdot \frac{I}{r}\left[\sin \theta_2-\sin \left(-\theta_1\right)\right]\)

⇒ \(\frac{\mu_0}{4 \pi} \cdot \frac{I}{r}\left(\sin \theta_2+\sin \theta_1\right)\)

So, \(B=\frac{\mu_0}{4 \pi} \cdot \frac{I}{r}\left(\sin \theta_1+\sin \theta_2\right)\)…(6)

This is the expression for the magnitude of the magnetic field at point P due to the current carrying wire.

Special cases:

For infinitely long conducting wire: Let the length of the wire below the point O be Ll and above this point, I². If these two lengths are much greater than r, i.e., and r<<I², the wire can be treated as of infinite length with respect to the point P. In that case, for the lowermost end of the wire,θ₁ ≈ 90° and for the uppermost end of the wire θ₂ ≈ 90°.

So from the equation (6), we can write,

⇒ \(B=\frac{\mu_0}{4 \pi} \cdot \frac{I}{r}\left(\sin 90^{\circ}+\sin 90^{\circ}\right)\)

∴ \(B=\frac{\mu_0}{4 \pi} \cdot \frac{2 I}{r}\)…(7)

For semi-infinitely long conducting wire: Let the lowermost end of the wire be O and the length of the wire = I. If r<<L, the wire can be called semi-infinitely long with respect to the point P. In that case, θ₁ = 0° and θ₂ = 90°.

So, from the equation (6), we can write

⇒ \(B=\frac{\mu_0}{4 \pi} \cdot \frac{I}{r}\left(\sin 0^{\circ}+\sin 90^{\circ}\right)\)

∴ \(B=\frac{\mu_0}{4 \pi} \cdot \frac{I}{r}\)…(8)

For a point on the extended part of the wire: If point P₁ is located according to, both θ₁ and θ₂ will be negative with respect to r which is the perpendicular distance between point P₁ and the extension of the wire AB.

So, the magnetic field at P₁,

⇒ \(B_1=\frac{\mu_0}{4 \pi} \cdot \frac{I}{r}\left[\sin \theta_1+\sin \left(-\theta_2\right)\right]\) [taking θ₂ negative in equation (6)]

⇒ \(\frac{\mu_0}{4 \pi} \cdot \frac{I}{r}\left(\sin \theta_1-\sin \theta_2\right)\)…..(9)

If we consider the point P₂ to lie on the extension of the wire AB, then θ₁ = 90° and θ₂ = 90°. Putting these in equation (9) we get, the magnetic field at the point P₂,

⇒ \(B_2=\frac{\mu_0}{4 \pi} \cdot \frac{I}{r}\left[\sin 90^{\circ}-\sin 90^{\circ}\right]=\frac{\mu_0}{4 \pi} \cdot \frac{I}{r}[1-1]=0\)

So, at any point lying along the length of a current-carrying conductor, the magnetic field due to that conductor will be zero.

Expressions in CGS or Gaussian system: In the above-mentioned equations (6), (7), and (8), if we substitute B→H, I→I, and μ₀→4π then we get

Magnetic intensity due to a straight conductor,

⇒ \(H=\frac{i}{r}\left(\sin \theta_1+\sin \theta_2\right)\)

Magnetic intensity due to an infinitely long conductor,

⇒ \(H=\frac{2 i}{r}\)

Magnetic intensity due to a semi-infinitely long conductor,

⇒ \(H=\frac{i}{r}\)

Magnetic Effect Of Current And Magnetism

Electromagnetism Numerical Examples

Example 1. The distance between two long straight conductors is 5m. Currents 2.5 A and 5 A are flowing through in the same direction. What will be the magnetic field at a themed point between them?
Solution:

According to the corkscrew rule, the magnetic fields at the mid-point due to die two conductors will be opposite in directions.

So, the relation, \(B=\frac{\mu_0}{4 \pi} \cdot \frac{2 I}{r}, \text { gives }\)

magnetic field at the mid-point due to the first conductor,

⇒ \(B_1=\frac{\mu_0}{4 \pi} \cdot \frac{2 I_1}{\frac{r}{2}}=\frac{\mu_0}{4 \pi} \cdot \frac{4 I_1}{r}\)

and magnetic field at the mid-point due to the second conductor,

⇒ \(B_2=\frac{\mu_0}{4 \pi} \cdot \frac{2 I_2}{\frac{r}{2}}=\frac{\mu_0}{4 \pi} \cdot \frac{4 I_2}{r}\)

⇒ \(I_2>I_1 \text { and hence } B_2>B_1\)

∴ The resultant magnetic field,

⇒ \(B=B_2-B_1=\frac{\mu_0}{4 \pi} \cdot \frac{4}{r}\left(I_2-I_1\right)\)

⇒ \(\frac{\mu_0}{4 \pi} \cdot \frac{4}{5}(5-2.5)=\frac{4 \pi \times 10^{-7}}{4 \pi} \times \frac{4}{5} \times 2.5\) [∵ r = 5m, I₁ = 2.5 A, I₂ = 5 A]

= 2 x 10^-7 T

Example 2. 5 A current Is flowing in two opposite directions through each of two parallel straight conducting wires kept 0.2 m apart. find who second wins) Determine the magnitudes and directions of the magnetic field at the points P, Q, and R lying on the plane containing the two wires.
Solution:

The magnetic field at P due to the first wire,

⇒ \(B_1=\frac{\mu_0}{4 \pi} \cdot \frac{2 \times 5}{0.1} \text { (upward) }\)

The magnetic field at P due to the second wire,

⇒ \(B_2=\frac{\mu_0}{4 \pi} \cdot \frac{2 \times 5}{0.3} \text { (downward) }\)

Since, B₁ > B₂, the resultant magnetic field will be upward from the plane.

∴ B_p = B₁ – B₂

⇒ \(\frac{\mu_0}{4 \pi}(2 \times 5)\left(\frac{1}{0.1}-\frac{1}{0.3}\right)\)

⇒ \(10^{-7} \times 10 \times 10 \times \frac{2}{3}\)

⇒ \(6.67 \times 10^{-6} \mathrm{~Wb} \cdot \mathrm{m}^{-2}\)

Similarly, the upward magnetic field at the point Q,

B_Q = 6.67 x 10^-6 Wb.m^-2

At point R, the magnetic field due to the first wire as well as for the second wire is downward. Hence, the resultant magnetic field will be the sum of these two fields.

So, \(B_R=2 B_1=2 \times \frac{\mu_0}{4 \pi} \cdot \frac{2 \times 5}{0.1}\)

= 2 x 10^-7 x 10 x 10

= 2 x 10^-5 Wb.m^-2

Example 3. An infinitely long conducting wire POQ is bent through right angles at O. If a current I . 0 is sent through this bent wire, what will be the magnitude of the magnetic field at point A at a distance r from each part of the wire?

Solution:

AM = AN = r

So, ∠OAM = ∠OAN = 45°

Now, the magnetic field at A due to PO is equal to that due to OQ both in magnitude and direction (downwards). Again, the points P and Q make angles at A relative to AM and AN. For infinitely long wires, each of these angles is close to 90°.

So the resultant magnetic field at A is

⇒ \(B=2 \times \frac{\mu_0}{4 \pi} \cdot \frac{I}{r}\left(\sin 90^{\circ}+\sin 45^{\circ}\right)\)

⇒ \(\frac{\mu_0}{4 \pi}: \frac{I}{r}(2+\sqrt{2}) \text { (directed downwards) }\)

Example 4. Through each of two wires POQ and P’O’Q’ an electric current I is passing. P The points Q, O, O’, and Q’ are collinear. Determine the magnetic field at the midpoint A of OO’.

Solution:

Point A lies along the length of the two parts OQ and O’Q’. Hence, no magnetic field exists at A due to these two parts.

The two parts PO and P’O’ are semi-infinite conducting wires with respect to point A. Hence, for each part, the magnetic field at \(A=\frac{\mu_0}{4 \pi} \cdot \frac{I}{r}\) and these two fields are downwards.

Hence, the downward magnetic field at the point A,

⇒ \(B=\frac{\mu_0}{4 \pi} \cdot \frac{I}{r}+\frac{\mu_0}{4 \pi} \cdot \frac{I}{r}=\frac{\mu_0}{4 \pi} \cdot \frac{2 I}{r}\)

Example 5. 5 A current is flowing through a long straight conducting wire. What is the magnitude of the magnetic field at a distance of 10 cm from the wire?
Solution:

We know, \(B=\cdot \frac{\mu_0}{4 \pi} \cdot \frac{2 I}{r}\) [for a straight infinite wire]

Here,I = 5 A, r = 10 cm = 0.1m, μ₀ = 4π x 10^-7 H.m^-1

∴ \(B=\frac{4 \pi \times 10^{-7}}{4 \pi} \times \frac{2 \times 5}{0.1}=10^{-5} \mathrm{~Wb} \cdot \mathrm{m}^{-2}\)

Application of Biot-Savart Law Circular Conductor:

1. Magnetic field at the center of a circular conductor: Let the current through a circular conductor of radius r be I.

Due to an element of length δ₁ of the conductor, the magnetic field at the center of a circle,

⇒ \(\delta B=\frac{\mu_0}{4 \pi} \cdot \frac{I \delta l \sin 90^{\circ}}{r^2}=\frac{\mu_0}{4 \pi} \cdot \frac{I \delta l}{r^2}\)

If the circumference of the circle is divided into a large number of such elements then, for each element, I and r remain the same and the angle between that element and the radius is 90°. Again sum of these elements = circumference of the circle = 2πr.

So, die magnetic field at the center of the circular conductor,

⇒ \(B=\frac{\mu_0}{4 \pi} \frac{I}{r^2} \sum \delta l=\frac{\mu_0}{4 \pi} \cdot \frac{I}{r^2} \cdot 2 \pi r=\frac{\mu_0 I}{2 r}\)…(1)

If the circular conductor of a single turn is replaced by a circular coil of N turns then,

⇒ \(B=\frac{\mu_0 N I}{2 r}\)….(2)

The direction magnetic field at the center of the circular conductor can be determined by the corkscrew rule. This direction is normally upward from the plane of the paper. If the conductor is in die form of a circular arc and if that arc makes an angle θ (rad) at the center, the magnetic field at the center of the circle

⇒ \(B=\frac{\mu_0 I}{2 r} \cdot \frac{\theta}{2 \pi}=\frac{\mu_0 I}{4 \pi r} \cdot \theta\)….(3)

The magnetic field on the axis of a circular conductor:

Let r = radius of a circular conductor, I = current through the conductor, and O is the center of the circular conductor and hence. P is any point on the axis (OP – x). An element of length δl is considered at the topmost point C of the conductor. The line segment CP is perpendicular to this element So, for the element of length 8/ at C, the genetic field at the point P is,

⇒ \(\left.\delta B=\frac{\mu_0}{4 \pi} \cdot \frac{I \delta l \sin 90^{\circ}}{u^2}=\frac{\mu_0}{4 \pi} \cdot \frac{I \delta l}{u^2} \text { [here } C P=u\right]\)

From the corkscrew rule, we see that the direction of δB is along PQ. The component of 8B along the axis is δBsinθ and perpendicular to the axis is δBcosθ. If an element of length δl is now considered at point D diametrically opposite to C on the circumference, the magnetic field at point P will be SB and its direction will be along PR. Naturally, its downward component δBcosθ neutralizes the previous component δBcosθ, but two components

δBsinθ each will be added together along the axis. In this way, if the whole circular conductor is considered, the algebraic sum of the components SBsinfl along the axis will be the resultant magnetic field due to the circular conductor at die point P.

∴ \(B=\sum \delta B \sin \theta=\frac{\mu_0}{4 \pi} \sum \frac{I \delta l}{u^2} \cdot \frac{r}{u}=\frac{\mu_0}{4 \pi} \sum \frac{I \delta l}{u^3} \cdot r\)

Now, due to the symmetry of the circular conductor, the magnitudes of I, r and u will be die same at every point on its circumference. Hence,

⇒ \(B=\frac{\mu_0}{4 \pi} \cdot \frac{I r}{u^3} \sum \delta l=\frac{\mu_0}{4 \pi} \cdot \frac{I r}{u^3} \cdot 2 \pi r=\frac{\mu_0 I}{2} \cdot \frac{r^2}{\left(u^2\right)^{3 / 2}}\)

So, \(B=\frac{\mu_0 I}{2} \cdot \frac{r^2}{\left(r^2+x^2\right)^{3 / 2}}\)….(4)

If a circular coil having N turns is taken instead of the circular conductor of single turn then,

⇒ \(B=\frac{\mu_0 N I}{2} \cdot \frac{r^2}{\left(r^2+x^2\right)^{3 / 2}}\)…..(5)

Now, at the center of the circle, i.e., at the point O, x = 0, and hence,

⇒ \(B=\frac{\mu_0 N l}{2 r}\), which is identical to the equation (2).

Expressions In COS or Gaussian system: In equations (1) to (5) above, substituting B→H, I→I, and μ0 → 4π we get,

magnetic intensity at the center of a circular conductor,.

⇒ \(H=\frac{2 \pi i}{r}\)

and magnetic intensity at the center of a circular coil having N-tums,

⇒ \(H=\frac{2 \pi N i}{r}\)

In the case of a conductor in the form of an arc of a circle, the magnetic intensity at the center of the conductor, H = \(\frac{i}{r}\)θ where θ is the angle made by the arc at the center.

Magnetic intensity at any point on the axis of a circular conductor,

⇒ \(H=2 \pi i \cdot \frac{r^2}{\left(r^2+x^2\right)^{3 / 2}}\)

In the case of a circular coil having N turns, the magnetic intensity at any point on its axis,

⇒ \(H=2 \pi N i \cdot \frac{r^2}{\left(r^2+x^2\right)^{3 / 2}}\)

Angle (θ) subtended by a current-carrying conductor of length 1 cm bent in the form of an arc of a circle of 1 cm radius is 1 rad. If the magnetic intensity at the center of the arc is 1 Oe, then substituting r = 1, θ = 1, H = 1 in the relation \(H=\frac{i \theta}{r}\), we get, i = 1.

Characteristics of the magnetic field of a circular conductor:

1. Direction of magnetic field: At any point on the axis of the conductor the direction of the resultant magnetic field is always along the axis. For the direction of current the magnetic field at the point P is along the axis and directed outward. But on the opposite side of the coil, at the point magnetic field is along the axis and directed inward, i.e., along Pf O. If the direction of current in the coil is reversed, at the point P magnetic field will be inward while at the point P’, it will be outward.

2. Magnitude of the magnetic field: From equation (5) we see that the magnitude of the field becomes maximum at the center of the circle, and decreases along the axis, on either side.

As a result, no uniform magnetic field is obtained at any point on the axis and thus a problem arises while constructing electrical instruments with a circular coil. This problem can be removed by using a Helmholtz double coil.

Helmholtz double coll: Two circular coils having the same radius (= r) are placed coaxially at a distance equal to their radius. If a direct current is passed through them in the same direction, the magnetic fields generated between them will have the same direction. Thus the resultant magnetic field remains almost uniform in between the coils.

Magnetic Effect Of Current And Magnetism

Electromagnetism Numerical Examples

Example 1. The radii of two concentric circular colls are 8 cm and 10 cm and the number of turns In them are 40 and 10, respectively. A 5A current Is passing through each of them in the same direction. Determine the magnetic field produced at the center of the two colls.
Solution:

Since current flows in the same direction through the two coils, the directions of magnetic fields at the center due to the coils will be the same. Hence, the resultant magnetic field will be obtained by adding these magnetic fields.

For the first coil, \(B_1=\frac{\mu_0 N_1 I}{2 r_1}\)

For the second coil, \(B_2=\frac{\mu_0 N_2 I}{2 r_2}\)

∴ The resultant magnetic field,

⇒ \(B=B_1+B_2=\frac{\mu_0 I}{2}\left(\frac{N_1}{r_1}+\frac{N_2}{r_2}\right)\)

⇒ \(\frac{4 \pi \times 10^{-7} \times 5}{2}\left(\frac{40}{8}+\frac{10}{10}\right)
\)

= 2π x 10^-7 x 5 x 6

= 1.885 x 10^-5 Wb.m^-2

Example 2. A current I is flowing through an Infinitely long wire PQRS. The wire is bent a J S at right angles so that the part QR becomes one-fourth of the circumference of a circle of radius r whose center is at O. Determine the magnetic field at O.

Solution:

With respect to point O, both the parts PQ and

RS are semi-infinite wires and hence, for each part magnetic field at point O will be,

⇒ \(B_1=B_2=\frac{\mu_0}{4 \pi} \cdot \frac{I}{r}\)

Again for a complete circular conductor, the magnetic field at its center = \(\frac{\mu_0 I}{2 r}\)

So. for the one-fourth part QR of the circle, the magnetic field
at O,

⇒ \(B_3=\frac{1}{4} \times \frac{\mu_0 I}{2 r}=\frac{\mu_0 I}{8 r}\)

Hence, the resultant magnetic field at O,

B = B₁ + B₂ + B₃

⇒ \(\frac{\mu_0 I}{4 \pi r}+\frac{\mu_0 I}{4 \pi r}+\frac{\mu_0 I}{8 r}\)

⇒ \(\frac{\mu_0 I}{4 r}\left(\frac{1}{\pi}+\frac{1}{\pi}+\frac{1}{2}\right)=\frac{\mu_0(4+\pi) I}{8 \pi r}\)

Example 3. Determine the magnetic field at point O due to the circuit.

Solution:

Point O lies on the same straight line with the two linear parts of the circuit Hence, no magnetic field acts at O due to those two parts.

For a complete circular conductor, the magnetic field at the tire center of the circle = \(\frac{\mu_0 I}{2 r}\)

Hence, for a semicircular conductor, the magnetic field at the center = \(\frac{\mu_0 I}{4 r}\)

The magnetic field at O due to the semicircular conductor of radius r is \(B_1=\frac{\mu_0 I}{4 r}\) and that due to the semicircular conductor of radius \(R \text { is } B_2=\frac{\mu_0 I}{4 R}\).

The corkscrew rule shows that Bj and B2 are oppositely directed. Since Bj > Bz, the resultant magnetic field at the point O (upward with respect to the page of the book),

⇒ \(B=B_1-B_2=\frac{\mu_0 I}{4}\left(\frac{1}{r}-\frac{1}{R}\right)=\frac{\mu_0 I(R-r)}{4 r R}\)

Example 4. What is the magnetic field produced at the center of a hydrogen atom due to the revolution of its electron in the first orbit (K-orbit)? The radius of the first orbit = 0.53 x 10^-10 m, the velocity of the electron in that orbit = 2.19 x 10s m s^-1
Solution:

Time period of revolution, \(T=\frac{2 \pi r}{v}\).

So, the circular loop formed due to the revolution of the electron carries an effective current

⇒ \(I=\frac{\text { charge }}{\text { time period }}=\frac{e}{T}=\frac{e v}{2 \pi r}\)

So, the magnetic field at the center of the atom,

⇒ \(B=\frac{\mu_0 I}{2 r}=\frac{\mu_0}{4 \pi} \cdot \frac{e v}{r^2}\)

⇒ \(10^{-7} \times \frac{\left(1.6 \times 10^{-19}\right) \times\left(2.19 \times 10^6\right)}{\left(0.53 \times 10^{-10}\right)^2}\)

= 12.47 T

Example 5. Calculate the magnetic induction at point O (center of the partial circular conductor).

Solution:

Magnetic induction at die center due to whole conductor, B = Magnetic induction due to (straight part AB + curved part BCD + straight part DE)

∴ \(B_{\text {total }}=B_{A B}+B_{B C D}+B_{D E}\)

BAB = 0 [∵ thePoint O is along AB]

⇒ \(B_{B C D}=\frac{3}{4}\) of the ma8netic due to the whole circle

⇒ \(\frac{3}{4} \times \frac{\mu_0}{4 \pi} \cdot \frac{2 \pi I}{r}\) [directed inwards and perpendicular to the plane of the conductor]

⇒ \(B_{D E}=\frac{\mu_0}{4 \pi} \cdot \frac{I}{r}\) [directed outwards and perpendicular to the plane of the conductor]

∴ \(B_{\text {total }}=\frac{3}{4} \frac{\mu_0}{4 \pi} \frac{2 \pi l}{r}-\frac{\mu_0}{4 \pi} \frac{l}{r}\)

⇒ \(\frac{\mu_0}{4 \pi}, \frac{I}{r}\left[\frac{3}{2} \pi-1\right]\) [directed inwards and perpendicular to the plane of the conductor]

Example 6. Two concentric but mutually perpendicular conducting coils are carrying current 3 A and 4 A, respectively. If the radius of each coil is 2x cm, what will be the magnetic induction at the center of the coils? (μ₀ = 4π x 10^-7 Wb A^-1.m^-1)
Solution:

⇒ \(r=2 \pi \mathrm{cm}=\frac{\pi}{50} \mathrm{~m}\)

Magnetic field at the center of a circular coil, \(B=\frac{\mu_0 I}{2 r}\)

∴ For the first coil,

⇒ \(B_1=\frac{\left(4 \pi \times 10^{-7}\right) \times 3}{2 \times \frac{\pi}{50}}=3 \times 10^{-5} \mathrm{~Wb} \cdot \mathrm{m}^{-2}\)

and for the second coil,

⇒ \(B_2=\frac{\left(4 \pi \times 10^{-7}\right) \times 4}{\ddots 2 \times \frac{\pi}{50}}=4 \times 10^{-5} \mathrm{~Wb} \cdot \mathrm{m}^{-2}\)

Since the coils are mutually perpendicular, By and B2 are also at right angles to each other. Hence, the resultant magnetic field,

⇒ \(B=\sqrt{B_1^2+B_2^2}=\sqrt{(3)^2+(4)^2} \times 10^{-5}\)

= 5 x 10^-5 Wb.m^-2

Example 7. Each of two long straight wires, passing through points A and B, carries a current directed vertically upwards with respect to the plane of the paper. The separation between them is r. Find out the magnetic field at a point P on that plane, which is at a distance r from each of the wires.

Solution:

Here PA = PB = r

Due to the current-carrying conductor, which passes through point A, the magnetic field intensity at P is,

⇒ \(B_1=\frac{\mu_0}{4 \pi} \frac{2 I}{P A}=\frac{\mu_0}{4 \pi} \frac{2 I}{r}\)

Similarly, due to the current-carrying conductor which passes through point B, the magnetic field intensity at point P is,

⇒ \(B_2=\frac{\mu_0}{4 \pi} \cdot \frac{2 I}{P B}=\frac{\mu_0}{4 \pi} \cdot \frac{2 I}{r} \text { (along } \overrightarrow{P B_2} \text { ) }\)

The resultant magnetic field intensity at point P is given by,

⇒ \(B=\sqrt{B_1^2+B_2^2+2 B_1 B_2 \cos 60^{\circ}}\)

⇒ \(\sqrt{\left(\frac{\mu_0}{4 \pi} \frac{2 I}{r}\right)^2\left[1^2+1^2+2 \cdot 1 \cdot 1 \cdot \frac{1}{2}\right]}\)

⇒ \(\frac{\mu_0}{4 \pi} \frac{2 I}{r} \sqrt{3} \text { (directed parallel to } \overrightarrow{B A} \text { ) }\)

Example 8. Two small identical circular loops marked (1) and (2) carrying equal currents are placed with their geometrical axes perpendicular to each other. Find the magnitude and direction of the net magnetic field produced at O. Also determine the field when the radius of the loop is very large as compared to the distance of the point from the center.

Solution:

Let R be the radius of each loop. Magnetic field at O due to loop 1,

⇒ \(B_1=\frac{\mu_0 I R^2}{2\left(R^2+x^2\right)^{3 / 2}} \text { (it acts along } \overrightarrow{O X} \text { ) }\)

Magnetic field at O due to loop 2,

⇒ \(B_2=\frac{\mu_0 I R^2}{2\left(R^2+x^2\right)^{3 / 2}} \text { (it acts along } \overrightarrow{O Y} \text { ) }\)

∴ The net magnetic field at O due to both the loops,

⇒ \(B=\sqrt{B_1^2+B_2^2}=\sqrt{2 B_1^2}=\sqrt{2} B_1 \cdot\left[∵ B_1=B_2\right]\)

⇒ \(=\sqrt{2} \frac{\mu_0 I R^2}{2\left(R^2+x^2\right)^{3 / 2}}=\frac{\mu_0 I R^2}{\sqrt{2}\left(R^2+x^2\right)^{3 / 2}}\)

This field acts at an angle of 45° with \(\vec{OX}\).

Now if R>>x, neglecting x₂/R₂ we get

⇒ \(B=\frac{\mu_0 I R^2}{\sqrt{2}\left[R^2\left(1+x^2 / R^2\right)\right]^{3 / 2}} \approx \frac{\mu_0 I R^2}{\sqrt{2} R^3}\)

∴ \(B \approx \frac{\mu_0 I}{\sqrt{2} R}\)

Example 9. Two circular coils of radii a and 2a having a common center, carry identical currents, but opposite directions. The number of turns of the second Conductor is 8. Show that magnetic field intensity at the center 3 times that glue to the smaller one. Also, find out the ChangeinWe previous result when current flows in the same direction throughout the coils
Solution:

Magnetic field intensity at the center O due to the smaller loop is,

⇒ \(B_1=\frac{\mu_0 i}{2 a} \text { (upwards) }\)

Similarly magnetic field intensity at the centre due to the bigger loop,

⇒ \(B_2=8 \times \frac{\mu_0 i}{2(2 a)}=\frac{2 \mu_0 i}{a}\) (downwards)

∴ The net magnetic field at O,

B = B₂-B₁

⇒ \(\frac{\mu_0 i}{a}\left(2-\frac{1}{2}\right)\)

⇒ \(\frac{3 \mu_0 i}{2 a}=3 B_1 \quad\left[∵ B_1=\frac{\mu_0 i}{2 a}\right]\)

Hence resultant field is 3 times that due to the smaller loop.

Now if the direction of the current is the same for both loops, the resultant field will be,

⇒ \(B=B_1+B_2=\frac{\mu_0 i}{2 a}+\frac{2 \mu_0 i}{a}=5 \frac{\mu_0 i}{2 a}=5 B_1\)

Hence the resultant field will be 5 times that due to the smaller loop if current flows in the same direction.

Example 10. A wire loop is formed by joining two semicircular wires of radii r₁ and r₂. If the loop carries a current I, find the magnetic field at the center O.

Solution:

The magnetic field at the point O due to the semicircular part MNP is,

⇒ \(\dot{B}_{M N P}=\frac{\mu_0}{4} \cdot \frac{I}{r_1} \text { (upwards) }\)

Similarly magnetic field at O due to the semicircular part GFE is,

⇒ \(B_{G F E}=\frac{\mu_0}{4} \cdot \frac{I}{r_2} \text { (upwards) }\)

As the point, O lies along the straight parts ME and PG of the loop, the magnetic field due to them at O is zero.

The so-net magnetic field at O due to the whole loop,

B = B₁ + B₂

⇒ \(\frac{\mu_0}{4} \cdot \frac{I}{r_1}+\frac{\mu_0}{4} \cdot \frac{I}{r_2}\)

∴ \(B=\frac{\mu_0}{4} I\left[\frac{1}{r_1}+\frac{1}{r_2}\right] \text { (upwards) }\)

Example 11. The radius and number of turns of a circular coil are 10 cm and 25 respectively. What should be the current through the coil that will produce a magnetic field of 6.28 x 10^-5Wb.m^-2 at its center?
Solution:

Radius of the coil, r = \(\frac{10}{2}\) = 5cm = 0.05m;

number of turns, N = 25, and magnetic field at the center of the coil

B = 6.28 x 10^-5Wb.m^-2.

Now, \(B=\frac{\mu_0 N I}{2 r}\)

∴ \(I=\frac{2 r B}{\mu_0 r}=\frac{2 \times 0.05 \times\left(6.28 \times 10^{-5}\right)}{\left(4 \times \pi \times 10^{-7}\right) \times 25}\)

= 0.2A

Example 12. The magnetic field due to a current carrying a circular loop of radius 3 cm at a point on the axis at a distance of 4 cm from the center is 54μT. What will be Its value at the center of the loop?
Solution:

Magnetic field on the axis of a circular conductor; \(B=\frac{\mu_0 I}{2} \frac{r^2}{\left(r^2+x^2\right)^{3 / 2}}\) and magnetic field at the centre (x = 0) is

⇒ \(B^{\prime}=\frac{\mu_0 I}{2 r}\)

∴ \(\frac{B^{\prime}}{B}=\frac{\left(r^2+x^2\right)^{3 / 2}}{r^3}\)

or, \(B^{\prime}=\frac{\left(r^2+x^2\right)^{3 / 2}}{r^3} B\)

⇒ \(\frac{\left(3^2+4^2\right)^{3 / 2}}{3^3} \times 54\)

∴ \(B^{\prime}=250 \mu \mathrm{T}\)

Magnetic Effect Of Current And Magnetism

Electromagnetism Ampere’s Circuital Law

Line integral or path integral of a vector: Let \(\vec{A}\) be a vector and 81 a very small line segment. This segment can be treated as a vector \(\delta \vec{l}\), where the magnitude of \(\delta \vec{l}\) is equal to the length of the segment and its direction is along the tangent to that segment. (\(\delta \vec{l}\) is shown in a magnified form),

Now, \(\vec{A} \cdot \delta \vec{l}=A \delta l \cos \theta\)

The sum of the above dot products along a finite line segment PQ can be expressed as an integral (using the symbol dl, instead of δl).

⇒ \(\lim _{n \rightarrow \infty} \sum_{i=1}^n \vec{A} \cdot \delta \vec{l}_i=\int_P^Q \vec{A} \cdot d \vec{l}=\int_P^Q A \cos \theta d l\)

It is called the line integral or path integral of vector \(\vec{A}\) along the path PQ. It is to be noted that the magnitude of vector \(\vec{A}\) may vary between point A to point B and if the direction of \(\vec{A}\) changes, θ will also change. The determination of the magnitude of a line integral is in general very complicated. However, due to different types of symmetry, the integral can often be determined easily.

Closed line integral:

A path that closes on itself is a closed path. To express the line integral of a vector along a closed path the symbol \(\oint\) is used.

For Example, the line integral of the vector \(\vec{A}\), called the circulation of \(\vec{A}\), is given by

⇒ \(\oint \vec{A} \cdot d \vec{l}=\oint A \cos \theta d l\)

Example:

The line integral of force vector work done:

If the force vector \(\vec{F}\) is taken as a special Example of vector \(\vec{A}\), according to the definition of work done we can write, work done by the force \(\vec{F}\) for displacement \(\overline{\delta l}\)

⇒ \(d W=\vec{F} \cdot \delta \vec{l}=F \delta l \cos \theta\)

So, tire total work done along the segment PQ.

⇒ \(W=\int_P^Q d W=\int_P^Q \vec{F} \cdot d \vec{l}=\int_P^Q F \cos \theta d l\)

Similarly, total work done by the force \(\vec{F}\) along a closed path,

⇒ \(W=\oint \vec{F} \cdot d \vec{l}=\oint F \cos \theta d l\)

If the force is conservative, the work done is zero. Naturally, the physical significance of the line integral of the force vector is this integral always indicates tire work done along a line.

Similarly, line integrals of different vectors in physics have different physical significances. For Example, the liter integral of tire electrostatic field \(\vec{E}\) around a closed path is zero because the electrostatic field is a conservation force field.

Statement of Ampere’s circuital law: The line integral of the magnetic field vector along a closed path in any magnetic field is equal to the product of the net current enclosed by the closed path and the permeability of vacuum, i.e.,

⇒ \(\oint \vec{B} \cdot d \vec{l}=\mu_0 I\)….(1)

Here, I = net current enclosed by the closed path.

Explanation:

If Ω is the closed path, it encloses current I, and as a result,

⇒ \(\oint_{X_1} \vec{B} \cdot d \vec{l}=\mu_0 I\)

On the other hand, if we consider the closed path X₂, it encloses no current and henceI = 0.

⇒ \(\oint_{X_2} \vec{B} \cdot d \vec{l}\)

2. If any closed path encloses a number of conductors, carrying currents in different directions, the algebraic sum of the enclosed currents is to be taken.

Currents in the parts AB, BC, and CD may be taken as I, -I, and I, respectively, and for the closed path X we can write,

⇒ \(\oint_X \vec{B} \cdot d \vec{l}=\mu_0(I-l+l)=\mu_0 l\)

Again, since an equal current, I flow through each of Tiro’s turns,

⇒ \(\oint_X \vec{B} \cdot d \vec{l}=\mu_0 N I \quad[N=\text { number of turns }]\)

CGS form of the law: Substituting B→H, I→i and μ₀ → 4π we can write,

for a closed path enclosing a single turn

⇒ \(\oint \vec{H} \cdot d \vec{l}=4 \pi i\)

and for a closed path enclosing N turns,

⇒ \(\oint \vec{H} \cdot d \vec{l}=4 \pi N i\)

Application of Ampere’s Circuital Law:

The magnetic field of a long straight wire:

Let a current I flow through a straight long conductor. We have to determine the magnetic field at the point P at a perpendicular distance r from the wire.

Taking the wire as an axis, a circular path is drawn through the point P having radius r in such a manner that the tire path lies in a plane perpendicular to the wire.

It is convenient to consider this circular path as Ampere’s closed path. For an element of length \(\delta \vec{l}\) on this closed path, the corkscrew rule shows that the magnetic field \(\vec{B}\) is parallel to \(\delta \vec{l}\) at every place, i.e., the angle between them is 0°.

Again, due to symmetry, the magnitude of \(\vec{B} \text { (i.e., }|\vec{B}|=B \text { ) }\) is the same at every point on the closed path.

So, \(\oint \vec{B} \cdot d \vec{l}=\oint_{B d l \cos \theta}\)

Since the current enclosedÿ by the closed path is I, from Ampere’s circuital law,

B – 2πr = μ₀I

or, \(B=\frac{\mu_0 I}{2 \pi r}=\frac{\mu_0}{4 \pi} \cdot \frac{2 I}{r}\)

Solenoid: If a long insulated conducting wire is wound tightly over the surface of a cylinder so that every circular turn is perpendicular to the axis of the cylinder, then this coll is called a solenoid.

The axis of the cylinder is the axis of the solenoid. Usually after making a solenoid the inner cylinder is removed. (A conductor with a coating of insulating material is called an insulated conductor.)

Magnetic lines of force: The solenoid is placed on a cardboard with its axis lying on the cardboard plane. Now some light iron filings are scattered over the cardboard and the cardboard is slightly tapped.

The iron filings arrange themselves along the magnetic lines of force. The magnetic lines of force.

Characteristics of the lines of force: The number density of magnetic lines of force inside the solenoid is very high, i.e., the magnetic field in that part is very strong.

The magnetic field outside the solenoid can be neglected in comparison. Moreover, the magnetic lines of force inside the solenoid are parallel to its axis. So, a strong and nearly uniform axial magnetic field is generated inside the solenoid

Magnetic field inside a long straight solenoid: Let the length of a long straight solenoid = L and its number of turns =N.

So, the number of turns per unit length of the solenoid, n = \(\frac{N}{L}\)A few number of turns of the solenoid is shown. Current through the solenoid = I.

Here, the rectangle abed is taken as the Ampere’s closed path whose side ab=L lies along the axis of the solenoid.

Let N be the number of turns enclosed by the rectangular path, \(\overrightarrow{d l}\) is a very small segment on this path, and \(\vec{B}\) is the magnetic field produced due to this part of the solenoid. According to Ampere’s circuital law,

⇒ \(\begin{array}{r} \oint_{a b c d} \vec{B} \cdot d \vec{l}=\int_a^b \vec{B} \cdot d \vec{l}+\int_b^c \vec{B} \cdot \overrightarrow{d l}+\int_c^d \vec{B} \cdot \overrightarrow{d l} \\ +\int_d^a \vec{B} \cdot d \vec{l}=\mu_0 N I \end{array}\) …(1)

As \(d \vec{l} \text { and } \vec{B}\) are in the same direction along ab so,

⇒ \(\int_a^b \vec{B} \cdot d \vec{l}=\int_a^b B d l \cos 0^{\circ}=B \int_a^b d l=B L\)….(2)

On the other hand, the magnetic field \(\vec{B}\) is perpendicular to the small segments \(d \vec{l}\) on the parts of me and da inside the solenoid. As the solenoid is an ideal one, \(\vec{B}\) = 0 for the side cd and also \(\vec{B}\) = 0 for the parts of me and da which are outside the solenoid.

Hence from equations (1) and (2) we get,

⇒ \(\oint_{a b c d} \vec{B} \cdot d \vec{l}=B L=\mu_0 N I\)…(3)

or, \(B=\mu_0 \frac{N}{L} \cdot I \quad \text { or, } B=\mu_0 n I\)….(4)

It is to be noted here that the magnitude of the magnetic field depends on the number of turns per unit length n but not on the total number of turns N of the solenoid. Hence, to increase the magnetic field it is not sufficient to increase the number of turns but it is also necessary to make the turns very close to each other so that the value of n increases.

CGS expression: Substituting B→H, I→I, and μ₀ → 4π in equation (4), we get, H = 4πni.

Toroid: A toroid is a long insulated conducting wire, wound on a donut-shaped core having a circular axis and uniform cross-section so that each turn is normal to the axis.

A toroid is nothing but a solenoid bent to close on itself. A long straight solenoid has two definite ends but a toroid is an endless solenoid.

The magnetic field of a toroid: Let the radius of the ring of a toroid = r and the total number of turns in it = N. Sd,’ the circumference of the circular axis of the toroid = 2nr, arid the number of turns per unit length of it, \(n=\frac{N}{2 \pi r}\). A few turns of a toroid are shown. Current through the toroid = I.

Here, the axis of the toroid is considered as the Ampere’s closed path. If any small part \(8 \vec{l}\) is taken on the axis, according to the corkscrew rule, the direction of the magnetic field \(\vec{B}\) is parallel to \(8 \vec{l}\) at every point, i.e., the angle between them is 0°. Now, due to symmetry, the magnitude of \(\vec{B}\) is the same at all points on the axis.

Hence,

⇒ \(\oint \vec{B} \cdot d \vec{l}=\oint B d l \cos 0^{\circ}=B \oint d l=B \cdot 2 \pi r\)

Again, the net current enclosed by the axis = current I through each of N turns = NI

So, according to Ampere’s circuital law,

⇒ \(B \cdot 2 \pi r=\mu_0 N I \quad \text { or, } B=\mu_0 \cdot \frac{N}{2 \pi r} I\)

or, B = μ₀nI….(5)

It should be noted that equation (5) is identical to equation (4). From this, it is concluded that if a solenoid is too long, whatever may be the shape, the magnitude of the magnetic field at any point on its axis will be B = μ₀nI.

CGS expression: Substituting B→H, I→i, and μ₀ → 4π in equation (5), we get, H = 4πni.

The core of a solenoid: In the above discussion, air is considered the core of a solenoid or a toroid. Thus, magnetic permeability is taken as μ₀. For any other core (like iron, nickel, etc.), the value of the magnetic permeability changes notably. In that case, Ampere’s circuital law takes the form

⇒ \(\oint \vec{B} \cdot d \vec{l}=\mu n I, \text { where } \mu\), is the permeability of the core.

Limitation of Ampere’s circuital law:

Maxwell proved that Ampere’s circuital law is valid only for steady current. If the enclosed current varies with time, on the right-hand side of equation (1), an additional term should be added.

By this correction, Maxwell arrived at his famous electromagnetic field equations. An elaborate discussion about it has been done in the chapter on Electromagnetic waves. We should remember that, Ampere’s law is not incorrect, it can only be called incomplete.

In our discussion, we consider the cases where electric current remains steady with time, and hence, the equations obtained from Ampere’s law are accurate.

Magnetic Effect Of Current And Magnetism

Electromagnetism Numerical Examples

Example 1. A solenoid with 7 turns per unit length Is carrying a current of 2.5 A. What is the magnetic Intensity Inside the solenoid?
Solution:

Turns per unit length, n = 7 cm^-1 =700 m^-1; if μ is the permeability of the medium inside the solenoid then, magnetic field B = μnl.

∴ Magnetic intensity,

H = \(\frac{B}{\mu}\) = nI

=700 x 2.5

= 1750 A m^-1

Example 2. The length of a solenoid is GO cm and Its total number of turns is 1250. If 2 A current is passed through It, what will be the magnetic field at any point on Its axis?
Solution:

Number of turns = 1250, length = 60 cm = 0.6 m

∴ Number of turns per unit length, n = \(\frac{1250}{0.6}\) m^-1

So, the magnetic field at any point on its axis,

B = μ₀nI

= 4π x 10^-7 x \(\frac{1250}{0.6}\) x 2 [μ₀ = 47T x 10^-7Wb.A^-1 m^-1]

= 5.23 x 10^-3 Wb/m²

Example 3. Two solenoids made of insulated conducting wires and of equal lengths are such that one is wound over another. The resistance of each of them is R and the number of turns per unit length is n. The solenoids are now connected in

1. Series
2. Parallel and the combination Is then connected with a battery of emf E.
3. If current flows through them in the same direction in both cases, determine the magnetic field along the axis of the solenoids in each case

Solution:

1. In case of series combination, equivalent resistance =2R and hence current through each solenoid, \(I_s=\frac{E}{2 R}\)

Hence, the resultant magnetic field along the axis,

⇒ \(B=B_1+B_2=\mu_0 n I_s+\mu_0 n I_s\)

⇒ \(2 \mu_0 n \cdot \frac{E}{2 R}=\frac{\mu_0 n E}{R}\)

in the case of parallel combination, the terminal potential difference across each solenoid = E.

So, current through each solenoid, Ip = \(\frac{E}{R}\).

⇒ \(B=B_1+B_2=\mu_0 n I_p+\mu_0 n I_p\)

⇒ \(2 \mu_0 n \cdot \frac{E}{R}=\frac{2 \mu_0 n E}{R}\)

Example 4. A long straight solid conductor of radius 5 cm carries a current of 2 A, which is uniformly distributed over its circular cross-section. Find the magnetic field at a distance of 3 cm from the axis of the conductor.
Solution:

Let us consider an internal point P at a distance r(= 3 cm) from the axis of the conductor. Imagine a circular path of radius r around the conductor, such that P lies on it. If R is the radius of the solid conductor then the current enclosed by the circular path,

⇒ \(I_1=\frac{I}{\pi R^2} \times \pi r^2=\frac{I r^2}{R^2}\)

Let B be the magnetic field at point P due to the current-carrying conductor. B acts tangentially to the circular path. So according to Ampere’s Circuital law,

⇒ \(\oint \vec{B} \cdot d \vec{l}=\mu_0 I_1\)

or, \(B \times 2 \pi r=\frac{\mu_0 I r^2}{R^2}\)

Here, I = 2 A, r = 3 cm = 0.03 m, R = 5 cm = 0.05 m

∴ \(B=\frac{10^{-7} \times 2 \times 2 \times 0.03}{(0.05)^2}=4.8 \times 10^{-6} \mathrm{~T}\)

Force on A Moving Charge In A Magnetic Field

Let \(\vec{B}\) = magnetic field at a point, q= electric charge of a particle, \(\vec{v}\) = velocity of the particle at the given point.

Magnetic force on the charged particle due to the magnetic field at that point,

⇒ \(\vec{F}=q \vec{v} \times \vec{B}\) ….(1)

If the angle between \(\vec{v} \text { and } \vec{B} \text { be } \theta\), from the cross product of two vectors,

⇒ \(F=|\vec{F}|=q \nu B \sin \theta\)…(2)

Naturally, if v = 0, F = 0, i.e., if a charged particle is at rest then no magnetic force acts on it.

Definition of magnetic field \(\vec{B}{/latex]:

Direction of [latex]\vec{B}{/latex]: If θ = 0° or 0 = 180°, F = 0. Thus no magnetic force acts on a charged particle that is moving parallel or antiparallel to the magnetic field. Hence, in a magnetic field, the direction (or its opposite direction) of a moving charged particle for which no magnetic force acts on it, defines the direction of [latex]\vec{B}{/latex].

Magnitude of [latex]\vec{B}{/latex]: If v and B remain perpendicular to each other, d = 90°. Thus, the magnitude of the magnetic force [latex]\vec{F}{/latex] becomes maximum. Expressing this maximum force by Fm, equation (2) can be written as,

⇒ [latex]F_m=q v B\)…(3)

If q = 1 and v – 1, then B = Fm. The maximum possible force exerted by a magnetic field on a particle of unit charge moving with unit velocity through the field defines this magnitude of the magnetic field.

Unit of B: From equation (3), \([B]=\frac{[F]}{[q][v]}\)

Hence the unit of B is,

⇒ \(\frac{\mathrm{N}}{\mathrm{A} \cdot \mathrm{s} \cdot\left(\mathrm{m} \cdot \mathrm{s}^{-1}\right)} \quad \text { or } \mathrm{N} \cdot \mathrm{A}^{-1} \cdot \mathrm{m}^{-1}\)

This unit is known as tesla (T) or Weber/metre2 (Wb.m”2).

Significance of the cross product:

Applying the rule of the cross product of two vectors in equation (1) we can conclude that \(\vec{F}\) is always perpendicular to the plane containing \(\vec{v} \text { and } \vec{B}\).

If a right-handed corkscrew rotated from the direction of \(\vec{v} \text { to } \vec{B}\), the direction of advancement of the screw-head indicates the direction of \(\vec{F}\).

Work done by the magnetic force is zero: \(\vec{F} \text { and } \vec{v}\) is always perpendicular to each other. Since the displacement of a particle is taken along the direction of its velocity, at any point in the magnetic field, the force \(\vec{F}\) and a small displacement \(\vec{ds}\) of the particle are perpendicular to each other. Then, for the magnetic force F and its displacement \(\vec{ds}\), work done,

⇒ \(d W=\vec{F} \cdot d \vec{s}=F d s \cos 90^{\circ}=0\)

Hence, the work done by the magnetic force on a moving charged particle in a magnetic field is zero. In other words, the magnetic force is a no-work force.

Again, we know that, work done on a free particle = change in the kinetic energy of the particle. Since magnetic force is a no-work force, the kinetic energy of a charged particle is not affected.

∴ AK = W =0

or, \(\frac{1}{2}\) mv² = constant (m = mass of the particle)

or, v = constant

So, when a magnetic force acts on a charged particle moving in a magnetic field, its speed as well as its kinetic energy remains unaffected.

Magnetic force In CGS system: in this system, charge q is expressed in esu [see the chapter: ‘Electric Fields’]. But to indicate the magnetic force, the current as well as the charge must be measured in emu.

1 emu of charge = 3 x 10¹⁰ esu of charge = c esu [ c = velocity of light in vacuum = 3 x 10¹⁰ cm.s^-1 ]

So, if the magnitude of an electric charge is q esu, in the electromagnetic unit, it will be \(\vec{q}{c}\) emu.

Hence, the above-mentioned SI equation (1) can be expressed in

CGS system as,

⇒ \(\vec{F}=\frac{q}{c} \vec{v} \times \vec{B}\)

In this equation, magnetic field \(\vec{B}\) is used but not the magnetic intensity \(\vec{H}\). Here c = 3 x 10¹⁰ cm.s^-1. F, q, v, and B are measured in dyn, esu, cm.s^-1, and gauss or G, respectively.

Note that, 1 C = 0.1 emu of charge = 3 x 10⁹ esu of charge.

Fleming’s left-hand rule: If a charged particle moves at right angles to the magnetic field, is if the angle between \(\vec{v}\) and \(\vec{B}\) be θ = 90°, then from it can be concluded that the three vectors \(\vec{v}, \vec{B} \text { and } \vec{F}\) are mutually perpendicular. This special case can be easily explained by Fleming’s left-hand rule.

Path of a Moving Charge in a Uniform Magnetic Field:

Uniform magnetic field: A magnetic field is said to be uniform if its magnitude and direction remain constant in a region. We know that a magnetic field is represented by magnetic lines of force. For a uniform magnetic field

Magnetic lines of force are parallel to each other because the direction of the magnetic field remains constant;

The magnitude of the magnetic field also remains unchanged and hence the number density of the lines of force at different points are equal, i.e., the lines of force are equispaced.

Moreover, to represent a uniform normal and upward magnetic field with respect to the plane of the page; equispaced marked points are used; on the other hand, to denote a downward uniform magnetic field similar marked points are used

In the region adjacent to the surface, the terrestrial magnetic field is assumed to be formed. Only in the vicinity of magnets or magnetic materials, are these lines of force distorted a little.

A uniform magnetic field is also generated between two strong opposite magnetic poles kept very close to each other.

Determination of the path of a charged partake: The charged particle is at rest: Since \(\vec{F}=q \vec{v} \times \vec{B}\) if \(\vec{v}\) = 0, \(\vec{F}\) = 0. Hence, in this case, no magnetic force acts on the particle and the charged particle remains at rest.

The charged particle enters with a velocity \(\vec{v}\) parallel to the magnetic field: If the velocity of the particle and the magnetic field are parallel to each other, then \(\vec{v} \times \vec{B}=0\). So, the magnetic force, \(\vec{F}=q \vec{v} \times \vec{B}=0\). Since no magnetic force acts in this case, the particle continues to move along a straight line.

The charged particle enters with a velocity \(\vec{v}\) perpendicular to the magnetic field:

Let a uniform magnetic field B be acting upward, perpendicular to the plane of the paper. A particle having charge +q enters that magnetic field at P with a velocity \(\vec{v}\) parallel to the plane of the paper.

Applying cross product rule, we see that the magnetic force \(\vec{F}\) ing on the particle will be normal to \(\vec{v}\) and along the direction \(\vec{PO}\).

As a result, the particle will be accelerated towards \(\vec{PO}\), and hence it will tend to deviate from its path.

When the particle reaches another point Q, the magnetic force \(\vec{F}\) will still act normally to \(\vec{v}\) and along \(\vec{QO}\).

In this way, magnetic force acting on the charged particle at every point of its path is always directed towards point O.

This force acts as a centripetal force on the particle, which therefore keeps revolving along a circular path centered at O and of radius r (r = PO = QO). Since the magnitude of velocity does not change under the influence of the magnetic field, the charged particle will have a uniform circular motion.

The radius of the circular path: Magnetic force, \(\vec{F}=q \vec{v} \times \vec{B}\).

Since the angle between v and B is 90°, the magnitude of the magnetic force,

⇒ \(F=|\vec{F}|=q v B \sin 90^{\circ}=q v B\)

If the mass of the charged particle is m,

⇒ \(\text { centripetal force }=\frac{m v^2}{r}\)

So, \(q v B=\frac{m v^2}{r} \quad

or, [latex]r=\frac{m v}{q B}\)…(1)

We see from the equation (1) that:

For a given charged particle (q = constant) and for a definite magnetic field (B = constant), the radius of the circular path is directly proportional to the momentum (mv) of the particle. This property is utilized in the measurement of the mass of a charged particle in a mass spectroscope.

If a given charged particle (q = constant) enters a magnetic field with a definite momentum (mv= constant) then the radius of the circular path is inversely proportional to the applied magnetic field (B).

The radius of the circular path is inversely proportional to its specific charge \(\frac{q}{m}\). For Example, the charge of a proton and electron is the same but the mass of a proton is 1836 times that of an electron.

Hence, the value of \(\frac{q}{m}\) for an electron is 1836 times higher. It means that if a proton and an electron enter a magnetic field with equal velocity, the electron revolves in a circular path of a much smaller radius.

It is clear that if the charge is -q instead of +q, the direction of uniform circular motion will be reversed.

Period of resolution and cyclotron frequency: Circumference of the circular path =2πr.

Since it is a uniform circular Otiort, time period,

⇒ \(T=\frac{2 \pi r}{\nu}=2 \pi \frac{\dot{m}}{q B}\)…(2)

The number of complete revolutions made in unit time, i.e., frequency of the circular motion,

⇒ \(n=\frac{1}{T}=\frac{1}{2 \pi}\left(\frac{q}{m}\right) B\) ….(3)

This frequency n is called cyclotron frequency.

Evidently, both T and n are independent of the radius of the path as well as of the velocity. From equation (1), it is clear that \(frac{r}{v}\) = constant. This property for the motion of a charged particle in a magnetic field is utilized in particle accelerators like cyclotrons.

The charged particle enters the magnetic field at an inclined path: Let the z-axis be chosen along the direction of a uniform magnetic field acting at a place. A particle of charge +q and mass m enters that magnetic field with velocity \(\vec{v}\) at the point P on the xy -plane.

If this velocity \(\vec{v}\) is inclined at an angle θ with the magnetic field \(\vec{B}\), the component of velocity along \(\vec{B}\), i.e., along z-axis = vcosθ and the component of velocity perpendicular to B, i.e., on the xy -plane = vsinθ.

Naturally, no magnetic force acts on the particle the component cost, and hence this corÿonent remains unchanged. So, the charged particle performs a uniform linear motion along the magnetic field.

Again, the component vsinθ produces a uniform circular motion. The radius o|this uniform circular motion can be obtained by using for v in equation (1),

⇒ \(r=\frac{m v \sin \theta}{q B}\)….(4)

The time period and frequency of the particle in uniform circular motion do not depend on the velocity of the particle; hence just like equations. (2) and (3), it can be written as

⇒ \(T=\frac{2 \pi m}{q B} \text { and } n=\frac{1}{T}=\frac{q B}{2 \pi m}\)….(5)

Due to the combination of linear motion parallel to the z-axis and uniform circular motion on the xy-plane, the charged particle follows a spiral or helical path.

The axis of this helical path is the z-axis. For each complete revolution of the particle, the distance covered along the z-axis, i.e., along the direction of the magnetic field is called the pitch of this helical motion.

So, pitch = time period x linear velocity

⇒ \(\frac{2 \pi m}{q B} \cdot v \cos \theta\)

⇒ \(\frac{2 \pi m}{q B} \cdot \frac{q B r}{m \sin \theta} \cdot \cos \theta\)

= 2πrcotθ

= circumference of the circular path x cotθ …(6)

The property of the helical motion of a charged particle in a magnetic field is utilized in the magnetic focussing of different equipment.

Cyclotron:

The cyclotron was developed in 1932 by Professor E O Lawrence at the BerkelyInstitute, taliformia is a powerful particle accelerator for accelerating positively charged particles such as protons, aaa particles, etc., to very high energies so that they can be used in disintegration experiments.

Description: It consists of two cylindrical shells of copper having semicircular cross sections with the same diameter and height. They are open towards their diameter and all other sides are closed.

The diameters of the two shells are much larger than their heights. They are arranged side by side in such a way that a small gap exists between their diameters. Each is called ‘dee’ on account of its shape like the letter D.

Two pole pieces of a strong electromagnet are placed above and below the dees in such a way that a uniform magnetic field -defects perpendicular to the plane of the dees.

An alternating potential of the order of 105 V and of high frequency (about 106 Hz) is applied between the dees. So the dees act as the two electrodes -of the source of potential.

An ion source S is located near the center of the dees and it supplies the positive ions to be accelerated. At the periphery of the dees, an auxiliary negative electrode deflects the accelerated ions onto the target to be bombarded.

The whole space inside the dee is evacuated to a pressure of about 10^-6 mm of mercury. If the ‘whole arrangement is seen horizontally.

Principle of action: Suppose that a positive ion (e.g., hydrogen ion H+ or helium ion He++ ) emerges from the ion sources S. Generally these ions are produced by bombarding gas molecules with high-velocity electrons.

The positive ion produced at S will be attracted to whichever dee happens to be negative at that moment. Due to this attractive force-velocity of the ion increases and it enters the dee. Inside the dee, there is no electric field.

Owing to the magnetic field which is perpendicular to the plane of the dee the ion describes a semicircular path. The radius of the path is obtained from equation (1) of the section.

The ion moves in this path at a constant speed. At the end of the semicircular path when the ion arrives at the gap between the dees, then the other dee should become negative. Then the ion is again attracted, its velocity increases and it enters the other dee. From equation

It is understood that due, to the increase of velocity v, the radius r of the semicircular path will also increase. But from the equation

Of section, it is seen that in spite of the increase in velocity and radius, the time taken to describe each of the semicircular paths remains the same.

In this way, then at last reaches the outer edge of the dee and comes out with a high velocity. To increase the velocity of the ion, the magnetic field (\(\vec{B}\)) may be increased

Resonance condition: It is obvious that the condition of proper operation of a cyclotron is that the time taken by the ion to traverse a semicircular path will be equal to half of the time period of the alternating potential. So the frequency of the applied alternating potential difference n₀ must be equal to the frequency of revolution of the ion, i.e., cyclotron frequency n.

⇒ \(\frac{1}{2 \pi}\left(\frac{q}{m}\right) B=n_0\)….(1)

This equation is called the resonance condition of the cyclotron. In practice, the value of n₀ is kept fixed and the magnetic field B is varied until the above condition is satisfied.

The kinetic energy of the accelerated particle: If v is the velocity of a charged particle of mass m and charge q and if it moves perpendicular to a magnetic field B, the radius of its circular path according to the equation (1) of the section is,

⇒ \(r=\frac{m v}{q B} \quad\)

or, \(v=\frac{q B r}{m}\)

If R is the radius of a dee of the cyclotron, then the velocity of the charged particle ejected from the outlet is given by,

⇒ \(v_0=\frac{q B R}{m}\)

So, the kinetic energy of, the particle,

⇒ \(E=\frac{1}{2} m \nu_0^2=\frac{q^2 B^2 R^2}{2 m}\)….(2)

The kinetic energy of the accelerated particle Disadvantages:

1. If the velocity of the charged particle is high enough to be displacement comparable to the velocity of light, the mass of the particle does not remain constant.

According to the theory of relativity, m increases with the increase of velocity. So the resonance condition according to equation (1) is violated and the cyclotron does not function.

For an electron, the relativistic increase of mass is much greater. So the electrons very quickly get out of step. Hence a cyclotron is not used for accelerating electrons.

2. To make the charged particle sufficiently fast, dees of very large diameter are to be taken. This diameter may even exceed 100m. The construction of an electromagnet of so large diameter is prohibitively expensive and technically complicated.

Remedies:

1. With the increase of mass due to relativity the frequency n₀ of the alternating source may also be diminished in such a way that the product mn₀ always remains constant. In that case, the resonance condition is never violated. The machine with this arrangement is called a synchro-cyclotron.

2. Both B and n₀ can be changed simultaneously in such a way that

1. The resonance condition is always satisfied
2. In spite of the increase in velocity of the charged particle, the radius r of its circular path remains unchanged. So a thin annular electromagnet of that radius will serve the purpose. Hence technical complications and expenses may be reduced sufficiently. This machine is called a synchrotron.

Path of a Charged Particle in a Uniform Electric Field:

If a particle of charge +q experiences a force \(\vec{F}\) in a uniform electric field, from definition,

the electric field, \(\vec{E}=\frac{\vec{F}}{q} \quad \text { i.e., } \vec{F}=q \vec{E}\)…(1)

If the mass of the particle is m, then acceleration,

⇒ \(\vec{a}=\frac{\vec{F}}{m}=\frac{q \vec{E}}{m}\)….(2)

Naturally, for negative charges, the acceleration (\(\vec{a}\)) will be opposite to \(\vec{E}\). Due to this acceleration, the charged particle acquires different kinds of motion in different cases.

1. The charged particle is initially at rest: Initial velocity u = o. Hence, after time t,

the velocity of the particle, \(v=a t=\frac{q E}{m} t\)

and displacement, \(s=\frac{1}{2} a t^2=\frac{q E}{2 m} t^2\)

2. The charged particle enters with a velocity \(\vec{u}\) along the electric held: Since the direction, of acceleration, is in the direction of \(\vec{E}\) after time t,

velocity of the particle, \(v=u+\frac{q E}{m} t\)

and the displacement, \(s=u t+\frac{1}{2} \frac{q E}{m} t^2\)

3. The charged particle enters with a velocity \(\vec{u}\) perpendicular to the electric field: Let the uniform electric field act parallel to the y-axis and the charged particle enters this electric field with a velocity \(\vec{u}\) at the point P along the x-axis

There is no component of E along the x-axis and hence the particle has no acceleration in this direction. So, along the x-axis, the particle possesses a uniform velocity u.

If the distance covered by the particle in time t is x, then t = \(\frac{x}{u}\). Again, along y-axis, initial velocity of the particle = 0, but acceleration, a = \(\frac{qE}{m}\).

So, in time t, displacement of the particle along the y-axis is,

⇒ \(y=\frac{1}{2} a t^2=\frac{1}{2} \frac{q E}{m}\left(\frac{x}{u}\right)^2=\frac{q E}{2 m u^2} \cdot x^2\)

or, \(x^2=\frac{2 m u^2}{q E} y\)….(3)

This is an equation of a parabola (in the form of x² = 4ay)

So, the path of the particle will be parabolic.

Path of Charged Particle in Crossed Electric Field and Magnetic Field:

Let a particle of charge +q and mass m enter an electromagnetic field at the point O along the z-axis with a velocity \(\vec{v}\),

Electric field \(\vec{E}\) at that point O is along the x-axis and magnetic field \(\vec{B}\) is along the y-axis, i.e.,\(\vec{v}\) \(\vec{E}\) and \(\vec{B}\) are mutually perpendicular. Now, the electric force acting on the particle along the x-axis = qE. The magnetic force acting on the particle = qvB, and the cross-product rule shows that this magnetic force acts along the negative direction of the x-axis.

So, here the electric and the magnetic fields are oppositely directed. Keeping the directions of i, and B unchanged, if these two forces are made equal in magnitude, the net force acting on the charged particle becomes zero. In this case, the charged particle continues its motion along the z-axis with its initial Velocity v without suffering any deviation. The condition for this situation is,

⇒ \(q E=q v B \quad \text { or, } \quad \nu=\frac{E}{B}\)

Suppose the electric field E and magnetic field \(\vec{B}\) are acting simultaneously perpendicular to each other.

Let a stream of charged particles enter the space in a direction perpendicular to both \(\vec{E}\) and \(\vec{B}\). If the speed of the charged particles is different, then only those particles whose speed is equal to the ratio \(\frac{E}{B}\) will pass through the hole S on the screen without any deflection.

All other particles will be deflected from, their path and will not reach point S. This arrangement is known as a velocity filter. In the figure only the green particle would pass through the hole S as its velocity is equal to \(\frac{E}{B}\).

This principle, known as the velocity selector, Is utilized In the determination of Q specific charge \(\frac{q}{m}\) of an electron by J J Thomson’s experiment and θ mass of the nucleus with the help of mass-spectrometer.

Lorentz Force:

In an electromagnetic field, if a charge q enters with velocity \(\vec{v}\), the forces acting on it are

Electric force, \(\vec{F}_e=q \vec{E} \quad[\vec{E}=\text { electric field }]\)

Magnetic force, \(\vec{F}_m=q \vec{v} \times \vec{B} \quad[\vec{B}=\text { magnetic field }]\)

These forces, \(\vec{F}_e \text { and } \vec{F}_m\) are called Lorentz electric force and Lorentz magnetic force, respectively. The resultant Lorentz
force acting on the particle (charge),

⇒ \(\vec{F}=\vec{F}_e+\vec{F}_m=q \vec{E}+q \vec{v} \times \vec{B}\) – q (\(\vec{F}\) x \(\vec{v}\) x \(\vec{B}\))….(1)

In CGS or Gaussian system:

⇒ \(\vec{F}_{e q}=q \vec{E}, \vec{F}_m=\frac{q}{c}(\vec{v} \times \vec{B})\)

So, the resultant Lorentz force, \(\vec{F}=q\left[\vec{E}+\frac{1}{c}(\vec{v} \times \vec{B})\right]\)

Remarks:

1. If the charged particle is at rest in an electromagnetic field, an electric force still acts on it, but as [lavec]\vec{v}[/latex] = 0 no magnetic, force acts.

2. For negatively charged particles (e.g., electron or negative ion) q is replaced by -q, and hence the direct|on of each force \(\left(\vec{F}_e \vec{F}_m \text { and } \vec{F}\right)\) will be reversed.

Magnetic Effect Of Current And Magnetism

Electromagnetism Numerical Examples

Example 1. An amagnetic field of 0.4 T is applied on a proton moving with a velocity of 5 x 10^-6 m s^-1. The magnetic field acts at an angle of 30º with the direction of velocity of the proton. What will be the acceleration of the proton? (mass ofa proton = 1.6.x 10^-27 kg)
Solution:

Force acting on the proton, F= qvBsinθ.

∴ Acceleration antiproton = \(\frac{\text { force acting }}{\text { mass of proton }}=\frac{q v B \sin \theta}{m}\)

Here, q = 1.6 x 10^-19 C, v = 5 x 10⁶ m.s^-1,

B = 0.40 T, θ = 30° and m = 1.6 x 10^-27 kg.

∴ Acceleration of the proton

⇒ \(\frac{1.6 \times 10^{-19} \times 5 \times 10^6 \times 0.40 \times \sin 30^{\circ}}{1.6 \times 10^{-27}}\)

= 10¹⁴ m.s^-2

Example 2. An electron (mass = 9 x 10^-31g, charge = 1.6 x 10^-19 C) enters a magnetic field with a velocity of 10⁶ m si and starts rotating in a circular path of radius 10 cm. What is the value of the magnetic field?
Solution:

The electron is rotating in a circular path i.e. the direction Of the velocity of the electron is perpendicular to the direction of the magnetic field. So the magnetic force = qvB, which provides the necessary centripetal force for the circular motion.

∴ \(q v B=\frac{m v^2}{r}\) [r = radius of the circular path = 10cm = 0.1 m]

or, \(B=\frac{m v}{q r}=\frac{\left(9 \times 10^{-31}\right) \times 10^6}{\left(2.6 \times 10^{-19}\right) \times 0.1}\)

= 5.6 x 10^-5 T

Example 3. Two particles of equal charge are accelerated by applying the same potential difference and then allowed to enter a uniform magnetic field normally. If the particles keep revolving along circular paths of radii R₁ and R₂, determine the ratio of their masses
Solution:

If the charge of each particle is q and the potential difference applied is V, kinetic energy acquired = qV

So, \(q V=\frac{1}{2} m_1 v_1^2=\frac{1}{2} m_2 v_2^2\)

or, \(\frac{v_1}{v_2}=\sqrt{\frac{m_2}{m_1}}\)

Again, the magnetic force required to revolve along a circular path = centripetal force.

∴ For the first particle,

⇒ \(q v_1 B=\frac{m_1 v_1^2}{R_1} \quad\)

or, \(q B=\frac{m_1 v_1}{R_1}\)

For the second particle, \(q B=\frac{m_2 v_2}{R_2}\)

∴ \(\frac{m_1 v_1}{R_1}=\frac{m_2 v_2}{R_2} \quad\)

or, \(\frac{m_1}{m_2}, \frac{v_1}{v_2}=\frac{R_1}{R_2}\)

or, \(\frac{m_1}{m_2} \cdot \sqrt{\frac{m_2}{m_1}}=\frac{R_1}{R_2} \quad\)

or, \(\sqrt{\frac{m_1}{m_2}}=\frac{R_1}{R_2}\)

or, \(\frac{m_1}{m_2}=\left(\frac{R_1}{R_2}\right)^2\)

Example 4. In a cyclotron, the frequency of alternating current is 12 MHz and the radius of its dee is 0.53 m.

1. What should be the operating magnetic field to accelerate protons?
2. What is the kinetic energy of the proton beam produced by the cyclotron? Given, the mass of the proton = 1.67 x 10^-27 kg and charge = +1.6 X 10^-19 C.

Solution:

1. Condition of resonance,

⇒ \(\frac{1}{2 \pi}\left(\frac{q}{m}\right) B=n_0 \quad \text { or, } B=\frac{2 \pi n_0 m}{q}\)

Here, n0 = 12 MHz = 12 x 10⁶ Hz = 12 x 10⁶ s^-1

So, \(B=\frac{2 \times 3.14 \times\left(12 \times 10^6\right) \times\left(1.67 \times 10^{-27}\right)}{1.6 \times 10^{-19}}\)

= 0.79 Wb.m^-2

2. Kinetic energy of proton,

⇒ \(E^{\prime}=\frac{q^2 B^2 R^2}{2 m}\)

= \(\frac{\left(1.6 \times 10^{-19}\right)^2 \times(0.79)^2 \times(0.53)^2}{2 \times\left(1.67 \times 10^{-27}\right)}\)

= \(1.34 \times 10^{-12} \mathrm{~J}\)

= \(\frac{1.34 \times 10^{-12}}{1.6 \times 10^{-19}} \mathrm{eV}\)

= 8375 x 10^-6 eV

= 8.375 MeV

Example 5. A beam of protons with velocity 4 x 10 sm.s^-1 enters a uniform magnetic field of 0.4 T at an angle of 60° to the magnetic field. Find the radius of the helical path of the proton beam and the time period of revolution. Also, find the pitch of the helix. Mass of proton = 1.67 X 10^-27 Vg
Solution:

Mass of the proton, m = 1.67 x 10^-27 kg velocity of the proton, v = 4 x 10⁵ m.s^-1; charge, q = 1.6 x 10⁴⁹ C.

∵ The component of velocity perpendicular to the field is vsinθ, the radius of the helical path,

⇒ \(r=\frac{m \nu \sin \theta}{q B}\)

= \(\frac{\left(1.67 \times 10^{-27}\right)\left(4 \times 10^5\right) \sqrt{3}}{\left(1.6 \times 10^{-19}\right) \times 0.4 \times 2}\)

= 9 x 10^-3 m

= 0.9 cm

During the time period of the revolution,

⇒ \(T=\frac{2 \pi r}{\nu \sin \theta}\)

= \(\frac{2 \times 3.14 \times 0.009 \times 2}{4 \times 10^5 \times \sqrt{3}}\)

= 1.63 x 10^-7 s

The pitch of the helix,

p = vcosθ x T =4 x 10⁵ x \(\frac{1}{2}\) x 1.63 x 10^-7

= 3.26 x 10^-2 m

= 3.26 cm

Magnetic Effect Of Current And Magnetism

Electromagnetism Action Of A Magnet On Current

Oersted’s experiment deals with the influence of a current-carrying conductor on a magnet. The magnet, in reaction, also exerts equal but opposite force on the current-carrying conductor.

So, it is possible to move a current-carrying wire in a magnetic field. This kind of motion is very important for practical purposes. This phenomenon is the subject matter of electrodynamics.

Force on a Current-Carrying Conductor in a Magnetic Field:

The Lorentz force acting on a charge q moving with velocity \(\vec{v}\) in a magnetic field \(\vec{B}\) is,

⇒ \(\vec{F}=q \vec{v} \times \vec{B}\)

When current flows in a circuit, the drift motion of free electric charges through the circuit is considered as the cause of that current. Naturally, the magnetic force acting on those free charges in a magnetic field acts as the force on the current-carrying conductor.

Let the current through a circuit = I; a V element of length \(\vec{dl}\) of that circuit is considered.

The magnetic field at the position of this element of the circuit = \(\vec{B}\). If dq amount of charge crosses dl length in time dt, current the circuit, \(I=\frac{d q}{d t}\)

Velocity of the charge, \(\vec{v}=\frac{d \vec{l}}{d t}\)

So, the magnetic force acting on the small part \(\vec{dl}\)

⇒ \(d \vec{F}=d q \vec{v} \times \vec{B}=d q \cdot \frac{d l}{d t} \times \vec{B}=\frac{d q}{d t} d \vec{l} \times \vec{B}\)

∴ \(d \vec{F}=I d \vec{l} \times \vec{B}\) ….(1)

If the angle between \(d \vec{l} \text { and } \vec{B}\) be θ, then

\(dF=|d \vec{F}|=I d l B \sin \theta=B I d l \sin \theta\)….(2)

So, the magnetic force acting on the whole circuit or on a finite part of it,

⇒ \(F=\int d F=\int B I d l \sin \theta\)…(3)

Generally, for different elements- of the circuit, both B and θ may change. Hence, in the absence of some special symmetry, the integral of equation (3) becomes very complicated.

Special cases:

The current carrying wire is at right angles with a uniform magnetic field: For any small part of the wire, θ = 90° or sinθ = 1. If the magnetic field is uniform, B = constant. If a current carrying wire of length is placed in this uniform magnetic field, according to equation (3) the force acting on the wire will be,

F = BIl….(4)

The current-carrying wire is placed parallel to the magnetic field: For any small part of the wire, 0 = 0° or sinθ = 0.

So, according to equation (3), the force acting on the wire is,

F = 0….(5)

Note that, the units of l or dl, I, B, and F are m, A, Wb.m^-2, and N, respectively.

Fleming’s left-hand rule:

If a long straight current-carrying conductor is kept at right angles with a uniform magnetic field, i.e., if for each element of the wire, the angle between \(\vec{dl}\) and \(\vec{B}\) is θ = 90°, then according to the rule of cross-product we can say from equation (1) that the three vectors \(\vec{dl}\), \(\vec{B}\) and \(\vec{dF}\) will be mutually perpendicular.

Fleming’s left-hand rule is a handy tool for determining the direction of the magnetic force acting on a current-carrying conductor, perpendicular to the magnetic field.

Statement: If the forefinger, middle finger, and thumb of the left hand are stretched mutually perpendicular to each other, such that the forefinger points to the direction of the magnetic field and the middle finger points to the direction of current, then the thumb points to the direction of force experienced by the conductor.

This left-hand rule is also known as the motor rule.

Barlow’s wheel: We can demonstrate the action of a magnet on an electric current with the help of this experimental arrangement.

Description: B is the Barlow’s wheel made of a thin copper plate having a number of sharp teeth. The wheel B is kept vertical such that its one sharp tooth remains in contact with mercury kept in a container placed in between two poles of a strong horse-shoe magnet (NS). Now the wheel and the mercury inside the container (M) are connected to a source of current E.

Working principle: The magnetic field acts from the JV-pole to the S-pole of the magnet parallel to the horizontal platform.

Since the tooth of the wheel remains in contact with mercury, the circuit is closed as soon as the battery is switched on. As a result, the current flows in a downward direction.

From Fleming’s left-hand rule, we see that a magnetic force acts on the current toward the right. Due to this force, the tooth will be deflected towards the right. As a result, the circuit breaks.

Hence magnetic force will no longer act on the wheel. But due to inertia of motion, the next tooth comes in contact with mercury and in a similar way will be deflected towards the right. Hence, a continuous anticlockwise motion will be observed in Barlow’s wheel.

Discussions:

1. To get an effective result, the wheel should be very light. Moreover, the magnetic field and the electric current should be very strong.

2. If the direction of either the magnetic field or the electric current is reversed, the rotation of the wheel will also be reversed; but if both of them are reversed then the rotation of the wheel will be in the same direction.

3. The machine which converts electrical energy into rotational energy (usually mechanical energy) is called a motor. So, Barlow’s wheel is a motor, although it has no practical use.

Torque on a Current Loop in a Uniform Magnetic Field:

PQRS is a rectangular conductor. Its length, PQ = RS =l and breadth, QR = PS = b

So, the area of the rectangular face of the conductor, <A= Tb. This rectangular conductor is placed in a magnetic field B in such a way that

1. PQ and RS 316 perpendicular to the magnetic field
2. The surface PQRS is parallel to the magnetic field.

If a current is sent through the conductor,

1. No force acts on the arms QR and PS as they are parallel to the magnetic field,

2. Magnitude of the force (\(\vec{F}\)) acting on each of the arms PQ and RS, F = BIl.

Applying Fleming’s left-hand rule, we see that these two equal forces are downward and upward, respectively and hence they constitute a couple.

Since the perpendicular distance between these two forces is QR = PS = b, the moment of the couple, i.e., the torque acting on the coil is,

⇒ \(\tau=B I l \cdot b=B I A\)….(1)

and due to this torque, the coil starts to rotate. Instead of a single turn, if the conducting coil has N turns, torque acting on it,

⇒ \(\tau\) = BNIA….(2)

Equations (1) and (2) show that the torque r depends only on area A of the coil. This means that the shape of the coil is not important. Instead of a rectangular coil, a coil of any other shape of equal area may be used to obtain the same torque.

Vector representation of torque: We know that any plane surface can be treated as a vector, such that the magnitude of the vector is equal to the magnitude of its area and its direction is along the normal to that plane.

So, the quantity A in equation (1) can be expressed as a vector \(\vec{A}\) the direct arbitrary position of the rotating loop under the influence of torque. If the angle between \(\vec{A}\) and the magnetic field \(\vec{B}\) is θ, from equation (1) we get,

⇒ \(\tau=B I A \sin \theta\)….(3)

The vector representation of the above equation is,

⇒ \(\vec{\tau}=I \vec{A} \times \vec{B}\)…..(4)

So, the torque acting on a conducting coil having N turns will be

⇒ \(\vec{\tau}=N I \vec{A} \times \vec{B}\)….(5)

Magnetic Effect Of Current And Magnetism

Electromagnetism Numerical Examples

Example 1. 2A current is flowing through a circular coil of radius 10 cm, made of insulated wire and having 100 turns.

1. If the circular plane of the conductor is kept at right angles to the direction of a magnetic field of 0.2 Wb.m^-2, determine the force acting on the coil.
2. If the conductor is placed parallel to the magnetic field, determine the torque acting on it.

Solution:

Here,N = 100, r = 10 cm = 0.1 m,

B = 0.2 Wb m^-2.

∴ Circumference of the circular loop,

L = 2πr

= 2 x I x 0.1

= 0.628 m

Area of the circular loop,

A = πr² = π x (0.1)²

= 0.0314 m²

1. The whole circumference of the conductor is normal to the magnetic field; hence magnetic force

= NILB

= 100 x 2 x 0.628 x 0.2

= 25.12 N

[It should be noted that, in this situation, if the circular plane is taken as a vector, its direction will be perpendicular to that plane, i.e., parallel to the magnetic field. As a result, no torque will act on the conductor.]

2. If the circular plane is taken as a vector, its direction will be perpendicular to the plane, i.e., normal to the magnetic field.

As a result, the torque acting on the conductor

= NIAB

= 100 x 2 x 0.0314 x 0.2

= 1.256 N.m

Example 2. The radius of a circular coil having 100 turns Is 5 cm and a current of 0.5 A is flowing through this coil. If it is placed in a uniform magnetic field of strength 0.001 T, then what torque will act on the coil when the plane of the coil is

1. Parallel to the magnetic field
2. Inclined at 30° with the magnetic field
3. Perpendicular to the magnetic field?

Solution:

Number of turns = 100

Current, I = 0.5A

Radius of the circular coil, r = 5 cm = 5 x 10^-2m

∴ Area of the coil, A = πr² = 3.14 x (5 x 10^-2)²m²

Magnetic field intensity, B = 0.0001 T

Now torque acting on the current carrying coil due to magnetic field,

⇒ \(\tau=n B I A \sin \theta\)

1. When the coil is parallel to the malefic field, then θ = 90°

So, \(\tau=n B I A \sin 90^{\circ}\)

= 100 x 0.001 x 0.5 x 3.14 x 25 x 10^-4

= 3.93 x 10^-4 N.m

2. When the coil makes an angle 30° with the field, then θ = 60°

So, \(\tau=n B I A \sin 60^{\circ}\)

⇒ \(=100 \times 0.001 \times 0.5 \times 3.14 \times 25 \times 10^{-4} \times \frac{\sqrt{3}}{2}\)

= 3.4 X 10^-4 N.m

3. When the coil is perpendicular to the field, θ = 0°.

∵ \(\tau=0\)

Example 3. On a smooth plane inclined at 30° with the horizontal, a thin current-carrying metallic rod is placed parallel to the horizontal ground. The plane is In a uniform magnetic field of 0.15 T along the vertical direction. For what value of current can the rod remain stationary? The mass per unit length of the rod is 0.30kg/m.
Solution:

Along the inclined plane, the component of the magnetic force acting on the conductor and the component of the weight of the conductor will bring equilibrium. So,

BIlcosd = mgsinθ….(1)

Here, force on the conductor BIl acts horizontally towards the right.

Now, from equation (1),

⇒ \(I=\frac{m g \sin \theta}{B l \cos \theta}=\frac{m g}{B l} \tan \theta\)

∵ \(\frac{m}{l}=0.30 \mathrm{~kg} / \mathrm{m}, g=9.8 \mathrm{~m} / \mathrm{s}^2\)

B = 0.15 T, θ = 30°

∴ \(I=\frac{0.30 \times 9.8}{0.15} \tan 30^{\circ}\)

⇒ \(=2 \times 9.8 \times \frac{1}{\sqrt{3}}=11.3 \mathrm{~A}\)

Magnetic Effect Of Current And Magnetism

Electromagnetism Action Of Current On Current

Like parallel currents: Let two straight parallel conductors PQ and RS be kept horizontally and I₁ and I₂ be the currents flowing through them in the same direction

According to the corkscrew rule, for current I₁, the magnetic field B at any point O on the wire RS will act downwards. If Fleming’s left-hand rule is applied at point O on wire RS, it is seen that the wire RS will experience a force F towards the wire PQ.

As Reactiion, the wire PQ also will experience the same force F towards the wire RS. So, we can conclude that two parallel currents attract each other.

Unlike parallel currents: Suppose I₁ and I₂ currents are flowing through the wires PQ and RS, respectively mutually opposite directions. Applying Fleming’s left-hand rule similarly, it is found that the wire RS experiences a force F away from the wire PQ. The wire PQ also experiences an equal but opposite force that acts on It away from the wire RS. So, we can conclude that two unlike parallel currents repel each other.

The magnitude of the force of attraction or repulsion: Let r be the perpendicular distance between two long, straight parallel conductors kept in a vacuum or air.

Due to current I₁ the magnetic field at a distance of r will be

⇒ \(B=\frac{\mu_0}{4 \pi} \cdot \frac{2 I_1}{r}\)

From the relation F = BIl, we get, the force acting on the unit length of the wire placed at r, carrying current I₂, which is

⇒ \(\left.F=\frac{\mu_0}{4 \pi} \cdot \frac{2 I_1 I_2}{r} \text { [Putting } B=\frac{\mu_0}{4 \pi} \cdot \frac{2 I_1}{r} \text { and } l=I_2\right]\)….(1)

Here, the unit of r is m, I₁, and I₂ is A, F is N.m^-1, and μ₀ = 4π x 10^-7 H.m^-1.

This equation gives the definition of international ampere. If I₁ = I₂ = 1 A and r = 1 m,

⇒ \(F=\frac{\mu_0}{4 \pi} \times \frac{2 \times 1 \times 1}{1}=\frac{4 \pi \times 10^{-7}}{4 \pi} \times 2\)

= 2 x 10^-7 N

1 ampere: Two long straight parallel conducting wires, having negligible cross sections and carrying equal currents are kept 1 m apart from each other. The steady direct current for which each wire experiences a force of 2 x 10^-7 N per unit length, is called 1 ampere.

Expressions in Gaussian system: Substituting \(I_1 \rightarrow i_1, I_2 \rightarrow i_2 \text { and } \mu_0 \rightarrow 4 \pi\) in equation (1) we get,

⇒ \(F=\frac{2 i_1 i_2}{r}\)

Here unit of r is cm, F is dyn, and i₁ and i₂ emu

Obliquo currents:

1. If currents through two oblique straight conductors lying in the same plane converge to or diverge from their point of intersection, they attract each other

2. If currents through two oblique straight conductors lying in the same plane are such that, one of them is directed towards and the other directed away from their point of intersection, they repel each other.

Experimental Demonstration Roget’s Vibrating Spiral:

Description: A long elastic spring made of insulated copper wire is suspended from a rigid support. A small spherical copper bob, attached at the lower end of the spring, just touches the mercury kept in a vessel. The fulcrum at the upper end of the spring and mercury in the vessel are joined to a battery.

Working principle: When the battery is switched on, a unidirectional parallel current flows through the turns of the spring and hence the turns attract each other, resulting in the contraction of the spring.

Naturally, the copper bob remains no longer in contact with mercury and hence the circuit is cut off. As soon as the bob leaves the surface of the mercury, the attraction between the turns of the spring no longer exists.

Hence, the spring elongates due to its own weight, the copper sphere touches the mercury again and closes the circuit. In this way, alternate compression and elongation of the spring goes on and hence the spring vibrates continuously.

This experiment proves that a number of unidirectional parallel currents attract each other.

Magnetic Effect Of Current And Magnetism

Electromagnetism Numerical Examples

Example 1. Two very long- conducting wires are kept at a distance of 4 cm from each other in a vacuum. Currents flowing through the wires are 25A and 5A, respectively. Find the length of each conductor, which experiences a force of 125 dynes.
Solution:

Here, I₁ = 25 A, I₂ = 5 A, r = 4 cm = 0.04 m

So, the force acting per unit length of the wire,

⇒ \(F=F=\frac{\mu_0}{4 \pi} \cdot \frac{2 I_1 I_2}{r}=\frac{4 \pi \times 10^{-7}}{4 \pi} \cdot \frac{2 \times 25 \times 5}{0.04}\)

= 625 x 10^-6 N.m^-5

For 125 dyn or 125 x 10^-5 N force, the effective length of each wire,

⇒ \(l=\frac{125 \times 10^{-5}}{625 \times 10^{-6}}=2 \mathrm{~m}\)

Example 2. Two long straight parallel conducting wires kept 0.5 m apart, carry 1 A and 3 A currents, respectively.

1. What is the force acting per unit length of tire two wires?
2. At what position in the plane of the wires, the resultant magnetic field will be zero?

Solution:

1. Force acting per unit length,

⇒ \(F=\frac{\mu_0}{4 \pi} \cdot \frac{2 I_1 I_2}{r}=10^{-7} \times \frac{2 \times 1 \times 3}{0.5}\)

= 1.2 x 10^-6 N.m^-1

According to the corkscrew rule, the magnetic fields produced between two wires are mutually opposite in direction. Let any point P in this region be at a distance x from the first wire.

∴ The magnetic field at point P due to the first wire

⇒ \(\frac{\mu_0}{4 \pi} \cdot \frac{2 I_1}{x}\)

and magnetic field at the point P due to the second wire

⇒ \(\frac{\mu_0}{4 \pi} \cdot \frac{2 I_2}{(0.5-x)}\)

If the resultant magnetic field at the point P is zero, then

⇒ \(\frac{\mu_0}{4 \pi} \cdot \frac{2 I_1}{x}=\frac{\mu_0}{4 \pi} \cdot \frac{2 I_2}{(0.5-x)} \quad\)

or, \(\frac{0.5-x}{x}=\frac{I_2}{I_1}=\frac{3}{1}\)

or, 3x = 0.5 – x

or, 4x = 0.5

or, x = \(\frac{0.5}{4}\)

= 0.125 m

So, at any point on the plane between the wires, 0.125 m away from the first wire, the resultant magnetic field will be zero

Example 3. Two long parallel conductors, kept at a distance d, carry currents I₁ and I₂, respectively. The mutual force acting between them. Now, the current one is doubled and its direction is also reversed. If the distance of separation between them is made 3d, what will be the force acting between the two conductors?
Solution:

For the unit length of the conductors,

⇒ \(F=\frac{\mu_0}{4 \pi} \cdot \frac{2 I_1 I_2}{d}\)

If the current in the first conductor is doubled and its direction is reversed, it will be -2I₁.

∴ In the second case

⇒ \(F^{\prime}=\frac{\mu_0}{4 \pi} \cdot \frac{2\left(-2 I_1\right) I_2}{3 d}\)

= \(-\frac{2}{3} \frac{\mu_0}{4 \pi} \cdot \frac{2 I_1 I_2}{d}\)

= \(-\frac{2}{3} F\)

Example 4. A long horizontal wire AB, which is free to move in a vertical plane and carries a steady current of 20 A, is in equilibrium at a height of 0.01 m over another parallel long wire CD, which is fixed in a horizontal plane and carries a steady current of 30A. Show that when AB is slightly depressed, it executes simple harmonic motion. Find its period of oscillations.

Solution:

Force per unit length on wire AB due to the magnetic field produced by wire CD

⇒ \(\frac{\mu_0}{4 \pi} \frac{2 I_1 I_2}{r}\)

where Ix = 20 A, I2 = 30 A and distance between AB and CD, r = 0.01m.

So force on the whole of wire AB due to wire CD,

⇒ \(\left.F=\frac{\mu_0}{4 \pi} \cdot \frac{2 I_1 I_2}{r} \times L \text { [where } L=\text { length of wire } A B\right]\)

Since wire AB is in equilibrium, the force F must balance its weight by acting downwards. This is possible only if the two wires repel each other and hence I1 and I2 are flowing in opposite directions.

Accordingly, \(F=\frac{\mu_0}{4 \pi} \cdot \frac{2 I_1 I_2}{r} \times L=M g\)…..(1) [where M=mass of wire AB]

If we consider that the wire is depressed by a small distance d so that the height of AB over CD becomes (r-d), then the force on wire AB increases to,

⇒ \(F_1=\frac{\mu_0 2 I_1 I_2}{4 \pi(r-d)} \times L=\frac{\mu_0 I_1 I_2}{2 \pi r}\left(1-\frac{d}{r}\right)^{-1} L\)

= \(=\frac{\mu_0 I_1 I_2}{2 \pi r}\left(1+\frac{d}{r}\right) L\)

Now on releasing the wire AB, it moves under the effect of restoring force f.

⇒ \(f=F_1-M g=F_1-F=\frac{\mu_0 I_1 I_2}{2 \pi r}\left[1+\frac{d}{r}-1\right] L\)

= \(\frac{\mu_0 I_1 I_2 L}{2 \pi r^2} d\)

Acceleration of the wire AB, \(a=\frac{f}{M}=\frac{\mu_0 I_1 I_2 L}{2 \pi r^2 M} d\)

Since, except d all other physical quantities are constants

Hence the motion of the wire is simple harmonic.

So time period,

⇒ \(T=2 \pi \sqrt{\frac{2 \pi r^2 M}{\mu_0 I_1 I_2 L}}=2 \pi \sqrt{\frac{2 \pi r^2}{\mu_o I_1 I_2 L} \times \frac{\mu_0 I_1 I_2 L}{2 \pi r g}}\) [putting the value of M from equation (1)]

∴ \(T=2 \pi \sqrt{\frac{r}{g}}=2 \pi \sqrt{\frac{0.01}{9.8}}\)

= 0.2s

Magnetic Effect Of Current And Magnetism

Electromagnetism Galvanometer

A galvanometer is an instrument used to detect and measure the current in a circuit.

A current measuring instrument can be constructed using the heating effect, the chemical effect, or the magnetic effect of current. However, to measure a direct current (dc), the magnetic effect is most advantageous for practical purposes.

The two types of galvanometers used widely in laboratories are:

1. Moving magnet galvanometer: The basic principle of this instrument is the action of electric current on a magnet.

Example: Tangent galvanometer, Since galvanometer, Helmholtz double coil galvanometer.

This type of galvanometer is rarely used due to its disadvantageous setup and low sensitivity. The discussion is out of our syllabus.

2. Moving coil galvanometer: The basic principle of this instrument is the action of a magnet on an electric current.

Example: D’Arsonval galvanometer, table galvanometer.

Moving Coil Galvanometer:

With a moving coil galvanometer, we can detect and measure even a very weak current (about 10^-9A) in a circuit. This galvanometer.

Description of a moving coil:

ABCD: A coil of insulated copper wire wound over a rectangular frame made of cane or aluminum.

I: A cylinder of soft iron. The axis of the cylinder coincides with the axis of the coil and the cylinder is fitted inside the coil.

NS: Two magnetic poles made of soft iron attached with a strong permanent magnet. The gap between the two poles is cylindrical and in this gap, the coil is placed in such a way that the axis of the gap coincides with the axis of the coil.

Due to this special shape of the magnetic poles, the magnetic field always remains parallel to the plane of the coil.

Now this coil is fitted with a mechanical system such that:

1. The circuit current to be measured can be sent directly through the coil
2. A very accurate measurement of the deflection of the coil due to the flow of current can be done. For this purpose, two types of mechanical arrangements are widely used

1. Suspended-coil galvanometer or D’Arsonval galvanometer: This type of arrangement.

W: Phosphor-bronze wire. The coil ABCD is hung by this wire.

M: A small mirror attached to the phosphor-bronze thread W. Ift he coil rotates through an angle θ, and the mirror also rotates through θ. So, if a ray of light falls on the mirror from a stationary source of light, the reflected ray rotates through an angle of 20. If the displacement of the reflected light spot on a scale, kept at a distance D from the mirror, is d then,

⇒ \(2 \theta=\frac{d}{D} \quad(\text { when, } d \ll D)\)

or, \(\theta=\frac{d}{2 D}\)….(1)

For this arrangement, it is also called a mirror galvanometer.

b This arrangement is shown. Two magnetic poles are not drawn here.

HH: Two hairsprings. Two ends of the coil are connected with these two hairsprings. The circuit current to be determined enters the coil through the two bearings P and Q attached to HH.

R: A pointer. With the rotation of the coil, it also rotates. When no current flows through the galvanometer, the pointer remains vertical. It can move on either side of this position depending on the direction of the current.

The deflection of the pointer gives an idea of the magnitude of the current. The tire value of the rotation is obtained from a graduated circular scale. Titus it is also called a pointer galvanometer.

Theory: Let the number of turns of the rectangular coil = N, the area of its surface A, the magnetic field parallel to the coil = D, and current through the coil = I. The torque acting on the coil, \(\tau=B N I A\) Due to torque, the coil rotates from its equilibrium position.

On the other hand, the phosphor-bronze thread in a suspended coil galvanometer or the hairsprings in a table galvanometer always tries to return the coil to its equilibrium position by virtue of its elastic property.

The value of this restoring torque \(\tau\)‘ is directly proportional to the tire angle of deflection of the coil. So, if the tire coil comes to rest at an angular displacement of θ

⇒ \(r=r^{\prime} \text { or, } B N I A=c \theta\)

or, \(I=\frac{c}{B N A} \cdot \theta\)….(2)

Here, c = restoring torque per unit twist = constant. From this equation, the tire value of the current can be obtained.

Using equation (1) for a suspended coil galvanometer we get,

⇒ \(I=\frac{c}{2 B N A D} d\)….(3)

Discussions:

1. Relation between current and angle of deflection: From equation (2), it is seen that current (I) is directly proportional to the angle of deflection (θ), i.e., a linear relation exists between them. Thus, the current measuring scale of this kind of galvanometer is always uniform.

2. Setting of the galvanometer: This galvanometer contains no magnetic needle and hence it is not necessary to set this galvanometer magnetic meridian plane. Rather, it can be placed in any position.

3. Galvanometer sensitivity: A moving coil galvanometer is very sensitive and current can be measured very accurately with it. Thus, this type of galvanometer is widely used for current measurement.

Suspended-type galvanometers are highly sensitive. The current through the coil for which the displacement of the light spot on a scale kept at a distance of 1 m from the mirror is 1 .mm, which is called the sensitivity of the galvanometer. Its unit is A mm^-1

The sensitivity of this kind of galvanometer is about 10^-9 A.mm^-1.So, the galvanometer will be more sensitive for a lesser value of sensitivity.

On the other hand, a table galvanometer is very durable in spite of our lower sensitivity. Different table galvanometers are so designed that their pointers give full-scale deflection for a current range of 50μA to 15 mA

4. Condition of sensitivity:

From equation (2) we get, \(\frac{\theta}{I}=\frac{B N A}{c}\)

So, for a given current I, to get greater deflection θ, the value of \(\frac{B N A}{c}\) should be greater.

1. Using a powerful magnet the value of B can be increased.
2. A rectangular coil of multiple turns (N) and of sufficiently large surface area (A) has to be used. However, with an increase of N and A, the resistance of the galvanometer also increases and the coil becomes very heavy.
3. A suspension thread should be chosen for which c is sufficiently small and hence a very thin wire should be used which can bear the weight of the coil. A suspension wire made of phosphor bronze serves this purpose

5. Limitation: The main disadvantage of a moving coil galvanometer is overload. If the current is more than its endurance limit and is passed through this galvanometer, its phosphor bronze wire or the hairsprings may get damaged permanently.

Magnetic Effect Of Current And Magnetism

Electromagnetism Use Of Galvanometer As An Ammeter Or A Voltmeter

Ammeter: The instrument which measures current in an electrical circuit is called an ammeter. An ammeter is always connected in series with a circuit so that the entire current in the circuit can pass through it, Again, its resistance should be very low so that when it is joined in a circuit, the main current in that circuit does not decrease appreciably.

So, an ammeter milliammeter or microammeter) is a low-resistance instrument connected in series with an electrical circuit. The resistance of an Ideal ammeter should be zero.

Voltmeter: The instrument used for measuring the potential difference between two points of an electrical circuit is called a voltmeter. To equalize the potential difference between those two points with the potential difference between the two ends of the voltmeter, it should be connected in parallel with those two points in the circuit.

Again, the resistance of the voltmeter should be very high so that no appreciable fraction of the main current passes through the voltmeter. So, a voltmeter (or millivoltmeter) is an instrument of high resistance connected in parallel with an electrical circuit. The resistance of an ideal voltmeter should be infinite.

Disadvantages of using a galvanometer: The basic instrument used for measuring current is a moving magnet galvanometer or a moving coil galvanometer. In spite of this, there are some problems with using these instruments directly in laboratory experiments.

1. Some time is spent on leveling and setting of these galvanometers.

2. The magnitude of the current cannot be obtained directly from any scale. To calculate the magnitude of current we have to use the galvanometer formula after measuring the angle of deflection.

3. Usually the resistance of a galvanometer is of intermediate order (in most of cases, from 100Ω to 500Ω). When this galvanometer is used to measure current or potential difference, the results turn out to be erroneous; because it is not an ideal one to use as an ammeter as its resistance is not too low, and again it is not an ideal one as a voltmeter as its resistance is not too high.

Transformation of Galvanometers:

To make a galvanometer fit for everyday use in the laboratory, it should be effectively converted into an ammeter or a voltmeter. Thus, a galvanometer is the primary instrument while an ammeter or a voltmeter is the secondary instrument.

To construct an ammeter or a voltmeter, a moving coil table galvanometer is generally used. For this

The resistance of the galvanometer, G;

The current I_G, for the full-scale deflection of the pointer in the circular scale, should be known beforehand.

Galvanometer to ammeter: Let the resistance of the galvanometer = G and the current required for the full-scale deflection of the pointer = I_G

Now, a low resistance in parallel, i.e., a shunt S is connected with the galvanometer. The whole system is kept in a box covered with glass in such a way that

The pointer and the circular scale remain visible from the outside,

The terminals A and B are outside the box. The points A and B are connected with the external circuit. If the main circuit current corresponds to a full-scale deflection of the galvanometer, the galvanometer is effectively converted into an ammeter suitable for measuring maximum current I. Note that, due to the parallel combination of G and S, the equivalent resistance of the ammeter thus formed becomes sufficiently small.

Galvanometer to ammeter Calculation:

V_X – V_Y = I_G.G = I_S.S [here, Is stands for shunt current]

So, \(S=\frac{I_G}{I_S} \cdot G\)

Again, \(I=I_G+I_S \quad \text { or, } \quad I_S=I-I_G\)

Hence, \(S=\frac{I_G}{I-I_G} \cdot G\)….(1)

So, to convert the galvanometer into an ammeter to measure a maximum current I, a shunt S as obtained from equation (1), should be connected in parallel with the galvanometer.

Range of the ammeter: In equation (1), I > I_G; hence when an ammeter is constructed from a galvanometer, its range always increases and never decreases. If this range becomes n times, \(\frac{I}{I_G}=n\); so from equation (1) we get,

⇒ \(S=\frac{1}{I / I_G-1} \cdot G=\frac{G}{n-1}\) ….(2)

Resistance of the ammeter: The equivalent resistance of G and S is the resistance (R) of the ammeter.

So, \(\frac{1}{R}=\frac{1}{G}+\frac{1}{S}=\frac{1}{G}+\frac{n-1}{G}=\frac{n}{G} \text { i.e., } R=\frac{G}{n}\)…(3)

Galvanometer to voltmeters: In this case, a high resistance R is connected in series with the galvanometer. The two points A and B are connected with the external circuit.

So, if the current in the external circuit is I_G, the galvanometer current will also be and as a result, the pointer gives full-scale deflection. Hence, In this condition, the potential difference (V_A – V_B) between the two points A and B will show full-scale deflection of the pointer.

So, the galvanometer will then be effectively converted into a voltmeter. Due to the series combination of G and R, the equivalent resistance of the voltmeter becomes sufficiently large.

Galvanometer to voltmeters Calculation:

V_A – V_B = V = I_G (G+R)

or, \(G+R=\frac{V}{I_G} \quad .. R=\frac{V}{I_G}-G\)….(4)

So, to convert the tire galvanometer into a voltmeter fit for measuring a maximum potential difference V, high resistance of magnitude R, as obtained from equation (4), should be connected in series.

Range of voltmeter: In equation (4), V > I_GG (= V_G), and hence if a galvanometer is converted into a voltmeter, the range of the voltage always increases and never decreases.If this range becomes n -times, then \(\frac{V}{V_G}=n\); So from equation (4) we get,

⇒ \(R=G\left(\frac{V}{I_G \cdot G}-1\right)=G\left(\frac{V}{V_G}-1\right)=G(n-1)\)…(5)

Resistance of the voltmeter: Naturally, the resistance of the newly formed voltmeter,

R_V = G + R = G + G(n-l) = nG….(6)

Differences between an ammeter and a voltmeter:

Theoretically moving magnet or suspended-coll galvanometers may be converted to an ammeter or voltmeter in a similar way. However due to the practical difficulties associated with such arrangements only moving coil table galvanometers are converted to be used as an ammeter or voltmeter.

Magnetic Effect Of Current And Magnetism

Electromagnetism Numerical Examples

Example 1. A galvanometer of resistance 10Ω gives lull scale deflection for a current of 10 mA. How can this galvanometer be used?

1. As an ammeter to measure current of range 0-2 A
2. As a voltmeter having a voltage range of 0-5 V?

Solution:

Resistance of the galvanometer, G = 10^-11; maximum current, I_G = 10 mA = 0.01 A

1. Connecting a shunt S in parallel to the galvanometer, if the instrument is used between points A and B, an ammeter of current range 0-1 will be obtained.

In the given question, I = 2 A.

Now, I_S.S= I_G.-G

or, \(S=\frac{I_G}{I_S} G=\frac{I_G}{I-I_G} \cdot G=\frac{0.01}{2-0.01} \times 10\)

= 0.0503Ω

This is the required value of the shunt resistance.

2. A voltmeter having a voltage range of zero to (V_A – V_G) is obtained if a resistance R is connected in series with the galvanometer and the instrument is used between points A and B.

In the given problem, V_A – V_B = 5V

Now, V_A – V_B = I_G (G + R)

or, \(R=\frac{V_A-V_B}{I_G}-G=\frac{5}{0.01}-10\)

= 500-10

= 490Ω

Example 2. Full-scale deflection occurs, in a moving coil galvanometer of resistance 36Ω when 100 mA current flows through it. What arrangement should be made to convert it into a voltmeter in the 0-5V range? Draw the necessary circuit diagram.

Solution:

Resistance of the galvanometer = 36Ω, maximum current, IG = 100 mA = 0.1 A.

Connecting a resistance R in series with the galvanometer and using the equipment in between the two points A and B, we shall get a voltmeter to measure voltage difference ranging from 0 to (V_A– V_B).

Here, V_A – V_B = 5V

Now, V_A – V_B = I_G(G+R)

or, \(R=\frac{V_A-V_B}{I_G}-G=\frac{5}{0.1}-36\)

= 50-36

= 14Ω

This is the required value of the resistance R.

Example 3. A millivoltmeter of range ‘0-50 mV’ and resistance 50Ω is to be converted into an ammeter of range 0-1 A. How can it be done?
Solution:

Resistance of the millivoltmeter, R = 50 Ω; maximum voltage, V = 50 mV = 0.05 V.

Connecting a shunt S in parallel with the millivoltmeter and using this combination in between points A and B, we shall get an ammeter to measure a maximum current I.

Here, I = 1 A.

Now, I_S. S = I_V.R

or, \(S=\frac{I_V}{I_S} \cdot R=\frac{I_V}{I-I_V} \cdot R\)

⇒ \(\frac{\frac{V}{R}}{I-\frac{V}{R}} \cdot R=\frac{V}{I-\frac{V}{R}}=\frac{0.05}{1-\frac{0.05}{50}}=\frac{0.05}{1-0.001}=0.05 \Omega\)

This is the required value of the shunt resistance.

Example 4. How would you convert a voltmeter that can measure up to 150 V to an ammeter that can measure current up to 8R The Resistance of the voltmeter is 300Ω.
Solution:

Resistance of the voltmeter, R = 300Ω; maximum voltage, V = 150 V.

∴ Maximum current through the voltmeter,

⇒ \(I_V=\frac{V}{R}=\frac{150}{300}=0.5\)

Connecting a shunt S in parallel with the voltmeter and using this combination in between the points A and B, we shall get an ammeter to measure a maximum current I.

Here, I = 8 A

Now, I_S.S = I_V.R

or, \(S=\frac{I_V}{I_S} \cdot R=\frac{I_V}{I-I_V} \cdot R=\frac{0.5}{8-0.5} \times 300=\frac{0.5}{7.5} \times 300\)

= 20Ω

This is the required value of the shunt resistance.

Example 5. A galvanometer of resistance 100Ω gives full-scale deflection for a current of 10 mA. What is the value of, the shunt to be used to convert it into an ammeter which can measure current up to 10 A?
Solution:

Resistance of the galvanometer, G= 100Ω; maximum current, I_G = 10 mA = 0.01 A. Connecting a shunt S in parallel with the voltmeter and using this combination in between the points A and B, we shall get an ammeter useful to measure a maximum current I.

Here, I = 10A and I_S.S = I_G.G

or, \(S=\frac{I_G}{I_S} \cdot G=\frac{I_G}{I-I_G} \cdot G=\frac{0.01}{10-0.01} \times 100=\frac{1}{9.99}\)

= 0.1001Ω

Example 6. A moving coil galvanometer of resistance 50Ω gives full-scale deflection for a current of 50 mA. How can this galvanometer used to convert it into a voltmeter that can measure voltage up to 200 V?
Solution:

Resistance of the galvanometer, G = 50Ω; maximum current, I_G= 50 mA = 0.05 A. Connecting a resistance R in series with the galvanometer and using the equipment in between the two points A and B, we shall get a voltmeter to measure the voltage difference ranging from 0 to ( V_A– V_B)

Here, V_A – V_B = 200V

Now V_A – V_B = IG(G+R)

or, \(R=\frac{V_A-V_B}{I_G}-G=\frac{5}{0.1}-36=14 \Omega\)

This is the required value of resistance R.

Magnetic Effect Of Current And Magnetism

Electromagnetism Conclusion

• Two like poles repel each other and two unlike poles attract each other.
• Repulsion is the conclusive test of magnetization.
• The region surrounding a magnet in which the influence of the magnet is felt is called the magnetic field of the magnet.
• A magnetic line of force in a magnetic field is defined as a continuous line such that the litigant drawn at any point on it gives the direction of the magnetic field at that point.
• The direction of deflection of a magnetic needle due to the flow of electric current can be determined by
  • Ampere’s swimming rule
  • Right-hand thumb rule.
• The right-hand thumb rule is also called Oersted’s rule. For determination of the direction of the magnetic field at a point near a current-carrying conductor)
  • Maxwell’s corkscrew rule
  • The right-hand grip rule is used. The ratio of the magnetic field (B) acting at a point by a substance to the magnetic permeability (μ) of that substance is called the magnetic intensity or magnetizing field (H) at that point.

Units of B and H:

Relation between SI and CGS units:

• 1A.m^-1 = 4π x 10^-3 Oe , 1Wb.m^-2 = 10⁴ G
• The current which, when flowing through a conducting wire of length 1 cm bent in the form of an arc of a circle of radius 1 cm, produces a magnetic field of 1 Oe at the center of the arc, is called an electromagnetic unit of current.
• 1 emu of current =10 A.
• Ampere’s circuital law: The line integral of the magnetic field vector along a closed path in a magnetic field is equal to the product of the total current enclosed by the closed path and the permeability of the vacuum.
• \(\text { i.e., } \oint \vec{B} \cdot \overrightarrow{d l}=\mu_0 I\)
• Here, I is the total current enclosed by the closed path.
• If a long, insulated conducting wire is wrapped over the surface of a cylinder in such a way that each circular turn is perpendicular to the axis of the cylinder, the coil thus formed is called a solenoid.
• If a long, insulated conducting wire is wound on a donut-shaped core of uniform cross-section having a circular axis such that each turn is perpendicular to its axis, the coil thus formed is called a toroid. A toroid is also called an endless solenoid.
• If a positively charged particle of magnitude 1 C moves with velocity 1m s^-1 in a magnetic field and experiences a maximum force of 1 N, then the magnitude of the magnetic field is called 1 T or 1 Wb.m^-2.
• The magnetic force experienced by a moving charged particle in a magnetic field is a no-work force.
• The speed and kinetic energy of a charged particle does not change when a magnetic force acts on it.
• If the magnitude and direction of a magnetic field throughout a certain region remain constant, the field is called a uniform magnetic field otherwise, it is a non-uniform magnetic field.
• A charged particle follows a circular path inside a uniform magnetic field when it enters the field perpendicularly. A charged particle follows a helical path inside a uniform magnetic field when it enters the tire field obliquely.

Fleming’s left-hand rule: If the tire forefinger of the middle finger and the thumb of the left hand are stretched mutually perpendicular to each other, such that the forefinger points the direction of the magnetic field and the tire middle finger points the direction of the current, the thumb points the direction of the force acting on tire conductor.

• Two like parallel currents attract each other and two unlike parallel currents repel each other.
• Two long, linear, and parallel conducting wires of negligible cross sections are kept 1 nr apart from each other. The steady direct current flowing in each wire for which both of them experience a force of 2 x 10^-7 N.m^-1 is called 1 A current.
• Galvanometers used in the laboratory are:
  • Moving magnet galvanometer-examples: Tangent galvanometer, sine galvanometer.
  • Moving coil galvanometer-examples: D’Arsonval galvanometer, table galvanometer.
• According to Biot-Savart law or Laplace’s law, die magnitude of magnetic field ⇒ \(\delta \vec{B}\) at a point due to a small element δ1 of a current carrying wire is,
• ⇒ \(\delta B \propto \frac{I \delta l \sin \theta}{r^2} \text { or, } \delta B=k \frac{I \delta l \sin \theta}{r^2}\)
• where r = distance of the point from the element and
• θ = angle between the current element \(\delta \vec{B}\) and position vector \(\vec{r}\).
• In SI, the conventional form of Biot-Savart law is,
• ⇒ \(\delta B=\frac{\mu_0}{4 \pi} \cdot \frac{I \delta l \sin \theta}{r^2}\)
• [μ0 = magnetic permeability of vacuum = 4π x 10^-7 H.m^-1]
• Generally, \(B=\mu H \quad \text { or, } \quad H=\frac{1}{\mu} B\)
• The magnetic field at any point near a straight conductor,
• ⇒ \(B=\frac{\mu_0}{4 \pi} \frac{I}{r}\left(\sin \theta_1+\sin \theta_2\right)\)
• Magnetic field due to an Infinite straight conductor,
• ⇒ \(B=\frac{\mu_0}{4 \pi} \cdot \frac{2 l}{r}\)
• The magnetic field at the center of a circular conductor of N turns,
• ⇒ \(B=\frac{\mu_0 I N}{2 r}[r=\text { radius of the circle }]\)
• The magnetic field at a point on the axis of a circular conductor of N times,
• ⇒ \(B=\frac{\mu_0 N I}{2} \cdot \frac{r^2}{\left(r^2+x^2\right)^{3 / 2}}\)
• [where, x = distance of the point from the center of the coil]
• Magnetic field due to a long, straight solenoid at any point on its axis,
• ⇒ \(B=\mu_0 n I\)
• [n = number of turns per unit length of the solenoid = \(\frac{N}{L}\)
• Magnetic field due to a toroid,
• ⇒ \(B=\mu_0 n I\)
• [n = number of turns per unit length of the toroid = \(\frac{N}{2 \pi r}\)]
• If N number of conductors carrying current I are enclosed by a closed loop then according to circuital law,
• ⇒ \(\oint \vec{B} \cdot d \vec{l}=\mu_0 N I\)
• Force on a charged particle moving with a velocity (\(\vec{v}\)) in a uniform magnetic field (\(\vec{B}\)) is
• ⇒ \(\vec{F}=q \vec{v} \times \vec{B}\)
• The radius of the circular path described by a moving charged particle entering a uniform magnetic field perpendicularly,
• ⇒ \(r=\frac{m v}{q B}\)
• [m = mass of the charged particle, v = velocity of the particle, q = charge of the particle and B = magnetic field]
• Time period of revolution, \(T=2 \pi \frac{m}{q B}\)
• Number of complete revolutions in the circular path, i.e., frequency of circular motion,
• ⇒ \(n=\frac{1}{T}=\frac{1}{2 \pi}\left(\frac{q}{m}\right) B\)
• This is known as cyclotron frequency.
• The kinetic energy of the charged particle ejected from the outlet of the cyclotron device = \(\frac{q^2 B^2 R^2}{2 m}\) [ R = radius of the dee of cyclotron]
• A charged particle, when it enters a magnetic field obliquely, follows a helical or spiral path. The pitch of this spiral or helical path
• = time period x linear velocity
• = circumference of the circular path x cotθ
• Force experienced by a particle of charge + q in a uniform electric field
• ⇒ \(\vec{F}=q \vec{E}\)
• For a charge q moving in an electromagnetic field with velocity \(\vec{v}\), the forces acting are,
  • Electric force, \(\vec{F}_e=q \vec{E} \quad[\vec{E}=\text { electric field }]\)
  • Magnetic force, \(\vec{F}_m=q \vec{v} \times \vec{B}[\vec{B}=\text { magnetic field }]\)
• Resultant Lorentz force, \(\vec{F}=q(\vec{E}+\vec{v} \times \vec{B})\)
• In the case of a current-carrying conductor in a magnetic field, the resultant magnetic force acting on the whole circuit or a finite part of the circuit,
• ⇒ \(F=\int d F=\int B I d l \sin \theta\)
• The torque acting on a rectangular conductor having N turns placed in a uniform magnetic field,
• ⇒ \(\tau=B N I A\) [I = current-strength, A = area of cross-section of the rectangular conductor]

Vectorial form:

• ⇒ \(\vec{\tau}=N I \vec{A} \times \vec{B}\)
• If the perpendicular distance between two long parallel conductors kept in vacuum or in air be r, the force of attraction or repulsion per unit length of the wire is,
• ⇒ \(F=\frac{\mu_0}{4 \pi} \cdot \frac{2 I_1 I_2}{r}\) [I₁ , I₂ are the currents in the two wires]
• In the case of a moving coil galvanometer, if the number of turns in the tire coil is N, the area of the plane of coil A, the magnetic field parallel to coil B, current the of the coil I, and the coil comes to rest at an angle θ, then
• ⇒ \(I=\frac{c}{B N A} \theta\)
• Here, c = restoring torque per unit deflection = constant
• Required shunt to be connected with a galvanometer in parallel to convert it to an ammeter,
• ⇒ \(S=\frac{I_G}{I-I_G} \cdot G\)
• [where S = shunt resistance, G = galvanometer resistance, IG = maximum galvanometer current, I = main current in circuit]
• If the range of an ammeter is to be increased n -times,
• ⇒ \(S=\frac{G}{n-1}\)
• To convert a galvanometer into a voltmeter for measuring a maximum voltage of V, the resistance to be connected in series with the galvanometer,
• ⇒ \(R=\frac{V}{I_G}-G\)
• If the range of the voltmeter is to be increased n -times, then
• R = G(n-l)

Units of some magnetic quantities:

• When two current-carrying loops make an angle θ with each other, then net magnetic field at their center,
• ⇒ \(B=\sqrt{B_1^2+B_2^2+2 B_1 B_2 \cos \theta}\)

• The value of the magnetic field at at point on the center of separation of two parallel conductors carrying equal currents in the same direction is zero.
• If a current-carrying circular loop is turned into a coil having n identical turns then the magnetic field at the center of the coil becomes n² times the previous electric field i.e., B_n = n²B₁ where B_n = magnetic field at the center of the coil having n turns; B₁ = magnetic field at the center of the circular loop.

Magnetic Effect Of Current And Magnetism

Electromagnetism Very Short Questions and Answers

Question 1. What is the nature of lines of force in a uniform magnetic field?
Answer: Parallel and equispaced

Question 2. In a uniform magnetic field, lines of force are equispaced – straight lines.
Answer: Parallel

Question 3. How is the direction of a magnetic field $ at a point related to the magnetic line of force passing through that point?
Answer:

⇒ \(\vec{B}\) is tangential to the line of force at that point

Question 4. A magnetic needle is kept below a very long conducting wire. If the current is sent through the wire from north to south, in which direction will the north pole of the needle be deflected?
Answer: Toward the east

Question 5. A magnetic needle is placed below a very long conducting wire. If the current is sent through the wire from east to west, in which direction will the north pole of that needle be deflected?
Answer: No deflection occurs

Question 6. How does the magnetic field at a point near a long straight current-carrying conductor vary with the current and the distance of the point?
Answer: Proportional, inversely proportional

Question 7. When 1 A current flows through a circular conductor, the magnetic: fields generated at its centre are 10^-7 T. For what value of current, will the magnetic field be 10^-6 T?
Answer: 10 A

Question 8. Which physical quantity has the unit Wb.m-2? Is it a scar lar or a vector quantity?
Answer: Magnetic field or magnetic induction \(\vec{B}\) it is a vector quantity

Question 9. What is the unit of magnetic permeability μ₀ of vacuum?
Answer: H.m^-1

Question 10. What is the magnetic field produced at a distance of 1 m from a long, straight conductor carrying 1 A current?
Answer: 2 x 10^-7 T

Question 11. The magnetic field at a distance of 1 m from a long, straight conductor is 10-7 T. What is the current through the conductor?
Answer: 0.5 A

Question 12. Consider the circuit where APB and AQB are semi-circles. What is the magnetic field at the center C of the circular loop?

Answer: Zero

Question 13. A solenoid carrying 1 A current has a length of 1 m and contains 10000 turns. What is the magnetic field on the axis of the solenoid?
Answer: 4π x 10^-3T

Question 14. What will be the nature of the path of a charged particle when it enters a uniform magnetic field 5 normally?
Answer: Circular

Question 18. An electron and a proton enter a uniform magnetic field perpendicularly with the same speed. How many times larger will be the radius of the proton’s path than that of the electron’s path? Given: the proton is 1840 times heavier than the electron
Answer: 1840

Question 19. Can a stationary charge produce a magnetic field?
Answer: No

Question 20. What is the magnitude of force experienced by a stationary charge placed in a uniform magnetic field?
Answer:

⇒ \(\vec{F}_m=q \vec{v} \times \vec{B}=0(… \vec{v}=0)\)

Question 21. Does any force act on a magnetic north pole ifit is brought near a negatively charged conductor at rest?
Answer: No

Question 22. Does any force act on a charge moving in a magnetic field?
Answer:

⇒ \(\text { yes, } \vec{F}=q \vec{v} \times \vec{B}\)

Question 23. A long straight wire is carrying a current. An electron starts its motion on a line parallel to the wire in a direction same as that of the current. What will be the direction of the force on the electron?
Answer: Away from the wire in a normal direction

Question 24. A charge q moves with velocity \(\vec{v}\) at an angle 6 to a magnetic field P. What is the force experienced by the particle?
Answer: qvBsinθ

Question 25. An electron moving with a velocity of 107m.s-1 enters a uniform magnetic field of 1 T along a direction parallel to the field. What would be its trajectory?
Answer: Undeviated straight line

Question 26. A certain proton moving through a magnetic field region experiences maximum force. When does this occur?
Answer: The velocity of the proton is perpendicular to the magnetic field

Question 27. An electron beam projected along the positive x-axis experiences a force, due to a magnetic field, along the positive y-axis. What is the direction of the magnetic field?
Answer: Positive z-axis

Question 30. What is the mutual action between two, unlike parallel currents?
Answer: Repulsion

Question 31. Two long and straight parallel wires carry a current 1 A each. If the distance between the two wires is 1 m, what will be the force acting per unit length on them?
Answer: 2 x 10^-7 N.m^-1

Question 32. What type of galvanometer is used to prepare an ammeter or a voltmeter in the laboratory?
Answer: Moving coil table galvanometer

Question 33. In the case of a moving coil galvanometer, what is the relation between the current I and the angle of deflection θ?
Answer: Proportional

Question 34. How is a galvanometer converted into an ammeter?
Answer: By connecting a rightly chosen low-resistance shunt in parallel to it

Question 35. How should a resistance be connected with a galvanometer to convert it into a voltmeter?
Answer: In series

Question 36. What is the nature of the magnetic field in a moving coil galvanometer?
Answer: Radial

Magnetic Effect Of Current And Magnetism

Electromagnetism Assertion Reason Type

Direction: Those questions have Statement 1 and Statement 2, Of the four choices given below, choose the one that best describes the two statements.

1. Statement 1 is true, Statement 2 is true; Statement 2 is a correct explanation for statement I.
2. Statement 1 is true, Statement 2 Is true; Statement 2 is not a correct explanation for Statement 1.
3. Statement 1 is true, Statement 2 is false.
4. Statement 1 is false, and Statement 2 Is true.

Question 1.

Statement 1: A magnetic needle that can rotate in a horizontal plane undergoes a deflection when current is passed through a conducting wire, placed above and parallel to it.

Statement 2: A magnetic field is developed around a current-carrying conductor.

Answer: 1. Statement 1 is true, Statement 2 is true; Statement 2 is a correct explanation for statement I.

Question 2.

Statement 1: A long straight conductor attracts iron filings when a current is passed through it.

Statement 2: A magnetic field is developed around a current-carrying conductor.

Answer: 4. Statement 1 is false, Statement 2 Is true.

Question 3.

Statement 1: I = 1 A and OA = 1 cm. The magnetic field developed at A is 10^-5 Wb.m^-2.

Statement 2: If I am the current flowing through a long straight conductor, then the magnetic field developed at a distance r from the conductor is \(B=\frac{\mu_0 I}{2 \pi r}\)

Answer: 2. Statement 1 is true, Statement 2 Is true; Statement 2 is not a correct explanation for Statement 1.

Question 4.

Statement 1: Neither the magnetic field vector \(\vec{B}\) nor the
magnetic intensity vector \(\vec{H}\) of a magnetic field depend on
the nature of the medium.

Statement 2: If the magnetic permeability of a medium is \(\mu \text { then } \vec{H}=\frac{1}{\mu} \vec{B}\)

Answer: 4. Statement 1 is false, Statement 2 Is true.

Question 5.

Statement 1: A galvanometer of resistance G is converted to an ammeter by increasing its range n times. The resistance of the ammeter is \(\frac{G}{n}\).

Statement 2: A shunt of resistance has \(\frac{G}{n-1}\) to be connected in parallel with a galvanometer in order to increase its range n times.

Answer: 1. Statement 1 is true, Statement 2 is true; Statement 2 is a correct explanation for statement I.

Question 6.

Statement 1: The velocity of injection of a charged particle, being accelerated in a cyclotron remains constant irrespective of the applied magnetic field.

Statement 2: In a cyclotron the charged particle is accelerated only due to the applied electric field because magnetic force is a no-work force.

Answer: 4. Statement 1 is false, Statement 2 Is true.

Question 7.

Statement 1: If an electron and a proton are projected with equal momentum in a uniform transverse magnetic field, then the curvature of their padis is equal.

Statement 2: The mass of a proton is much higher than that of an electron.

Answer: 2. Statement 1 is true, Statement 2 Is true; Statement 2 is not a correct explanation for Statement 1.

Question 8.

Statement 1: A charged particle moves perpendicular to a magnetic field. Its kinetic energy remains constant, but momentum changes.

Statement 2: A magnetic force acts on the charged particle.

Answer: 2. Statement 1 is true, Statement 2 Is true; Statement 2 is not a correct explanation for Statement 1.

Magnetic Effect Of Current And Magnetism

Electromagnetism Match The Column

Question 1. Match the electromagnetic quantities in column A with the rules or formulae in column B.

Answer: 1-B, 2-C, 3-D, 4-A

Question 2. The BC of the conductor is semicircular. Match the values of the magnetic field developed at O given in column 2 with the corresponding segments given in column 1.

Answer: 1-A, 2-D, 3-B, 4-C

Question 3. A galvanometer of resistance 100Ω shows full-scale deflection for a current of 10 mA. Match the resistances in column 2 necessary to convert the galvanometer to an ammeter or voltmeter as given in column 1.

Answer: 1-D, 2-A, 3-C, 4-B

Question 4. Current in each of the two conductors parallel to each other and at a distance r is I (column 1). The magnetic fields at a point on the same plane and equidistant from the two conductors are given in column 2 (μ₀ = 4π x 10^-7 H.m^-1)

Answer: 1-A, 2-C, 3-D, 4-B

Question 5. A current-carrying wire. An electron at a point P moves with a velocity \(\vec{v}\). The direction of \(\vec{v}\) is given in column A and its deflection in column B. Match them.

Answer: 1-B, 2-C, 3-D,

Electric Energy And Power Introduction

In the preceding chapters, different types of electrical circuits, their constructions and uses have been discussed. The subject of discussion of this chapter is the measurement of electrical energy consumed in different electrical circuits and the power of different effective elements of a circuit. The practical idea about electrical power and energy is essential for systematic planning and control of the following subjects

1. How much electrical energy is to be generated in a generating station,
2. Howdie energy is to be distributed in different areas,
3. How the loss of energy is to be minimized,
4. What types of electrical appliances are to be used in the residence office and factory and how they should be used so that, necessary work will be obtained at minimum cost?

Electric Energy And Power Electrical Work

From the discussion in statical electricity, we know that, if Q amount of charge flows through a section under the potential difference V, then the amount of electrical work done is given by,

W = Q x V….(1)

In this equation, if Q is measured in coulomb'(C) and V in volt (V), then the unit of IV is the joule (I), i.e., the relation of the unit is

I = C x V

Read and Learn More Class 12 Physics Notes

Definition: The amount of work done to carry a 1-coulomb charge through a potential difference of 1 volt is called 1 joule.

Besides this, we know that, if current I flow through a conductor for time t, then the amount of charge flowing through it is given by,

Q = It…..(2)

So, combining equations (1) and (2) wo haw.

The electrical work done in the conductor,

W = VIt…..(3)

This work is the amount of electrical energy consumed by the conductor.

Active and Passive Electrical Devices:

The electrical work done by an electrical source in a circuit, l.e., the electrical energy sent in a circuit is used up generally in two ways:

Electrical energy is converted only to treat energy in the connecting wires and in some devices of the circuit.

In most of the cases, this heat energy has no use, i.e., this energy is dissipated. These devices are called passive devices of the circuit and their resistances are called passive resistances.

But in our daily lives, heat energy is utilized in many ways; e.g., electric heaters, electric irons, etc. are very useful appliances.

Yet, the resistance of diesel appliances will also be called passive resistance in conformity with the present discussion.

Electrical energy may be converted to other forms of energy.

For Example, from electrical energy, mechanical energy is obtained in an electric fan and chemical energy is obtained during the charging of a storage cell.

These types of devices are called active devices. However, in these appliances, some amount of heat is evolved. So, it may be said that the active devices possess some passive resistance also.

Let V be the potential difference between the points ll and C in the circuit. If current 1 flows through the circuit for time t, then electrical work,

IV = Vlt

In section AB of the circuit, there is an active electrical device (e.g., an electric fan) and in tire section DC there is a passive device whose resistance is R . This R also Includes (he resistance that generates heat in the active device.

Now, V_A – V_B = V₀

and, V_B – V_C = V’

V_AC = V_A – V_C

= (V_A– V_B) + (V_B– V_C)

= V₀ + V’

= V (say)

∴ V = V₀ + V’

or, Vlt = V₀lt+ V’lt

Again, in the section BC, according to Ohm’s law,

V’ = IR

or, V’t = IR- It

= I²Rt

So, Vlt= V₀It+I²Rt….(4)

This equation (4) can be expressed in the following form:

Electrical work in the circuit = transformed active energy + exothermic energy

If energy and heat are expressed in the same unit (e.g., in SI, both have the same unit joule or), the amount of exothermic energy and heat produced (H) will be equal

i.e., in that case, H = I²Rt

But in many cases, calorie (CGS) is used as the unit of heat. In that case, Joule’s law regarding the relation between heat and work, i.e., W = JH is to be used. In that case,

J = mechanical equivalent of heat or Joule’s equivalent
of heat

= 4.2 J.cal^-1

= 4.2 x 10⁷ erg.cal^-1

In this condition, if exothermic work or energy is W’, then heat produced,

⇒ \(H=\frac{W^{\prime}}{J}\)

i.e., \(H=\frac{I^2 R t}{J}\)….(5)

If work or energy is expressed in joule (J) and heat in calorie (cal), then J = 4.2 J.cal^-1

i.e., \(\frac{1}{J}=\frac{1}{4.2}\)

So, the practical form of the equation (5) is

⇒ \(H=\frac{t^2 R t}{4.2}=0.24 I^2 \mathrm{Rt}\)….(6)

the expense of electrical work Is known as the soulful effect or Joule heating.

Three laws regarding Joule effect are offlined easily from equation (5) or (6):

⇒ \(H \propto I^2\); when R and t are constants;

⇒ \(H \propto R\); when I and t are constants;

⇒ \(H \propto t\); when I and R are constants.

Mechanical equivalent of heat:

Prom equation (5), we have,

⇒ \(J=\frac{I^2 R t}{H}\)

Now, if I = 1 , R = 1 and t = 1

we have, J = \(\frac{1}{H}\)

From the tills relation, the mechanical equivalent of heat Is defined in current electricity.

Definition: The reciprocal of the heat produced in one second in a conductor of unit resistance, for the passage of unit current through it, is called the mechanical equivalent of heat.

In electricity, J = 4.2 J.cal^-1 means:

The amount of heat produced in 1Ω resistance, for the passage of 1 A current for 1 s through It = \(\frac{1}{4.2}\) cal = 0.24 cal.

Electric Energy And Power Numerical Examples

Example 1. 2A current was sent through a coil of resistance 100Ω for 30 min. Determine the amount of heat produced, the quantity of charge passed, and the amount of work done.
Solution:

Here, I = 2A, R = 100Ω t = 30min = 30 x 60s

∴ The amount of heat produced,

⇒ \(H=\frac{I^2 R t}{J}=\frac{(2)^2 \times 100 \times 30 \times 60}{4.2}\)

= 1.71 x 10⁵ cal

The quantity of charge flowing,

Q = It

=2 x 30 x 60

= 3600C

The amount of work done,

W = QV = QIR

= 3600 x 2 x 100

= 7.2 x 10s J

Example 2. Two separate circuits are made with resistances r₁ and r₂ connected to the same storage battery. What should be the internal resistance (r) of the storage battery for which an equal amount of heat is produced in the external circuits?
Solution:

For the first connection, current Ix = \(I_1=\frac{E}{r+r_1}\) [E = emf of the battery]

So, the heat produced in resistance rx in time t,

⇒ \(H_1=\frac{I_1^2 r_1 t}{J}=\frac{E^2 r_1 t}{\left(r+r_1\right)^2 J}\)

Similarly, the heat produced at the same time in the resistance r₂,

⇒ \(H_2=\frac{E^2 r_2 t}{\left(r+r_2\right)^2 J}\)

According to the question, H₁ = H₂

∴ \(\frac{E^2 r_1 t}{\left(r+r_1\right)^2 J}=\frac{E^2 r_2 t}{\left(r+r_2\right)^2 J}\)

or, \(r_1\left(r+r_2\right)^2=r_2\left(r+r_1\right)^2\)

or, \(r_1\left(r^2+2 r r_2+r_2^2\right)=r_2\left(r^2+2 r r_1+r_1^2\right)\)

or, \(r_1 r^2+r_1 r_2^2=r_2 r^2+r_2 r_1^2\)

or, \(r^2\left(r_1-r_2\right)=r_1 r_2\left(r_1-r_2\right)\)

or, \(r^2=r_1 r_2\)

or, \(r=\sqrt{r_1 r_2}\)

Example 3. A heating coll of resistance 5Ω is connected to a cell. The internal resistance of the cell is 20Ω. Calculate the value of the shunt to be introduced, so that, the energy consumed in the heating coil will be \(\frac{1}{9}\)th of the previous value.
Solution:

Let E be the emf of the cell and S be the shunt.

Here, the internal resistance of the cell, r = 20Ω, and external resistance, R = 5Ω.

In the absence of the shunt, current flowing in the circuit,

⇒ \(I_1=\frac{E}{R+r}\)

In time t, energy consumed in the resistance R,

⇒ \(W_1=I_1^2 R t=\frac{E^2 R t}{(R+r)^2}\)

Now, if shunt S is connected, current flowing in the circuit,

⇒ \(I_2=\frac{E}{\frac{R S}{R+S}+r}=\frac{E(R+S)}{R S+r(R+S)}\)

∴ Current flowing in the resistance R,

⇒ \(I_R=I_2 \cdot \frac{S}{R+S}=\frac{E S}{R S+r(R+S)}\)

∴ In time t, energy consumed in the resistance R

⇒ \(W_2=I_R^2 R t=\frac{E^2 S^2 R t}{[R S+r(R+S)]^2}\)

According to the question,

⇒ \(W_2=\frac{W_1}{9} \quad \text { or, } W_1=9 W_2\)

∴ \(\frac{E^2 R t}{(R+r)^2}=9 \cdot \frac{E^2 S^2 R t}{[R S+r(R+S)]^2}\)

or, \(\frac{1}{(R+r)^2}=\frac{9 S^2}{[R S+r(R+S)]^2}\)

or, \(\frac{1}{R+r}=\frac{3 S}{R S+r(R+S)} \text { or, } R+r=\frac{R S+r R+r S}{3 S}\)

or, \(S=\frac{r R}{2(R+r)}=\frac{20 \times 5}{2(20+5)}=\frac{20 \times 5}{2 \times 25}=2 \Omega\)

Example 4. The rate of energy consumed in 5Ω resistance is 10 J.s^-1. What will be the rate of energy consumed in 4Ω resistance?
Solution:

In the adjoining if, V_A-V_B= V, then,

⇒ \(I_1=\frac{V}{5} \text { and } I_2=\frac{V}{4+6}=\frac{V}{10}\)

∴ \({Energy consumed in 1 \mathrm{~s} in 4 \Omega resistance}{Energy consumed in 1 \mathrm{~s} in 5 \Omega resistance}\)

⇒ \(\frac{I_2^2 \cdot 4}{I_1^2 \cdot 5}=\frac{\left(\frac{V}{10}\right)^2 \cdot 4}{\left(\frac{V}{5}\right)^2 \cdot 5}=\frac{1}{5}\)

So, energy consumed in 1s in 4 resistance

= \(\frac{1}{5}\) x energy consumed in 5Ω resistance

= \(\frac{1}{5}\) x 10

= 2J

i.e., the rate of energy consumed in 4Ω resistance is 2 J.s-1

Example 5. Water boils in an electric kettle for 10 minutes after being switched on. How will you modify the heating coil to boil water in 5 minutes using the same source
Solution:

For the source of power of the same emf, V = constant.

As an equal amount of heat is produced in the two cases, we have

⇒ \(\frac{V^2 t_1}{R_1}=\frac{V^2 t_2}{R_2} \text {, because heat produced } \propto \frac{V^2 t}{R}\)

∴ \(R_2=R_1 \frac{t_2}{t_1}=R_1 \times \frac{5 \mathrm{~min}}{10 \mathrm{~min}}=\frac{1}{2} R_1\)

This means that a heating coil with half the resistance compared to the initial one, is to be used.

Electric Energy And Power Electrical Power

If an electrical source sends current I for time t, in a circuit under potential difference V, then the electrical work done in the circuit,

W = Vlt

We know that the rate of work done with respect to time is called power.

∴ \(P=\frac{W}{t}=\frac{V I t}{t}\)

or, P = VI …(1)

i.e., electrical power = potential difference x current

Unit of power: The units of current and potential difference are volt (V) and ampere (A) respectively. Again, the unit of work is joule (J) and the unit of power is joule/second (J/s) or watt (W).

So, from equation (1) we have,

watt = volt x ampere ….(2)

Definition: If the potential difference at the two terminals of an electrical appliance is 1 volt and the current passing through it is 1 ampere, then the power of the appliance is 1 watt.

We have seen that the relation between the electrical work done in an electrical circuit or in a portion of the circuit is electrical work = active energy + exothermic energy

or, Vlt = V₀It + I2Rt

For t= 1s; VI = V₀I+I2R

Equation (3) expresses that, electrical power applied = active power + power consumed in the production of heat

The active energy/power is of little concern to us. We are only interested in the latter, for which we shall find an alternative expression.

An alternative expression of the power of passive resistance: If V’ is the terminal potential difference and I is the current flowing through a passive resistance R, then according to Ohm’s law,

V’ = IR

The power consumed in this resistance is,

P’ = I²R [equation (3)]

= IR.I

= V’I, as already shown.

Since, \(I=\frac{V^{\prime}}{R}\)

∴ \(P^{\prime}=\left(V^{\prime} \cdot \frac{V^{\prime}}{R}\right)=\frac{V^{\prime 2}}{R}\)

So, \(P^{\prime}=V^{\prime} I=I^2 R=\frac{V^{\prime 2}}{R}\) …(4)

In most electrical circuits, there is no active device. All the resistances are passive. So, the entire power applied is consumed in the form of heat. In that case, the applied potential difference and the potential difference across the passive resistance (or resistances) become equal to each other.

That is, in this condition, V = V’ and P = P’. So equation (4) can be written as,

⇒ \(P=V I=I^2 R=\frac{V^2}{R}\)…(5)

It is evident from equations (4) and (5) that, the power consumed in the external resistance can be expressed in three different ways. P = VI is the general expression of power and it is applicable in all types of electrical appliances.

On the other hand, the relations \(P=I^2 R \text { or, } P=\frac{V^2}{R}\) are used as the expressions for power consumed in exothermic devices.

Wastage of energy in the transmission of electric power. Suppose, electrical power is to be transmitted over long distances from a generating station. Usually, a long line wire is used for this purpose.

Now, as current flows through this wire, a huge amount of heat is generated due to the Joule effect, i.e., a large amount of electrical energy is wasted as heat during transmission. So, the most important condition for transmitting the electrical power from one place to another is that the wastage of energy should be reduced as much as possible.

A solution to minimize wastage of energy: From the expression of the power P = VI, it can be easily understood that the higher the value of V, the lower will be the value of I.

Again, from equation (3), it is seen that, if the value of I is small, the value of power consumed in the production of heat (I2R) will also diminish, i.e., the wastage of energy will also diminish. So, a high-voltage power transmission allows for lesser resistive losses over long distances in wiring.

In actual practice, electrical energy is transmitted along the long line wire at a high voltage, 11000 V or more. At the place where the supply is required, the high voltage is stepped down to 220 V or 440 V by a step-down transformer.

Horsepower or hp: This is a large unit of power. From the discussion of mechanical energy and power, we know,

1 hp = 746 W

Generally, a horsepower unit is used to express the power of those machines (motor, pump, etc.), where electrical energy is transformed into mechanical energy.

Power Consumed in an Electrical Circuit:

Suppose, in an electrical circuit, the emf of the battery is E. If the internal resistance of the battery is r and R is the equivalent resistance of all the resistances used in the external circuit, current in the circuit, \(I=\frac{E}{R+r}\)

∴ Power consumed in the circuit, \(P_0=I^2(R+r)=\frac{E^2}{R+r}\)

Maximum power in the external circuit: A part of the total power, in the circuit equivalent to I²r, is consumed due to the internal resistance of the battery and it cannot be utilized in any other way. The best part of the power is spent in the external circuit and is utilized for running different electrical appliances connected in the external circuit. Since the equivalent resistance of the external circuit is R, the available power in the external circuit,

⇒ \(P=I^2 R=\frac{E^2 R}{(R+r)^2}\)

In case of constant emf and internal resistance of the battery (E = constant, r = constant), the condition of availability of maximum power in the external circuit is given by,

⇒ \(\frac{d P}{d R}=0 \quad \text { or, } \frac{d}{d R}\left[\frac{E^2 R}{(R+r)^2}\right]=0\)

or, \(E^2 \cdot \frac{1(R+r)^2-2(R+r) R}{(R+r)^2}=0\)

or, (R + r)²-2R²-2Rr = 0

or, r²-R² = 0

or, R = r

i.e., maximum power is obtained in the external circuit, if the equivalent resistance of the external circuit is equal to the internal resistance of the battery. The value of maximum power is

⇒ \(P_{\max }=\frac{E^2 R}{(R+R)^2}=\frac{E^2}{4 R}=\frac{E^2}{4 r}\)

Electric Energy And Power Numerical Examples

Example 1. A 20Ω resistor can dissipate a maximum of 2kW power as heat without being damaged. Should this resistor be connected directly across a 300 V dc source of negligible Internal resistance?
Solution:

If the resistor is connected directly with a 300Vdc source, its power would be,

⇒ \(P=\frac{V^2}{R}=\frac{(300)^2}{20}\)

= 4500W

= 4.5W

Since, this power is greater than 2 kW, so the resistor should not be directly connected to a 300V dc source.

Example 2. Three resistors of equal resistances which en connected in series across voltage sources, dissipate 1.00 watatt of power. What would be the power dissipated, if the resistors are connected in parallel across the same source of emf
Solution:

If each resistance is r, then the equivalent resistance in series, R₁ = 3r; and the equivalent resistance in parallel

⇒ \(R_2=\frac{r}{3}\)

In both combinations, the potential difference across the two ends is equal to V (say).

∴ In case of series combination, power, \(P_1=\frac{V^2}{R_1}\) and in case of parallel combination, power, \(P_2=\frac{V^2}{R_2}\)

∴ \(\frac{P_1}{P_2}=\frac{R_2}{R_1} \quad \text { or, } P_2=P_1 \cdot \frac{R_1}{R_2}=100 \times \frac{3 r}{\frac{r}{3}}\)

= 900W

Example 3. The coil of a heater connected to a 200V line consumes a power of 100 W. The coil Is divided Into two equal parts. The two parts are combined In parallel and connected to a 200 V line. What will be the power consumed by the new combination?
Solution:

If R is the resistance of the coll, then power summed in the potential difference V, \(P=\frac{V^2}{R}\)

In the first case, \(100=\frac{(200)^2}{R}\)

In the second case, the resistance of each of the two equal parts = \(\frac{R}{2}\)

So, the equivalent resistance in the parallel combination

⇒ \(\frac{\frac{R}{2} \times \frac{R}{2}}{\frac{R}{2}+\frac{R}{2}}=\frac{R}{4}\)

So, power consumed,

⇒ \(P_2=\frac{(200)^2}{\frac{R}{4}}=4 \times \frac{(200)^2}{R}\)

= 4 x 100

= 400W

Example 4. The power consumed in the circuit. Is 150 W. What is the value of it?

Solution:

Equivalent resistance,

⇒ \(r=\frac{R \cdot 2}{R+2}=\frac{2 R}{R+2}\)

So, power, \(P=\frac{V^2}{r}=\frac{V^2(R+2)}{2 R}\)

or, \(\frac{R+2}{R}=\frac{2 P}{V^2} \quad\)

or, \(1+\frac{2}{R}=\frac{2 \times 150}{(15)^2}=\frac{4}{3}\)

or, \(\frac{2}{R}=\frac{4}{3}-1=\frac{1}{3}\)

or, R = 6Ω

Example 5. With a ceil of emf 1.5 V and of internal resistance of 0.1Ω when connected with a resistor and an ammeter of negligible resistance in series, the ammeter shows a 2.0 A steady current. Find

1. The rate of energy dissipated within the cell
2. The power consumed in

Solution:

1. The rate of energy dissipated within the cell

= El = 1.5 x 2.0

= 3W

2. \(I=\frac{E}{R+r}\); R = resistance of the resistor

or, \(R=\frac{E}{I}-r\)

= \(\frac{1.5}{2}-0.1\)

= 0.65Ω

∴ Power consumed in the resistor,

= (2.0)² x 0.65

= 2.6 W

Example 6. A balanced Wheatstone bridge has resistances 100Ω, ion, 500Ω, and 50Ω respectively in its four arms. Determine the ratio of powers consumed in its different arms.
Solution:

Wheatstone Bridge has been shown.

Let \(I_1=\frac{V}{P+Q} \text { and } I_2=\frac{V}{R+S}\)

∴ \(\frac{I_1}{I_2}=\frac{R+S}{P+Q}=\frac{500+50}{100+10}=5 \quad \text { or, } I_1=5 I_2\)

∴ The required ratio

⇒ \(I_1^2 P: I_1^2 Q: I_2^2 R: I_2^2 S\)

⇒ \(\left(5 I_2\right)^2 \times 100:\left(5 I_2\right)^2 \times 10: I_2^2 \times 500: I_2^2 \times 50\)

= 2500: 250: 500: 50

=50: 5: 10: 1

Example 7. A factory requires a power of 90kW. The energy is transmitted to the factory through a 2.5 n line wire. If 10% of the power generated is lost in transmission, calculate

1. The transmission line current
2. The potential difference at the power generating station
3. The potential drop due to line resistance

Solution:

Here, \(\begin{aligned}
\frac{\text { power of the generating station }}{\text { power required for the factory }} & =\frac{100}{(100-10)} \\
& =\frac{100}{90}
\end{aligned}\)

or, power of the generating station = 90 x \(\frac{100}{90}\) = 100 kW

Power lost =100-90

= 10kW

= 10000 W

This power is lost in the transmission line.

So, according to the equation, P = I²R

Current in the transmission line,

⇒ \(I=\sqrt{\frac{P}{R}}=\sqrt{\frac{10000}{2.5}}=\sqrt{4000}\)

= 63.25A(approx.)

According to the equation, P = VI,

potential difference at the generating station,

⇒ \(V=\frac{P}{I}=\frac{10000}{63.25}=1581 \mathrm{~V}(\text { approx })\)

Potential drop due to line resistance,

V’ =IR

= 63.25 x 2.5

= 158.1V (approx.)

Example 8. Electrical energy is transmitted at the rate of 2.2 MW through the line wire. The resistance of the line wire is 25Ω. Calculate the percentage of heat dissipation of the electrical energy for each line voltage

1. 22000V
2. 110kV.

Solution:

VI = 2.2 MW

= 2.2 x 106W

1. \(V=22000 \mathrm{~V} ; I=\frac{V I}{V}=\frac{2.2 \times 10^6}{22000}=100 \mathrm{~A}\)

∴ Percentage of heat dissipation = \(\frac{I^2 R}{V I} \times 100 \%\)

⇒ \(\frac{(100)^2 \times 25}{2.2 \times 10^6} \times 100 \%\)

= 11.4%

2. \(V=110 \mathrm{kV}=110000 \mathrm{~V} ; I=\frac{V I}{V}=\frac{2.2 \times 10^6}{110000}=20 \mathrm{~A}\)

∴ Percentage of heat dissipation = \(\frac{I^2 R}{V I} \times 100 \%\)

⇒ \(\frac{(20)^2 \times 25}{2.2 \times 10^6} \times 100 \%\)

0.45%

[Obviously, it is seen that the dissipation of energy becomes less if the line voltage is high.]

Electric Energy And Power Commercial Units Of Electrical Work And Electrical Energy

Watt. hour (W.h): Work done or electrical energy consumed in 1 hour by an electrical appliance having power 1W is called 1W.h.

∴ lW h = lWx lh = 1W x 3600 s =3600J

It is obvious that, W.h = W x h

= V x A x h

i.e., if 1 A current flows for lh under the potential difference of IV, the amount of electrical energy spent will be 1W.h.

kiloWatt. hour (kW.h): Work done or electrical energy spent in lh by an electrical appliance having power lkW is called lkW h or 1 BOT unit.

Obviously, this is a bigger unit of electrical energy. The electricity supplied to us for our use is measured by this unit. Many times, kW h or BOT unit is simply called unit

lkW.h = 1000 W h = 1000 W x lh

= 1000 W X 3600 s

= 3.6 X 10⁶ W.S

= 3.6 X 10⁶J

⇒ \(\mathrm{kW} \cdot \mathrm{h}=\frac{\mathbf{W} \times \mathbf{h}}{1000}=\frac{\mathrm{V} \times \mathrm{A} \times \mathrm{h}}{1000}\)

For Example, if an electric bulb marked 100 W glows for 20h then, the amount of electrical energy spent = \(=\frac{100 \times 20}{1000}=2 \mathrm{~kW} \cdot \mathrm{h}\)

= 2 unit

So, in this case, the value of the electric meter will rise up by 2 units.

Electric Energy And Power Numerical Examples

Example 1. In a house, there are 10 lamps of 40 W each, 5 fans of 80 W each, and a TV set of 80 W. They run for 6 hours a day. Find the consumption of electrical energy in a month of 30 days. What is its value in BOT unit
Solution:

Total power = (10 x 40) + (5 x 80) + 80

= 880 W.

Total time = 6 x 30

= 180 h

= 180 x 3600 s

Energy consumed in a month = 880 x 180 x 360

= 5.7 x 108J

Its value in BOT unit = \(\frac{\mathrm{W} \times \mathrm{h}}{1000}=\frac{880 \times 180}{1000}\)

= 158.4

Example 2. In an evening college, there are 100 bulbs of 60 W each, 80 bulbs of 100 W each, and 70 fans of 100 W each. They run for 5h, 4h and 4h respectively per day. If each kW.h costs ₹ 0.50, calculate the electric bill of the college for a month
Solution:

The electrical energy consumed by 60 W bulb

= \(\frac{(60 \times 100) \times 5 \times 30}{1000}\)

= 900kW.h

Electrical energy consumed by 100 W bulbs

⇒ \(\frac{(100 \times 80) \times 4 \times 30}{1000}\)

= 960kW.h

Electrical energy consumed by fans

⇒ \(\frac{(100 \times 70) \times 4 \times 30}{1000}\)

= 840 kW.h

∴ Cost of electrical energy = (900 + 960 + 840) x 0.50

= ₹ 1350

Example 3. In a house, there are 20 lamps of 60W each, and 10 fans that operate in 0.5A current. If the main power supply is 220V, the expense per kW h is 50 paise and each application runs 6 h per day then calculate the electric bill of
Solution:

November month has 30 days.

∴Total time = 30 x 6h = 180h ;

50 paise = ₹ 0.5

Power of each fan =220 x 0.5 = 110W

∴ Electrical energy

⇒ \(\frac{(60 \times 20+110 \times 10) \times(30 \times 6)}{1000}\)

⇒ \(\frac{2300 \times 30 \times 6}{1000}\)

= 23 x 18 kW.h

∴Electric bill of November = ₹(23 x 18) x 0.5

= ₹ 207

Example 4. There are six 40 W and two 100 W lamps, four 40 W fans, and a 1000 W electric heater in the house. If in April, each lamp runs for 5 hours a day, each fan for 15 hours a day, and the heater for 2 hours a day, what will be the electric bill for that month? It may be supposed that the main supply voltage is 200 V and the cost of each BOT unit = ₹ 1.50. Which one of the given three fuse wires of rating 5 A, 10 A, and 15 A will be appropriate for connection in the main switch?
Solution:

Number of days in April = 30

Electrical energy consumed by the lamp

⇒ \(\frac{(6 \times 40+2 \times 100) \times 5 \times 30}{1000}\)

= 66 BOT unit

Electrical energy consumed by the fans

⇒ \(\frac{(4 \times 40) \times 15 \times 30}{1000}\)

= 72 BOT unit

The electrical energy consumed by the heat

⇒ \(=\frac{1000 \times 2 \times 30}{1000}=60 \mathrm{BOT} \text { unit }\)

∴ Monthly expenditure on electricity

= (66 + 72 + 60) X 1.50 = ₹ 297

Again, the total power of the appliances

= (6 x 40) + (2 x 100) + (4 x 40) + 1000

= 1600 W

If all the appliances run simultaneously, according to the relation, P = V x I,

we have, \(I=\frac{P}{V}=\frac{1600}{200}=8 \mathrm{~A}\)

So, a 10 A fuse will be appropriate for connection in the main switch
because a 5 A fuse will melt and there is no need for a 15 A fuse

Example 5. The power of a small electric motor is \(\frac{1}{8}\) HP. If it is connected to a 220 V supply line, how much current will it j draw? If the motor runs for 80 hours, what will be the cost? j (Each BOT unit costs 70 paise)
Solution:

Power, P = \(\frac{1}{8}\) HP = \(\frac{1}{8}\) x 746 W

According to the relation, P = V x I,

we have, \(I=\frac{P}{V}=\frac{1}{8} \times 746 \times \frac{1}{220}=0.424 \mathrm{~A}\)

Number of BOT units = \(\frac{1}{8} \times 746 \times 80 \times \frac{1}{1000}\)

∴ \(\text { Cost }=\frac{1}{8} \times 746 \times 80 \times \frac{1}{1000} \times 70 \text { paise }\)

= ₹ 5.22 (approx.)

Example 6. A heating coil of resistance 100 fl is connected for 30 min to 220 V. By this time, determine

1. Amount of charge flowing,
2. Amount of electrical energy consumed
3. Amountofheatgenerated. Determine the cost of consumed electrical energy if 1 kW.h costs ₹1.

Solution:

1. Amount of charge flowing,

⇒ \(Q=I t=\frac{V}{R} t=\frac{220}{100} \times 30 \times 60=3960 \mathrm{C}\)

2. Amount of electrical energy consumed,

W = Q.V

= 3960 x 220

= 8.712 x 105 J

3. Amount of heat generated,

⇒ \(H=\frac{W}{J}=\frac{8.712 \times 10^5}{4.2}=2.074 \times 10^5 \mathrm{cal}\)

= 2.074 x 105 cal

4. Cost of consumed electrical energy = \(\frac{8.712 \times 10^5}{3600 \times 1000} \times 1\)

= ₹ 0.24

= 24 paise

Electric Fuse:

Every metal, such as copper, aluminum, etc. used as the element of connecting wires in electrical circuits and electrical applications has a definite melting point.

Due to a short circuit or any other reason, if an excessively high current flows in the circuit, the temperature of the wire may rise to its melting point.

As a result, the whole circuit may be damaged and may cause fire. A fuse inserted in the circuit may prevent an accidental fire. It is prepared from an alloy (e.g., 3 parts lead and 1 part tin) having a comparatively low melting point.

This fuse is placed in an insulating box and is connected in series with the main circuit. Before the current of the main circuit reaches the danger point, the temperature of the fuse reaches its melting point.

As a result, the fuse melts down and the main circuit is disconnected. Hence, there is no chance of damage to the main circuit

Circuit breaker:

This is an automatic switch system that disconnects the circuit before being damaged due to short-circuit or overload. At the very moment the circuit is disconnected, the switch of the breaker moves to the ‘off’ position.

After replacing the defective instrument for which the circuit was broken, the switch is made ‘on current flows as usual through the circuit. In different types of circuit breakers, different physical phenomena are used.

For Example, in the circuit through which high current flows, a bimetallic strip bends due to heating and disconnects the circuit; or if a high current flows through an electromagnet, it attracts a soft-iron plate of the circuit and thus the circuit is cut off.

The main difference between a fuse wire and a circuit breaker is that, the fuse wire is burnt off and needs to be replaced with a new wire every time the fuse melts.

On the other hand, a circuit breaker is not at all damaged and hence there is no need to replace it. Once the fault has been repaired, the contacts must again be closed to restore power to the interrupted circuit.

The circuit breaker which is used in low voltage lines (0-1000 V) is generally called a Mechanical or Miniature Circuit Breaker (MCB).

Nowadays, in place of fuse wire, MCB is widely used. The switch gear is generally used as a circuit breaker in the supply line of high voltage (11000 V or more) from the generating station.

Highest safe current: If l is the length of the fuse wire, r is its radius, and \(\rho\) its specific resistance, then the resistance of the wire is given by,

⇒ \(R=\rho \frac{l}{\pi r^2}\)

So, heat generated in the wire per second is,

⇒ \(H=\frac{l^2 R}{J}=\frac{l^2 \rho l}{J \pi r^2}\)

This heat is used in two ways:

1. One part increases the temperature of the tire wire,

2. The other part radiates to the surroundings from the outer surface of the wire.

Thus, a situation arises when the total heat generated in the wire, radiates to the surroundings. So, the temperature does not increase further and reaches a fixed value. If this temperature lies just below the tire melting point of the material of the fuse wire, then the tire current flowing in the tire wire at this stage is called the tire’s highest safe current.

Now, the outer surface area of the tire wire = 2πrl.

If h is the tire amount of radiant heat from the unit area of the tire wire per second, then the total radiant heat from the outer surface of the wire per second is, H’ = 2πrlh.

So, for the highest safe current I,

⇒ \(H=H^{\prime}\)

or, \(\frac{I^2 \rho l}{J \pi r^2}=2 \pi r l h \quad\)

or, \(I^2=\frac{2 \pi^2 J h}{\rho} \cdot r^3\)

i.e., \(I^2 \propto r^3 \text { or, } I \propto r^{3 / 2}\)

It is to be noted that, the expression for highest safe current I is independent of the length I of the fuse wire.

Electric Energy And Power Rating Of Electrical Instruments

Voltage rating: Bach and every electrical instrument depends on potential differences. Each instrument is marked with a fixed value of potential difference, indicating the range of potential difference within which it can be run safely. This is called voltage rating.

For Example, the voltage rating of each domestic appliance is 220 V. This means that, if the potential difference between two ends of the appliance is 220 V, then its efficiency becomes maximum.

The efficiency of the appliance decreases if for any reason the supply voltage Is diminished, i.e., the electric lamp glows dimly, the electric fan moves comparatively slowly, etc.

On the other hand, if the voltage exceeds 220 V, there is a chance of damage to the appliance.

This is to note that, the appliances heaving the same voltage rating should be connected in parallel while used simultaneously in a circuit. In that case, the potential difference across each application remains the same.

For this reason, daily appliances are all connected in parallel combination with a 220 V supply line.

Watt rating: Different electrical appliances, even running at the same voltage may not consume the same electrical energy in unit time.

Hence, the power of the appliance, i.e., the electrical energy consumed per second should be mentioned.

This is called the watt rating of electrical appliances. Thus 220V-100W on the body of a lamp or 220V-2000W on the body of a heater refers to the voltage and watt ratings of the appliance.

Significance of ratings:

1. Maximum capability: Voltage rating indicates at what voltage, the appliance does maximum work, and watt rating indicates the rate of consumption of electrical energy at that voltage.

For Example, 220V-100W indicates diet, if the voltage at the two ends of the lamp is 220 V, it will glow with maximum brightness, consuming energy at the rate of 100 W

2. Calculation of current and resistance: It is possible to calculate the current and resistance of the appliance when it operates with maximum efficiency.

Significance of rating Example:

⇒ \(I(\text { current })=\frac{P(\text { power })}{V(\text { potential difference })}=\frac{100 \mathrm{~W}}{220 \mathrm{~V}}=0.45 \mathrm{~A}\)

⇒ \(R(\text { resistance })=\frac{V}{I}=\frac{220 \mathrm{~V}}{0.45 \mathrm{~A}}=489 \Omega\)

Safety of supply line: Whether an electrical appliance is safe or not for a supply line, is known from its rating. Consider the case of a 220 V-2000 W electric heater.

Current drawn by the heater = \(\frac{2000}{20}\) = 9.1 A (approx.)

So, this heater cannot be used in the house, where the meter and fuse are installed for 5 A current. If it is used, the supply line will be damaged.

Power in a combination of more than one electrical device:

Parallel combination: Suppose, and R₂ are the resistances of two electrical appliances. Their watt ratings are P₁ and P₂ respectively and the voltage rating of both of them is V.

If they are connected in parallel to an electric source of voltage V, then the voltage at the two ends of both the appliances will be V. The two appliances will function at full power and the total power consumed by the circuit will be,

P = P₁ + P₂

For any number of electrical appliances connected in parallel, P = P₁+P₂ + P₃+…. obviously, total power will be greater than that of each appliance.

Household electrical appliances are usually connected in
parallel.

Now, \(P_1=\frac{V^2}{R_1} \text { and } P_2=\frac{V^2}{R_2}\)

∴ \(\frac{P_1}{P_2}=\frac{R_2}{R_1}\)….(1)

For Example, if two electric lamps of 220 V-100 W and 220 V-60 W are connected in parallel and the combination is joined to a 220 V line, the 100 W lamp will glow brighter and the total power of the circuit will be, P – 100 + 60 = 160 W.

Series combination: Now the two appliances are nected in series and this combination is connected to the same electric source. The same current, say I, will flow through both of them. So, their powers will be respectively,

P’1 = I²R₁ and P’2 = I²R₂

So, \(\frac{P_1^{\prime}}{P_2^{\prime}}=\frac{R_1}{R_2}\) …..(2)

From equations (1) and (2) we have,

⇒ \(\frac{P_1^{\prime}}{P_2^{\prime}}=\frac{P_2}{P_1}\)

So, if \(P_1>P_2, P_1^{\prime}<P_2^{\prime}\)

i.e., in series combination, the appliance having a higher watt rating will consume less power.

For Example, if two electric lamps having the power of 100 W and 60 W are connected in series, the power of the 100 W lamp will be comparatively lower and it will glow with less brightness.

Again, according to,

⇒ \(I=\frac{V}{R_1+R_2}\)

So, the total power of the circuit,

⇒ \(P^{\prime}=I^2\left(R_1+R_2\right)=\frac{V^2}{R_1+R_2}\)

∴ \(\frac{1}{P^{\prime}}=\frac{R_1+R_2}{V^2}=\frac{R_1}{V^2}+\frac{R_2}{V^2}=\frac{1}{P_1}+\frac{1}{P_2}\)

For any number of electrical appliances connected in series,

⇒ \(\frac{1}{P^{\prime}}=\frac{1}{P_1}+\frac{1}{P_2}+\frac{1}{P_3}+\cdots\)

Obviously, the total power P’ will be less than that of each application.

For Example, if two electric lamps of 220 V-100 W and 220 V-60 W are connected in series and joined to a 220 V supply line, the 60 W lamp will glow brighter and the total power of the circuit will be less than 60 W. Here

⇒ \(\frac{1}{P^{\prime}}=\frac{1}{100}+\frac{1}{60} \quad\)

or, \(P^{\prime}=\frac{100 \times 60}{100+60}\)

= 37.5 W^

Rating of a resistor: No resistor can endure excess current over a certain designated value depending upon its nature. If the limit is exceeded, the excessive heat generated in the resistor will bum it. We may say that, as soon as the power consumption due to the production of heat crosses a certain limit, the resistor gets damaged. This maximum power tolerance is known as the watt rating of the resistor.

For Example, a resistor with a rating lW-100Ω means a 100Ω resistance and of maximum power tolerance 1W. We know, power,

P = VI = IR.I = I²R

So maximum current tolerance of the resistor Is,

⇒ \(I=\sqrt{\frac{P}{R}}=\sqrt{\frac{1}{100}}=\frac{1}{10} \mathrm{~A}\)

Under this condition, the potential difference between the two ends of the resistor is,

⇒ \(V=I R=\frac{1}{10} \times 100\)

= 10V

Therefore, before joining a circuit, the rating of a resistor should be known. In that case, it can be connected in such a way that the current should always remain below

⇒ \(\frac{1}{10}\) A through it

Electric Energy And Power Numerical Examples

Example 1. A 220V-60W electric bulb is connected To a 220 V line. What is the resistance of the filament of the bulb, when it is turned on?
Solution:

Power, \(P=V I=V \cdot \frac{V}{R}=\frac{V^2}{R} \quad[∵ V=I R]\)

∴ \(R=\frac{V^2}{P}=\frac{220 \times 220}{60}\)

= 806.67Ω

Example 2. The resistance of a hot tungsten filament is about 10 times that in its normal state. What will be the resistance of a 100 W-200 V tungsten lamp in its normal state?
Solution:

Resistance of the lamp when it is hot,

⇒ \(R=\frac{V^2}{P}=\frac{(200)^2}{100}=400 \Omega\)

So, resistance of the lamp in its normal state = \(\frac{400}{10}=40 \Omega\)

Example 3. The main meter of a house is marked 10 A-220 V. How many 60 W electrical amps can be used safely in this line?
Solution:

Maximum power of the household line,

P = VI

= 220 X 10

= 2200 W

Maximum number of 60 W lamp = \(\frac{2200}{6}\) = 36.67

So, a maximum of 36 lamps can be used at a time.

Example 4. A 220V-100W electric lamp fuses above 150 power. What should be the maximum tolerable voltage for the lamp?
Solution:

Power, \(P=\frac{V^2}{R}\)

If we ignore the change or resistance to the Increase of power,

We have, \(P \propto V^2\)

So, for two different powers, \(\frac{P_1}{P_2}=\frac{v_1^2}{v_2^2}\)

i.e., \(\cdot V_2=V_1 \sqrt{\frac{P_2}{P_1}}=220 \times \sqrt{\frac{150}{100}}\)

= 269.4 V

So, the maximum tolerable voltage for the lamp Is 269.4 V.

Example 5. In order to run a 60V-120W lamp safely in a 220V dc line, a resistor of what minimum magnitude should be placed in series with it?
Solution:

When the lamp runs safely In the given dc line, the maximum current is,

⇒ \(I=\frac{120}{60}=2 \mathrm{~A}\)

So, this 2 A current will also flow through the resistance placed in series.

Again, the potential difference at the two ends of the resistance

= 220-60

= 160V.

∴ The minimum value of the resistance = \(\frac{160}{2}\)

= 80Ω

Example 6. Draw a household circuit having a 1200 W toaster, a 1000 W oven, an 800 W heater, and a 1500 W cooler. The circular has a heavy-duty wire and a 20 A circuit breaker. Will the circuit breaker trip, if all the appliances are operated simultaneously in a 220 V supply voltage?
Solution:

The household circuit. When all the appliances are operated simultaneously, the total po is, P = 1200 + 1000 + 800 + 1500

= 4500 W.

Now, P = VI

or, \(I=\frac{P}{V}=\frac{4500}{220}\)

= 20.45A

But the circuit breaker is rated for 20 A.

So, if all the appliances are operated simultaneously, the circuit breaker will trip.

Example 7. Two lamps of 200 W and 100W arc connected in series in 200 V mains. Assuming the resistance of the two lamps remains unchanged, calculate the power consumed by each of them.
Solution:

⇒ \(P=\frac{V^2}{R} \quad \text { or, } R=\frac{V^2}{P}\)

So, the resistance of the first lamp, \(R_1=\frac{200 \times 200}{200}=200 \Omega\)

and resistance of the second lamp, \(R_2=\frac{200 \times 200}{100}=400 \Omega\)

Now, if we connect the two lamps in series with 200 V mains, current flowing through each of them,

⇒ \(I=\frac{V}{R_1+R_2}=\frac{200}{200+400}=\frac{1}{3} \mathrm{~A}\)

∴ Power of the first lamp, \(P_1=I^2 R_1=\frac{1}{3} \times \frac{1}{3} \times 200=22.2 \mathrm{~W}\)

Power of the second lamp, \(P_2=I^2 R_2=\frac{1}{3} \times \frac{1}{3} \times 400=44.4 \mathrm{~W}\)

Example 8. Two electric bulbs each designed to operate with a power of 500 W In a 220 V line are connected in series in a 110 V line. What will be the power generated by each bulb?
Solution:

Power, \(P=\frac{V^2}{R}\)

So, the resistance of each bulb,

⇒ \(R=\frac{V^2}{P}=\frac{(220)^2}{500}=\frac{484}{5} \Omega\)

When connected in series, the potential difference at the two ends of each bulb = \(\frac{110}{2}\)

=55V.

∴ \(\frac{(55)^2}{\frac{484}{5}}=\frac{55 \times 55 \times 5}{484}\)

= 31.25W

Example 9. If the supply voltage drops from 220 V to 200 V , what would be the percentage reduction in heat produced by a 220 V-1000 W heater? Neglect the change of resistance. If the change of resistance is taken into consideration, would the reduction of heat produced be smaller or larger than the previously calculated value?
Solution:

We know, \(P=\frac{V^2}{R}\)

So, \(R=\frac{V^2}{P}\)

If the variation of resistance is ignored,

In the first case, \(R=\frac{V_1^2}{P_1}\)

In the second case, \(R=\frac{V_2^2}{P_2}\)

i.e., \(\frac{V_1^2}{P_1}=\frac{V_2^2}{P_2} \text { or, } \frac{P_2}{P_1}=\frac{V_2^2}{V_1^2}\)

or, \(\frac{P_2-P_1}{P_1}=\frac{V_2^2-V_1^2}{V_1^2}=\frac{\left(V_2+V_1\right)\left(V_2-V_1\right)}{V_1^2}\)

= \(\frac{(200+220)(200-220)}{220 \times 220}\)

or, \(\frac{P_2-P_1}{P_1}=-\frac{420 \times 20}{220 \times 220}\)

= -0.1736 (approx.)

So, the power of the heater will reduce by 17.36%; i.e., the percentage reduction of heat produced is 17.36%. In the above calculation, the decrease of resistance with voltage drop was ignored. But actually, with a decrease in the temperature of the coil, its resistance also decreases. Hence according to the relation P = \(P=\frac{V^2}{R}\), power Increases, i.e., heat supplied by the heater also increases. So, the reduction of heat produced will be lower than 17.36%.

Example 10. The emf of the cell, E = 20V. The rating of each resistance R₁ and R₂, is 1W-100Ω. What should be the minimum value of the resistance R in the circuit? Also, find its minimum watt rating

Solution:

According to the formula,

P = I²R

or, \(I=\sqrt{P / R}\)

The maximum current that can flow through each of the resistance R1, and R2 is,

⇒ \(I_1=I_2=\sqrt{\frac{1}{100}}=\frac{1}{10} \mathrm{~A}\)

∴ The maximum current that can flow through the resistance R,

⇒ \(I=I_1+I_2=2 \times \frac{1}{10}=\frac{1}{5} \mathrm{~A}\)

On die other hand, equivalent resistance of R1 and R2.

⇒ \(\frac{R_1 R_2}{R_1+R_2}=\frac{100 \times 100}{100+100}=50 \Omega\)

Now, \(I=\frac{E}{R+50} \quad\)

or, \(\frac{1}{5}=\frac{20}{R+50}\)

or, R + 50 = 100

∴ R = 100-50 = 50Ω

i.e., to keep the value of I lower than \(\frac{1}{5}\) A, R sliould be greater than 50Ω. Power consumption for the minimum value of R,

⇒ \(P=I^2 R=\left(\frac{1}{5}\right)^2 \times 50=2 \mathrm{~W}\)

Example 11. If a 220 V-1000 W lamp is connected to 110 V line then what will be the power consumed by it?
Solution:

Resistance of the lamp, \(R=\frac{V_1^2}{P}=\frac{(220)^2}{1000} \Omega\)

∴ Power consumed by the lamp when it is connected to 110V
line,

⇒ \(P_2=\frac{V_2^2}{R}=\frac{(110)^2 \times 1000}{(220)^2}=250 \mathrm{~W}\)

Example 12. 2.2 kW power is supplied through a line of 10Ω resistance under a 22000 V voltage difference. What is the rate of heat dissipation in the line?
Solution:

⇒ \(I=\frac{P}{V}=\frac{2.2 \times 10^3}{22000} \quad\left[∵ 2.2 \mathrm{~kW}=2.2 \times 10^3 \mathrm{~W}\right]\)

= 0.1 A

∴ Rate of heat dissipation = I²R

= (0.1 )2 x 10

= 0.1 W

Example 13. The potential difference between the two ends of an electric lamp is decreased by 1%. Neglecting the change in its resistance, calculate the percentage increase or decrease in the power of the lamp.
Solution:

We know power, \(P=\frac{V^2}{R}\)

∴ InR = 21nV-InR

By differentiating \(\frac{d P}{P}=2 \cdot \frac{d V}{V}[∵ R=\text { constant, } d R=0]\)

According to die problem, \(\frac{d V}{V}=-1 \%=-\frac{1}{100}\)

∴ \(\frac{d P}{P}=-\frac{2}{100}=-2 \%\)

i.e., the power of the lamp will be decreased by 2%.

Example 14. 15 kW power is supplied through a line of 0.5Ω resistance under 250 V potential difference. Find the efficiency of the supply in percentage.
Solution:

Power, P = 15 kW

= 15000 W

So, currently in the supply line.

⇒ \(I=\frac{P}{V}=\frac{15000}{250}=60 \mathrm{~A}\)

Power loss due to inactive resistance in the line

= I²R = (60)² x 0.5

= 1800W;

Therefore, total power in line = 15000 + 1800

= 16800 W

∴ \(\text { Efficiency }=\frac{\text { effective power }}{\text { total power }} \times 100 \%\)

⇒ \(\frac{15000}{16800} \times 100 \%=89 \%\)

Example 15. Two incandescent lamps (25 W, 120 V) and (100 W, 120 V) are connected in series across a 240 V supply. Assuming that the resistances of the lamps do not vary with current, find the power dissipated in each lamp after the connection.
Solution:

The resistance of the lamp is 25 W-120 V,

⇒ \(R_1=\frac{120^2}{25}=576 \Omega\)

The resistance of the lamp \(100 \mathrm{~W}-120 \mathrm{~V}, R_2=\frac{120^2}{100}=144 \Omega\)

When the lamp is connected in series, the equivalent resistance of the circuit,

R’ = 576 + 144 = 720Ω

∴ Current flowing through the circuit = \(=\frac{240}{720}=0.33 \mathrm{~A}\)

∴ Power dissipated in die first bulb =(0.33)²x576 = 62.72 W

∴ Power dissipated in die second bulb = (0.33)² x 144

= 15.68 W

Electric Energy And Power Synopsis

• The amount of work done to carry a 1-coulomb charge through a potential difference of 1 volt is called 1 joule.
• The reciprocal of the heat produced in one second in a conductor of unit resistance, for the passage of unit current through it, is called the mechanical equivalent of heat.
• If in a resistor, electrical energy is totally transformed into heat energy, the resistor is then called a passive resistor.
• Electrical power is to be transmitted from one place to another at high voltage and low current. This is the most important condition to minimize heat loss.
• If the potential difference at the two terminals of an electrical appliance is 1 volt and the current passing through it is 1 ampere, then the power of the appliance is l watt.
• Work done or electrical energy spent in 1 hour by an electrical appliance having power 1W is called 1Wh. Work done or electrical energy spent in lh by an electrical appliance having power lkW is called lkW-h or 1BOT unit.
• In writing 220V-100W on the body of an electric lamp or 220V-2000W on the body of an electric heater, both voltage rating and watt rating are preferred.
• If Q charge flows through a section under the potential difference V, then the amount of electrical work done,
• W = QV.
• The relation between joule (I), coulomb (C), and volt (V) is J = C x V [∵ W= QV]
• The heat evolved in a conductor of resistance R due to current I flowing through it for time t,
• \(H=\frac{I^2 R t}{J}\)
• where, J = mechanical equivalent of heat
• = 4.2 x 10⁷ erg/cal (In CGS)
• = 4.2 J/cal
• = 1 (In SI)
• i.e., \(H=0.24 I^2 R t=0.24 \frac{V^2 t}{R}\) [∵ V = IR]

• Current in a total circuit,
• I = \(\frac{E}{R+r}\) , where, E = emf of the cell, R = external resisR + r tance, r = internal resistance
• Power consumption in the circuit, \(P_0=I^2(R+r)=\frac{E^2}{R+r}\)
• Output power, \(P=I^2 R=\frac{E^2 R}{(R+r)^2}\)
• lW.h = 3600J, lkW.h = 3.6 x 106 J
• \(1 W \cdot h=\frac{W \times h}{1000}=\frac{V \times A \times h}{1000}\)
• For a parallel combination of a number of electrical appliances, equivalent power, P = P₁ + P₂ + P₃+….
• For a series combination of a number of electrical appliances, equivalent power,
• \(P^{\prime}=\frac{1}{\frac{1}{P_1}+\frac{1}{P_2}+\frac{1}{P_3}+\cdots}\)
• If a heating coil of resistance R, consumes power P, when voltage V is applied to it, then by keeping V constant if it is cut in n equal parts then the resistance of each part will be R/n and the power consumed by each part P’ = nP.
• Power consumed by a n equal resistors is parallel is n₂ times that of power consumed in series if V remains the same i.e., P_p = n₂P_s.

Electric Energy And Power Very Short Questions and Answers

Question 1. A current I flows through a potential drop V across a conductor. What is the rate of heat production?
Answer: VI

Question 2. A small heating element connected to a 10 V dc supply draws a current of 5 A. Find the electric power supplied to the heater.
Answer: 50W

Question 3. A 220 V – 1000 W electric heater is connected in parallel with a 220 V – 60 W electric lamp, and the combination is fed by a 220 V main. Now, if the lamp is replaced by another 220 V – 100 W lamp, what will be the change in the rate of heat generation in the heater?
Answer: No change

Question 4. Two wires having resistances R and 2R are connected in series. If current is allowed to pass through the combination, what will be the ratio of power consumed in the two resistances?
Answer: 1: 2

Question 5. Two wires having resistances R and 2R are connected in parallel. If current is allowed to pass through the combination, what will be the ratio of power consumed in the two resistances?
Answer: 2: 1

Question 6. A 240 V- 1000 W lamp and a 240 V- 100 W lamp—which of these two has a thinner filament?
Answer: 240 V – 100 VV lamp

Question 7. Two resistances, each of magnitude 2Ω, are connected in series and a potential difference of 2V is applied at the two terminals of the combination. What is the power of the combination?
Answer: 1W

Question 8. Two resistances each of magnitude 2Ω are connected in parallel and a potential difference of 2V is applied at the two ends of the combination. What is the power of the combination?
Answer: 4W

Question 9. What is the resistance of an electric bulb marked 220 V-100 W in an incandescent state?
Answer: 484Ω

Question 10. An electric lamp is marked 240V -60 W. What is the resistance of the lamp in an incandescent state?
Answer: 960Ω

Question 11. A 220V -60 W electric lamp is connected to a 220V supply line. Determine the resistance of the lamp in an incandescent state.
Answer: 8.67Ω

Question 12. Which one of two electrical appliances, rated 100 W-200 V and 60 W-200 V, would have a higher resistance?
Answer: 60 W-220 V

Question 13. What is the largest voltage you can safely put across a 98Ω, 0.5 W resistor?
Answer: 7V

Electric Energy And Power Fill In The Blanks

1. A 2Ω resistance is connected to a source of constant emf. Another 2Ω resistance is connected in parallel to the previous one. The power consumed in the circuit becomes double

2. Two resistances are connected in series. If the current is made to pass through the combination, power consumed in the larger resistance will be higher

3. Two resistances are connected in parallel. If the current is made to pass through the combination power consumed in the larger resistance will be lower

4. A 220 V-100 W lamp and a 220V-60W lamp are connected in parallel. If the current is made to pass through the combination, the brightness of the first lamp will be higher

5. A 220 V-100 W lamp and a 220V-60W lamp are connected in series. If the current is made to pass through the combination, the brightness of the first lamp will be lower

Electric Energy And Power Assertion Reason Type

Direction: These questions have Statement 1 and Statement 2. Of the four choices given below, choose the one that best describes the two statements.

1. Statement 1 is true, Statement 2 is true; Statement 2 is a correct explanation for Statement 1
2. Statement 1 is true, Statement 2 is true; Statement 2 is not a correct explanation for Statement 1
3. Statement 1 is true, Statement 2 is false
4. Statement 1 is false, Statement 2 is true

Question 1.

Statement 1: Electric current is distributed in different branches of a circuit in such a way, that the total heat evolved in the circuit is the lowest.

Statement 2: The transformation of electrical energy into heat energy in a circuit is less probable than its transformation into other forms of energy.

Answer: 3. Statement 1 is true, Statement 2 is false

Question 2.

Statement 1: An external circuit can draw a maximum power of 9 W from a source of emf 6 V and internal resistance 1 n.

Statement 2: The condition, for which an external circuit of resistance R draws the maximum power from a source of internal resistance r, is R = r.

Answer: 1. Statement 1 is true, Statement 2 is true; Statement 2 is a correct explanation for Statement 1

Question 3.

Statement 1: The power consumed would be 50 W by each of two 200V-100W lamps when their series combination is driven by a potential difference of 200V.

Statement 2: If P is the power consumed by a series combination of some electrical devices of power P₁, P₂, P₃,…., then, \(\frac{1}{P}=\frac{1}{P_1}+\frac{1}{P_2}+\frac{1}{P_3}+\cdots\)

Answer: 4. Statement 1 is false, Statement 2 is true

Question 4.

Statement 1: A fuse wire of diameter 0.5 mm can withstand a maximum current of I A. For a current of 8 A, a fuse wire made of the same alloy should have a diameter of 2 mm.

Statement 2: The radius r of a fuse wire and the maximum safe current I that may pass through it are related as \(I \propto r^{3 / 2}\)

Answer: 1. Statement 1 is true, Statement 2 is true; Statement 2 is a correct explanation for Statement 1

Question 5.

Statement 1: The coil resistance of a 200 V-100 W electric fan is 20Ω. A power of 5 W is lost as heat when the fan rotates at its maximum speed.

Statement 2: If, for an electrical device the current is I and the terminal potential difference is V, the power consumed = VI

Answer: 2. Statement 1 is true, Statement 2 is true; Statement 2 is not a correct explanation for Statement 1

Electric Energy And Power Match The Columns

Question 1. Each of the two cells in the circuit has emf 5 V and internal resistance 2Ω. Match the following two columns for this circuit.

Answer: 1-B, 2-D, 3-A, 4-C

Question 2. The cell in the circuit has negligible internal resistance. The resistance values are in Ω. Match the following two columns for this circuit.

Answer: 1-C, 2-A, 3-D, 4-B

Question 3. The ratings of a few electrical elements are given in Column A, and Column B shows the values of maximum safe currents. Match the two columns.

Answer: 1-B, 2-A, 3-D, 4-C

Kirchhoff’s Laws And Electrical Measurement Kirchhoff’s Laws

Kirchhoff formulated the following two laws which enable us to find the distribution of current in complicated electrical circuits (or networks of conductors).

Kirchhoff’s First Law or Kirchhoff’s Current Law (KCL):

Statement: In an electrical circuit 9or network of wires) the algebraic sum of currents through the conductors as meeting at a point is zero, i.e., \(\Sigma i=0\)

Kirchhoff’s Current Law Explanation of the law: Let us consider the number of wires connected at point A. Currents i₁, i₂, i₃, and i4 flow through these wires in the directions. Here currents i₁ and i₂ are approaching point A while currents i₃ and i₄ are leaving point A. Taking current entering point A as positive current while current leaving point A as negative current and applying the first law we can write,

i₁ + i₂ – i₃ – i₄ = 0…(1)

Kirchhoff’s Current Law Discussion:

Node analysis: Any connecting point in an electrical circuit is called a node. By applying the first law of Kirchhoff, the analysis of an electrical circuit is called node analysis.

Conservation of electric charge: If we write equation (1) in the form i.e., i₁ + i₂ = i₃ + i₄, we can say that the sum of the currents approaching a connecting point = the sum of the currents leaving the point. If the current flows for time t we have

Read and Learn More Class 12 Physics Notes

⇒ \(i_1 t+i_2 t=t_3 t+i_4 t \quad \text { or, } q_1=q_2=q_3+q_4\)

i.e., the sum of the charges approaching the connecting point = the sum of the charges leaving the point.

The significance of the above equation is discussed below:

There cannot be any accumulation of charge at any connecting point in an electrical circuit

The charge cannot be created or destroyed, i.e., the total charge will remain constant

Kirchhoff’s Second Law or Kirchhoff’s Voltage Law (KVL):

Statement: The algebraic sum of the product of the current and resistance in any closed loop of a circuit is equal to the algebraic sum of electromotive force acting in that loop, i.e.,

⇒ \(\Sigma i r=\Sigma e\)

Explanation of the law: A closed-loop ACBDA within an electrical circuitIn the loop, current i₁ is in a clockwise direction while current- i₂ is in an anticlockwise direction. Taking clockwise current as positive and anticlockwise current as negative, i₁ becomes positive and i₂ becomes negative.

Again, the electromotive force of the sources that send currents in clockwise and anticlockwise directions in the closed loop are taken as positive and negative respectively.

So emf e₁ becomes negative and emf e₂ becomes positive. Hence, for the closed-loop ACBDA,

i₁r₁-i₂r₂ = -e₁ + e₂ ….(2)

We can write by applying Kirchhoff’s first law to the point A

i-i₁-i₂ = 0 on i₂ = i – i₁

So, equation (2) can be written as

i₁r₁ -(i-i₁)r₂

= –₁+ e₂

on i₁(r₁ + r₂) = e₂– e₁ + ir₂

or, \(i_1=\frac{\left(e_2-e_1\right)+i r_2}{r_1+r_2}\)….(3)

Knowing the values of die quantities of equation (3) we can determine i₁ and i₂.

Kirchhoff’s Voltage Law Discussion:

Mesh analysis: A complicated circuit formed by a number of adjacent loops resembles a mesh which can be analyzed one by one by applying Kirchhoff’s second law.

S3 Conservation of energy: We know that inside a source of electricity, other forms of energy are converted into electrical energy, and in an external circuit, electrical energy is converted into other forms of energy.

The amount of electrical energy developed inside a source of electricity to send a one-coulomb charge in a circuit is the electromotive force of the source. Again the amount of electrical energy spent in an external circuit due to the flow of one coulomb charge is the potential difference of the external circuit According to the second law of Kirchhoff,

⇒ \(\Sigma i r=\Sigma e \text { i.e., } \Sigma e=\Sigma v\) [potential difference, v = ir]

i.e., electrical energy developed in any circuit due to unit charge = electrical energy spent in the circuit

So, Kirchhoff’s second law obeys the law of conservation of energy

Kirchhoff’s Laws And Electrical Measurement Kirchhoff’s Laws Numerical Examples

Example 1. Two cells, one of emf 1.2 V and internal resistance 0.5Ω, the other of emf 2 V and internal resistance 0.1Ω are connected in parallel and the combination is connected in series with an external resistance of 5Ω. What is the current through this resistor?
Solution:

Applying Kirchhoff’s first law to point A we have,

I₁ + I₂ – I = 0

or, I₂ = I-I₁ …..(1)

Applying Kirchhoff’s second law to the loop ACBDA we have,

-I₁ x 0.5 + 12 x 0.1 = -1.2 + 2

-I₁ x 0.3 -5-(I-I₁) x 0.1 = 0.8

or, 0.1 I – 0.6I₁ = 0.8 …….(2)

For the loop ADBFA,

-I₂ x 0.1 – I x 5 = -2

or, (I-I₁) x 0.1 + 5I = 2….(3)

Now, multiplying equation (3) by 6 and subtracting equation (2) from it we have,

⇒ \(30.5 I=11.2 \quad \text { or, } I=\frac{11.2}{30.5}=0.367 \mathrm{~A}\)

Example 2. Determine the current flowing through the resistor of resistance 200Ω and the potential difference across its ends.

Solution:

Applying Kirchhoff’s first law to point B we have,

I-I₁-I₂ = 0…..(1)

Applying Kirchhoff’s second law to the loop ABCA we have,

100I₁ + 200I = 10

or, 10I₁ + 20I = 1….(2)

For the loop ADBA,

-200I-150I₂ = -30

200I+ 150(I-I₁) = 30

350I- 150I₁ = 30

or – 15I₁ + 35I = 3….(3)

From the equations (2) and (3) we get,

⇒ \(130 I=9 \text { or, } I=\frac{9}{130} \mathrm{~A}=69.2 \mathrm{~mA}\)

⇒ \(V_{A B}=\frac{9}{130} \times 200=\frac{180}{13} V \cdot 13.84\) V

Example 3. Twelve equal wires, each of resistance r ohm, are connected so as to form the frame of a cube. An electric current enters this cube at one corner and leaves from the diagonally opposite corner. Calculate the total resistance between the two corners.
Solution:

Let ADGCHEFB be the frame of the cube formed by twelve equal wires each of resistance r. Let the total current entering at the comer A and leaving the diagonally opposite comer B be 6x.

Therefore, the 12 wires of the cubic framework (mesh) will have equivalent resistance,

⇒ \(R=\frac{V}{I}=\frac{V}{6 x}\) …..(1)

As the resistance of each wire is the same, the current 6x is divided at A into three equal parts, one along AD, the other along AE and the third along AC.

At points D, E, and C the current is again divided into two equal parts. At points F, G, and H the currents combine so that the current in each of the arms FB, GB, and HB is 2x.

The currents at point B again combine. If V is the potential difference across A and B, then taking the path ACHBA and applying Kirchhoff’s second law we get,

2x.r + x.r + 2x.r = V

or, 5xr = V

From equations (1) and (2) we have,

6xR = 5xr

or, R = \(\frac{5}{6}\)r

Node Voltage Loop Current:

Node voltage: During the analysis of any circular we can consider a potential for every connecting point i.e., node. This is called node voltage. For analysis circuit with the help of node voltage the following rules are adopted:

1. The potential of any connecting point of the circuit may be taken as zero, because for the calculation of current the potential difference between two points Is required whatever may be the Individual potential of the two points. In the potential of C Is taken ns zero.

2. If there is no source, of electricity or resistance (or condenser or inductance) between any two points, the potential of the two points is equal.

3. The potential of the negative terminal of a source of electricity is less than that of the positive terminal by the amount equal to the emf of the source. For Example, in the potential of A = V; the potential of B = (V- 5) V.

4. Current through any resistance is determined by applying the formula \(\frac{V}{R}\) = I. For Example, in the current through BC

⇒ \(I=\frac{V_B-V_C}{10}=\frac{(V-5)-0}{10}=\frac{V-5}{10}\)

5. After calculating the currents, Kirchhoff’s first law can be applied at various connecting points.

Loop Current: During the analysis of any circuit a separate current for every closed loop may be considered. This is called loop current.

For analysis of a circuit with the help of loop current the following rules are adopted:

1. Loop current is to be shown with a definite direction. For Example, in current i1 for the loop, ABCD is shown in the clockwise direction

2. For the calculation of the current flowing in every branch of a loop the current in the adjacent loop is to be taken into account. For Example, in the current through AB, CD, and DA = i₁, but current through BC = i₁– i₂.

3. Next Kirchhoff’s second law can be applied to the loop.

Kirchhoff’s Laws And Electrical Measurement Numerical Examples

Example 1. Determine the current flowing through the resistance of 5Ω as shown In the circuit. 

Solution:

Potentials of different connecting points are taken in the following way:

V_A = V; V_B = V-1.2;

V_C = V – 2 and V_D = 0

Current along \(B D=\frac{(V-1.2)-0}{0.5}=\frac{V-1.2}{0.5} \mathrm{~A}\)

Current along \(C D=\frac{(V-2)-0}{0.1}=\frac{V-2}{0.1} \mathrm{~A}\)

Current along \(A D=\frac{V-0}{5}=\frac{V}{5} \mathrm{~A} .\)

Applying Kirchhoff’s first law to the connecting point D we have,

⇒ \(\frac{V-1.2}{0.5}+\frac{V-2}{0.1}+\frac{V}{5}=0\)

or, \(\frac{10(V-1.2)+50(V-2)+V}{5}=0\)

or, 10V- 12 + 50V- 100 + V = 0

or, \(61 V=112 \text { or, } V=\frac{112}{61} \mathrm{~V}\)

∴ Current through resistance of \(5 \Omega=\frac{V}{5}=\frac{112}{61 \times 5}=0.367 \mathrm{~A}\)

Example 2. The resistances of the ammeter and the voltmeter are 10Ω and 900Ω respectively. What are the readings of the ammeter and the voltmeter?

Solution:

In the circuit, currents flowing in the three loops have been shown as i₁, i₂, and i₃ (clockwise).

From the second law of Kirchhoff’s, we get, for the first loop,

100i₁ + 100(i₁– i₂) = 0

or, 2i₁– i₂ = 0 …..(1)

for the second loop,

100(i₂-i₁) + 10i₂ + 100(i₂– i₃) = 12

or, – 100i₁ + 210i₂ – 100i₃ = 12…(2)

fol the third loop,

100(i₃ -i₂) + 900 i₃ = 0.

or, 10i₃-i₂ = 0 ….(3)

From (1) and (3),

i₂ = 2i₁ = 10i₃

i₁ = 5i₃

Substituting these values in equation (2) we get,

-100 x 5i₃ + 210 x 10i₃ – 100i₃ = 12

or, 1500i₃ = 12

or, \(i_3=\frac{12}{1500}=\frac{4}{500} \mathrm{~A}\)

∴ Reading of voltmeter

= \(i_3 \times 900=\frac{4}{500} \times 900\)

= 7.2 V

and reading of ammeter

= \(i_2=10 i_3=10 \times \frac{4}{500}\)

= 0.08 A

Kirchhoff’s Laws And Electrical Measurement Potential Divider Or Potentiometer

Potential Divider Or Potentiometer Description: A wire of uniform resistance is connected to two binding screws A and B. A slider or jockey J can move along the wire, remaining in contact with it. J is connected with the binding screw C. This arrangement is generally called a potential divider or potentiometer

Suppose, the length of the resistor AB = L; resistance = R, the

length of the portion AJ = l; its resistance = r.

Since the resistance of the resistor is uniform,

⇒ \(\frac{L}{l}=\frac{R}{r} \quad \text { or, } \quad r=\frac{l}{L} \cdot R\)….(1)

Uses:

Determination of the emf of a cell: Suppose, a source of emf E₀ is connected between the two ends A and B of the potentiometer wire of resistance R. The cell whose emf E is to be measured is connected between the points A and C through a galvanometer G. It is to be noted that the positive pole of the cell is connected at A.

Now, suppose that the current in the circuit ACJA due to the effective emf VAC is ix and the current in that circuit due to the cell = i2. These two currents flow in opposite directions.

The slider J is shifted from point to point and placed at a position for which the current through the galvanometer is zero.

Clearly at that condition i₁ = i₂. If R₀ is the total resistance of the circuit ACJA, then,

⇒ \(i_1=\frac{V_{A C}}{R_0} \quad \text { and } i_2=\frac{E}{R_0}\)

So for zero galvanometer current, \(\frac{V_{A C}}{R_0}=\frac{E}{R_0}\)

It is evident that the current l passes through AJ. So,

⇒ \(V_{A J}=\frac{E_0}{R} \cdot \frac{l}{L} \cdot R_i=\frac{l}{L} \cdot E_0 \quad \text { i.e., } V_{A C}=\frac{l}{L} E_0\)….(2) [length of 1 AB = L, length of AJ =l]

or, \(E=V_{A C}=\frac{l}{L} E_0\)…(3)

If we already know the values of E₀ and L then by noting the value of l from the position of the slider, the emf E of the cell can be determined from the equation (3).

Comparison of the emfs of two cells: From equation (3) we can write, \(\frac{E}{E_0}=\frac{l}{L}\). So the ratio of the emfs of two cells can be determined with a potentiometer even though the emf of each cell may not be known.

Determination of the internal resistance of a cell: For the length of the potentiometer wire AJ₁ = l₁ (when the slider is placed at the point J₁, if the galvanometer current is zero, according to the equation (3),

⇒ \(E=\frac{l_1}{L} E_0\)

Now, a known resistance R is to be connected parallel to the cell. Under this condition, the potentiometer circuit measures the potential difference V across R. If the galvanometer current is zero for the length of the potentiometer wire AJ₂ = l₂ (when the slider is placed at J₂

⇒ \(V=\frac{l_2}{L} E_0\)…(4)

Therefore, \(\frac{V}{E}=\frac{l_2}{l_1}\left[\text { since, } E=\frac{l_1}{L} E_0\right]\)

Again if r is the internal resistance of the cell, E is the emf and I is the current in the circuit due to the cell, then

E = I(R + r) and V = IR

⇒ \(\text { i.e., } \frac{V}{E}=\frac{R}{R+r}\)

So, \(\frac{l_2}{l_1}=\frac{R}{R+r} \quad \text { or, } R+r=\frac{l_1}{l_2} \cdot R \quad \text { or, } r=\left(\frac{l_1}{l_2}-1\right) R\)

Therefore, to determine the internal resistance r of the cell, we have no need to know the values of L, E_0, or E.

Measurement of the potential difference between two points of an electric circuit: The potential difference between any two points of a circuit car i be measured by using a potentiometer in the same way as the emf of a cell is measured.

One of the two points (M and N) of an electrical circuit across which the potential difference is to be measured, is connected to the end A of the potentiometer wire, and the other point is connected to the jockey J through a galvanometer G.

The two poles of the battery of emf E₀ are connected to the ends A and B of the potentiometer wire in such a tray that the current i₁ in the part AM due to the batten’ of emf E₀ and the cu rent i₂ due to the external circuit are opposite to each other.

Now changing the position of the jockey J from point to point we observe the position of the jockey. for which the galvanometer current is zero. In this position i₁ = i₂. Let the total resistance of the circuit MNJA be R₀

Then \(\left.i_1=\frac{V_{A J}}{R_0} \text { [length of } A B=L \text {, length of } A J=l\right]\)

and \(i_2=\frac{V_{M N}}{R_0} \quad \text { So, } V_{M N}=V_{A J}=\frac{l}{L} E_0\)….(5)

If the values of E0 and L are known, then by knowing the value of I from the position of J, the value of VMN i.e., the potential difference between M and N can be determined from the equation (5).

Condition of effectiveness: In this arrangement, the potentiometer behaves as a voltmeter. So like any voltmeter, the resistance of the potentiometer should be very high.

So the condition of measurement of potential difference by a potentiometer is that the resistance of the potentiometer wire between the two points A and J must be many times greater than the resistance between the two points M and N of the experimental circuit If it is not so, then as soon as the potentiometer is connected, the equivalent resistance between the points M and N of the experimental circuit will be reduced to a large extent and much error will be found in the measured value of VMN.

Kirchhoff’s Laws And Electrical Measurement Potential Divider Or Potentiometer Numerical Examples

Example 1. The length of the wire of a potentiometer is 100 cm and the emf of a standard cell connected to it is E volt. While measuring the emf of a battery having an Internal resistance of 0.5Ω. the null point is obtained at a length of 30 cm. Determine the emf of the battery.
Solution:

When the annuli point is obtained, no .current flows through the battery. So there is no internal potential drop. In this case, the internal resistance of the battery has no influence. So if E’ is the emf of the battery, then

⇒ \(\frac{E}{100}=\frac{E^{\prime}}{30} \quad \text { or, } \frac{E}{E^{\prime}}=\frac{10}{3}\)

or, E’ = 0.3E

Example 2. A potential difference of 220 V is applied at the two ends of a rheostat of 12000Ω. A voltmeter of resistance 6000 fl is connected between points A and D. If point D divides AB in the ratio of 1:4 what will be the reading of the voltmeter?

Solution:

⇒ \(\frac{A D}{B D}=\frac{1}{4} \quad \text { or, } \frac{A D+B D}{B D}=\frac{1+4}{4} \text { or, } \frac{A B}{B D}=\frac{5}{4}\)

or, \(B D=\frac{4}{5} A B=\frac{4}{5} \times 12000=9600 \Omega\)

∴ AD = AB – BD

= 12000-9600

= 2400Ω

If the voltmeter is connected between A and D, the equivalent resistance of the circuit

= \(9600+\frac{2400 \times 6000}{2400+6000}\)

= \(9600+\frac{24 \times 6000}{84}\)

⇒ \(\left(9600+\frac{2 \times 6000}{7}\right) \Omega\)

∴ Current \(I=\frac{220}{9600+\frac{2 \times 6000}{7}} \mathrm{~A}\)

∴ \(V_{B D}=I \times B D=\frac{220}{9600+\frac{2 \times 6000}{7}} \times 9600\)

⇒ \(\frac{220}{1+\frac{2 \times 6000}{7 \times 9600}}=\frac{220}{1+\frac{5}{28}}=\frac{220 \times 28}{33}=\frac{560}{3} \mathrm{~V}\)

⇒ \(V_{A D}=220-\frac{560}{3}=\frac{100}{3}\)

= 33.3V

Since the voltmeter is connected between points A and D, its reading will be 33.3 V.

Example 3. In a potentiometric arrangement, a cell is connected to the potentiometer wire of 60 cm in length to make the 1 deflection zero in the galvanometer. Now if the cell is shunted with a 6Ω resistor, a null point is found in the 50 cm length of the wire. What is the internal resistance of the cell?
Solution:

Here, shunt R = 6Ω, l₁ = 60cm l₂ = 50cm.

∴ Internal resistance of the cell

⇒ \(r=\left(\frac{l_1}{l_2}-1\right) R\)

= \(\left(\frac{60}{50}-1\right) \times 6\)

= \(0.2 \times 6\)

= \(1.2 \Omega\)

Example 4. A length of potentiometer wire of 188 cm balances the emf of a cell In the n circuit and its length of 135 cm when the cell tins a conductor of resistance 8Ω connected between Its terminals. Find the Internal resistance of
Solution:

Let the emf of the cell = E, and Its internal resistance =r. When a resistance R Is connected between Its terminals (here, R = 8Ω), the coll is subjected to an internal drop of potential Then, the terminal potential difference, the cell.

V=E-Ir

= \(E-\frac{E r}{R+r}\)

= \(E\left(1-\frac{r}{R+r}\right)\)

= \(E \frac{R}{R+r}\)

From the given data,

⇒ \(\frac{E}{V}=\frac{155}{135} \quad\)

or, \(\frac{R+r}{R}=\frac{155}{135} \quad\)

or, \(1+\frac{r}{R}=\frac{155}{135}\)

∴ \(r=R\left(\frac{155}{135}-1\right)=8 \times \frac{20}{135}\)

= 1.185Ω

Kirchhoff’s Laws And Electrical Measurement Wheatstone Bridge

Description: Four resistances P, Q, S, and R form the four arms AB, BC, CD, and DA respectively of a quadrilateral ARCD.

A battery B Is connected between A and C and a galvanometer of resistance G Is connected between B and D, This complete arrangement Is called a Wheatstone bridge circuit. In this circuit the first arm AB has resistance P, the second arm HC has resistance Q, the third arm AD has resistance it, and the fourth arm DC has resistance S.

Working principle: The resistances P, Q, R, and S for the four arms of the bridge are so selected that the galvanometer deflection is zero i.e., galvanometer current IG = 0. This Is called the null condition or balanced condition of the bridge.

In the null condition, the relation of the resistances is:

⇒ \(\frac{P}{Q}\) = \(\frac{R}{S}\) ….(1)

Of the four resistances, if three are known, the fourth resistance can be determined from the above relation. Usually, the unknown resistance is placed in the fourth arm i.e.,.in place of S.

Proof of the relation P/Q = R/S: In the null condition, IG = 0. So, the potential difference between the two ends of the galvanometer,

V_B – V_D = IG.G = 0 [G = resistance of galvanometer]

or, V_B = V_D

For I_G = 0, suppose current in resistance P = current in resistance Q = I₁ and current resistance R = current resistance S = I₂.

∴ V_A-V_B = I₁P; V_A-V_D = I₂R

and V_B-V_C = I₁Q; V_D-V_C = I₂S

Now, Since V_B = V_D,

∴ I₁P = I₂R…(2)

and I₁Q = I₂S….(3)

Now, by dividing equation (2) by equation (3) we have,

⇒ \(\frac{P}{Q}\) = \(\frac{R}{S}\)

On the basis of this relation, the arms P and Q are called ratio arms.

Discussion:

1. Wheatstone bridge is used for measurement of ordinary resistances (from 1Ω to about 1000Ω).

2. If a low resistance (less than 1Ω) is placed in the position of S, the resistances of the connecting wires become nearly equal to S. So the value of S becomes erroneous due to the resistance of the connecting wires. Therefore, for measurements of low resistances Wheatstone bridge is not used.

3. Its high resistance (higher than 1000Ω) Is placed In the position of .S’, and a very small current will pass through it. So in the null condition, the value of I₁ Is much greater than I₂ he., almost the entire current passes through P and Q. In that case, even when the bridge Is unbalanced, no appreciable current flows through the galvanometer, and the galvanometer deflection stays at zero.

Then, It becomes difficult to Identify the actual null condition. In other words, the sensitivity of the bridge decreases. So, Wheatstone bridge Is not used for measurement of high resistances.

4. If the bridge Is sensitive enough, even In slightly off-balance conditions the galvanometer deflection is visible distinctly.
The conditions of sensitiveness of this bridge are:

1. The galvanometer must be very sensitive. Even for the passage of a very feeble current through the galvanometer, the deflection of the pointer should be distinct.
2. The four resistances P, Q, R, and S as well as the resistance of the galvanometer G should be of almost equal value i.e., P ≈ Q ≈ R ≈ S ≈ G.

It is to be noted that, the null condition does not depend on galvanometer resistance G, but the sensitivity of the bridge relies on G. Current can flow through the galvanometer easily if G has a lesser value.

On the other hand, if the number of turns of the coll of the galvanometer is increased, the pointer of the galvanometer deflects more, and G also increases. So a galvanometer is designed for optimum sensitivity.

Various types of galvanometers are available based on which circuit it is connected. In the Wheatstone bridge, if galvanometer resistances are nearly equal to the resistances of four arms i.e., P ≈ Q ≈ R ≈ S ≈ G bridge becomes most sensitive.

5. A galvanometer is a very sensitive instrument for the detection and measurement of current. While making adjustments for the balanced condition of the bridge, there is a possibility of high current passing through the galvanometer. Thus the galvanometer may get damaged. For this reason, in the beginning, it is judicious

1. To connect a high resistance in series with the galvanometer, or
2. To connect a shunt in parallel with the galvanometer.

Later when the bridge comes close to the balanced condition,

1. The resistance can be removed or a shunt can be added to it, or
2. The shunt which is connected with the galvanometer may be removed.

6. The balanced condition does not depend on the emf of the battery, its internal resistance, or any resistance connected with the battery. However, generally, a cell of low emf (e.g., Leclanche cell) Is used. Otherwise, due to the flow of large currents, the resistances of P, Q, R, and S may increase due to the heating effect.

7. If the position of the battery and the galvanometer interchanged, the battery exists between B and D and the galvanometer between A and C. Then Q, S, P, and R become the first, second, third, and fourth arms respectively of the bridge. So in this case the null condition is

⇒ \(\frac{Q}{S}=\frac{P}{R}\).

Obviously, this condition is identical to the condition \(\frac{P}{Q}=\frac{R}{S}\). So it can be said that the balanced condition of the Wheatstone bridge remains unchanged in spite of the interchange of the positions of the battery and the galvanometer. For this reason, the diagonals AC and BD of the bridge are called mutually conjugate.

8. The sensitivity of the bridge is not the same in the two mutually conjugate positions. The bridge becomes more sensitive if the galvanometer is connected between the junction of the two smaller resistances and the junction of the two larger resistances.

9. When the equivalent resistance of the bridge is calculated in a balanced condition, the resistance of the galvanometer i.e., the resistance of the arm BD should not be taken into consideration.

Practical applications of Wheatstone bridge: For measurement of various electrical quantities in the laboratory, the Wheatstone bridge and some of its modified forms are extensively used. Two familiar forms of Wheatstone bridge for measurement of resistances in the laboratory are

1. Post office box and
2. Metre bridge. (Discussion on post office box is beyond our present syllabus.)

Kirchhoff’s Laws And Electrical Measurement Wheatstone Bridge Numerical Examples

Example 1. Resistances of the ratio arms of a Wheatstone bridge are 100Ω and 10Ω respectively. An unknown resistance is placed in the fourth arm and the galvanometer current becomes zero when 153Ω resistance is placed in the third arm. What is the value of unknown resistance?
Solution:

Here, P = 100Ω, Q = 10Ω and R = 153Ω

Since, \(\frac{P}{Q}=\frac{R}{S} \quad\)

∴ \(S=\frac{R Q}{P}=\frac{153 \times 10}{100}, 15.3 \Omega\)

Example 2. Five resistances are connected. What is the effective resistance between points A and B?

Solution:

⇒ \(\text { Here, } \frac{2 \Omega}{3 \Omega}=\frac{4 \Omega}{6 \Omega}\)

So, the circuit between points A and B is a balanced Wheatstone bridge. So no current flows through the resistance of 7Ω. In this condition to calculate the effective resistance, the resistance of 7Ω will be ignored.

Equivalent resistance of 2Ω and 3Ω = 2 + 3 = 5Ω

Equivalent resistance of 4Ω and 6Ω = 4 + 6 = 10Ω

Now at points A and B these 5Ω and 10Ω resistances are connected in parallel.

So, the effective resistance = \(\frac{5 \times 10}{5+10}=\frac{50}{15}=\frac{10}{3} \Omega\)

Example 3. In a Wheatstone bridge, a resistance and an unknown resistance are placed in the third arm and fourth arm respectively. Current through galvanometer becomes zero when the ratio of resistances of first and second arms is 3: 2. Find the value of unknown resistance.
Solution:

⇒ \(\text { Here, } R=15 \Omega \text { and } \frac{P}{Q}=\frac{3}{2}\)

since \(\frac{P}{Q}=\frac{R}{S} \quad\)

∴ \( S=R \cdot \frac{Q}{P}=15 \times \frac{2}{3}=10 \Omega\)

Example 4. The resistances of the four arms of a Wheatstone bridge are 100Ω, 10Ω, 300Ω, and 30Ω respectively. A battery of emf 1.5 V and negligible internal resistance is connected to the bridge. Calculate the current flowing through each resistance.
Solution:

Here, P = 100Ω,

Q = 10Ω, R = 300Ω, S = 30Ω and E = 1.5 V

Since, \(\frac{100}{10}=\frac{300}{30}\)

i.e., \(\frac{P}{Q}=\frac{R}{S}\), the bridge Is in a balanced condition.

∴ Current in the resistance P = current in the resistance Q

⇒ \(\frac{V_{A B}}{P+Q}=\frac{1.5}{100+10}=\frac{1.5}{110}\)

= 0.0136A

Again, current in the resistance R = current in the resistance S

⇒ \(\frac{V_{A B}}{R+S}=\frac{1.5}{300+30}=\frac{1.5}{330}\)

= 0.0045A

Example 5. Every resistance is of magnitude r. What is the equivalent resistance between A and B?

Solution:

The Wheatstone bridge circuit is the equivalent circuit.

Here, P = Q- R = S = r

∴ \(\frac{P}{Q}=\frac{R}{S}\)

i.e., the circuit is in a balanced condition. So for the calculation of equivalent resistance, the value of CD is of no use.

Now, resistance along the path ADB = r+ r = 2r and resistance along the path ACB = r+ r = 2r

Since ADB and ACB are connected in parallel, the equivalent resistance between the points A and B is

⇒ \(R^{\prime}=\frac{2 r \times 2 r}{2 r+2 r}=\frac{4 r^2}{4 r}=r\)

Example 6. Each resistance in the given circuit in is of value R. Calculate the equivalent resistance of the circuit with respect to the points A and JB.

Solution:

The equivalent circuit.

Now proceeding in the same way as done in Example 1, it can be shown that equivalent resistance =R.

Example 7. A coil of wire is kept in melting ice. Its resistance measured by a Wheatstone bridge is 5Ω. If the coil is heated to I00°C and another wire of resistance 100Ω is connected in parallel to it, the balanced condition of the bridge remains unchanged. Determine the temperature coefficient of resistance of the wire of the coil.
Solution:

If the resistance of the coil at 100°C is R, then the equivalent resistance of the parallel combination of R and 100Ω is obviously 5Ω

∴ \(\frac{R \times 100}{R+100}=5 \quad\)

or, 95 R=500

or, \(R=\frac{100}{19} \Omega\)

Again, if or be the temperature coefficient of resistance of the wire of the coil, then

R = R₀(l + αt)

or, \(\alpha=\frac{\frac{R}{R_0}-1}{t}=\frac{\frac{100}{19 \times 5}-1}{100}=\frac{5}{9500}\)

= 5.26 x 10^-4 °C^-1

Example 8. An electrical circuit. Calculate the potential difference across the resistance 400Ω as will be measured by the voltmeter V of resistance 400Ω.

Solution:

The equivalent circuit.

The equivalent resistance between B and C = \(\frac{400 \times 400}{400+400}\)

= 200Ω

This is a Wheatstone bridge circuit, where

⇒ \(\frac{\text { resistance of the side } A B(P)}{\text { resistance of the side } B C(Q)}=\frac{100}{200}\)

= \(\frac{\text { resistance of side } A D(R)}{\text { resistance of side } D C(S)}\)

So, the bridge is in a balanced condition, i.e., no current flows in the arm BD. Under this condition the reading of the voltmeter, V is

⇒ \(V_B-V_C=I Q=\frac{V_A-V_C}{P+Q} \times Q\)

⇒ \(\frac{10}{100+200} \times 200=\frac{20}{3}\)

= 6.67V

Example 9. ABCD is a Wheatstone bridge in which the resistance of the arms AB, BC, CD, and DA are respectively 2Ω,4Ω,6Ω, and 8&&. Points A and C are connected to the terminals of a cell of emf 2 V and have negligible internal resistance. Points B and D are connected to a galvanometer of resistance 50Ω. Using KirchhofFs laws find the current flowing through the galvanometer
Solution:

In the circuit of let V_C = 0; then V_A = 2V;

again, if V_B = V₁ and V_B-V_D = V, then V_D=V₁-V

Now, applying Kirchhoff’s 1st law at junction B, we get

⇒ \(\frac{V_1-2}{2}+\frac{V_1}{4}+\frac{V}{50}=0\)

or, \(\frac{50 V_1-100+25 V_1+2 V}{100}=0\)

or, 75V₁ + 2V = 100…..(1)

Again applying Kirchhoff’s 1st law at junction D, we get

⇒ \(\frac{V_1-V-2}{8}+\frac{V_1-V}{6}+\frac{-V}{50}=0\)

or, \(\frac{75 V_1-75 V-150+100 V_1-100 V-12 V}{600}=0\)

or, 175V₁ – 187V = 150…(2)

Doing (1) x 7- (2) X 3 we get

575 V = 250

or, \(V=\frac{250}{575} \mathrm{~V}\)

∴ \(I_G=\frac{V}{50}=\frac{250}{575 \times 50}=\frac{1}{115}\)

= 0.0087A

= 8.7mA

Example 10. We determine the value of the fourth resistance to be 8Ω with the help of a Wheatstone bridge with three known resistances 100Ω, 10Ω, and 80Ω respectively. If the emf of the cell and its internal resistance are 2 V and 1.1Ω respectively, find the current passing through the cell.
Solution:

The bridge is in a balanced condition because \(\frac{100}{10}=\frac{80}{8}\) galvanometer current is zero. In this case, the effective circuit. Total resistance of the circuit

⇒ \(R=1.1+\frac{(100+10)(80+8)}{(100+10)+(80+8)}\)

= \(1.1+\frac{110 \times 88}{198}\)

= 50Ω

∴ Current through the cell, i.e., the main current of the circuit,

I = \(\frac{2}{50}\) = 0.04A

Metre Bridge:

It is another practical form of the Wheatstone Bridge. It also helps in measuring ordinary resistances very easily.

From the null condition of Wheatstone bridge, we have,

⇒ \(S=R \cdot \frac{Q}{P}\)…(1)

1. In the meter bridge, the steps that are followed are mentioned below.
2. The unknown resistance is placed in the fourth arm S.
3. The magnitude of R, i.e., the resistance of the third arm is
   kept constant.
4. By increasing or decreasing the ratio \(\frac{Q}{P}\), the bridge is brought to a balanced condition.
5. Now by using equation (1) the unknown resistance S is calculated.

Description:

The meter bridge circuit. It consists of a one-metre-long thin uniform wire AB made of manganin or German silver. The wire is stretched and fixed to the two points a and f of two copper strips ab and fe and is placed along a meter scale M over a wooden board. There is another copper strip cd.

The gaps be and de is called the left gap and the right gap, respectively. A jockey J is connected to the midpoint of the copper strip cd through a galvanometer G and its sharp end touches the wire AB and can be moved along it when required.

A resistance box R is inserted in the left gap (third arm of the bridge). The resistance S to be determined is introduced in the right gap (fourth arm of the bridge). A cell E is connected to the bridge through a commutator C.

Working principle: An equivalent meter bridge circuit. Obviously, it is a Wheatstone bridge circuit.

The direction of current through the wire AB can be reversed with the help of the commutator C. The left portion of the wire with respect to the position of the jockey is the first arm of the bridge and the right portion of the wire is the second arm, i.e., resistance of the portion AJ of the wire =P and resistance of the portion JB, of the wire = Q.

A suitable resistance R is introduced in ‘the resistance Now the moved along the wire till a null point is reached. At this balanced condition, let the length of AJ =l cm.

Length of JB = (100-l) cm [∵ AB = 1 m = 100 cm]

If ρ is the resistance per unit length of the wire then,

⇒ \(P=\rho l \text { and } Q=\rho(100-l)\)

∴ \(\frac{Q}{P}=\frac{100-l}{l} \text { and } S=R \cdot \frac{Q}{P}=R \cdot \frac{100-l}{l}\)…..(2)

Knowing l from the scale, S can be calculated from the equation (2). Taking different resistances from box R the experience is repeated.

Discussion:

Material of the wire: The wire of the meter bridge is generally made of alloys like manganin, German silver, etc. The resistance of a uniform manganin wire of length 1 m and diameter 0.5 mm is about 2Ω. The bridge becomes sensitive enough for this magnitude of resistance of the wire. On the other hand, the resistance of a copper wire of the same length and diameter is about 0.1Ω. The bridge is not at all sensitive if such a low resistance is used. For this reason, copper wire is never used in a meter bridge.

Thermoelectric effect:

The copper strips at the ends A and B of the wire come in contact with manganin or German silver. Now, if current passes through the junctions of two dissimilar metals, heat is evolved at one junction and is absorbed at the other due to the thermoelectric effect.

So, the wire AB does not remain in thermal equilibrium and hence its resistance per unit length, ρ is not uniform. Therefore, when we apply equation (2) to calculate the unknown resistance, an error occurs.

To remove this deficit is necessary to reverse the direction of the current through the wire AB. Then, heat will be absorbed at the end where it evolved originally and vice versa.

In this way, the thermal equilibrium of the wire AB will be maintained. To reverse the direction of current in the wire AB, a commutator C is used.

End error:

Though the resistance of the copper strips is very small, it is finitely non-zero.

The bridge wire is soldered at the two ends A and B with the copper strips. So some resistance exists in these two places.

The 0 and 100 cm marks of the meter stale (M) may not coincide accurately with the ends ll and B of the bridge wire. Hence it is assumed that some additional resistance exists at the two ends of the bridge wire.

This is called end resistance. So when we measure an unknown resistance, we face some end errors.

End correction: The end resistance may be assumed to be equivalent to the resistance of a certain length of the bridge wire. Suppose, the end resistance of end A = resistance of Ax cm of the bridge wire and end resistance of end B = resistance of Az cm of the bridge wire

i.e., resistance of l₁ length of the bridge wire = resistance of (l₁ + λ₁) length and resistance of l₂ length of the bridge wire = resistance of (l₂ + λ₂) length

In the position of the null point be l₁ cm then we have from equation (2),

⇒ \(\frac{S}{R}=\frac{\left(100-l_1\right)+\lambda_2}{l_1+\lambda_1}\)…(3)

Now, by interchanging the position of R and S, if the null point is obtained at l₂ cm, then we have

⇒ \(\frac{R}{S}=\frac{\left(100-l_2\right)+\lambda_2}{l_2+\lambda_1}\) …..(4)

If we use two known resistances R and S, we can find out λ₁ and λ₂ by solving equations (3) and (4). Next, using the values of λ₁ and λ₂, the unknown resistance can be measured accurately.

Position of the null point: The bridge becomes sensitive if the position of the null point lies between 40 cm and 60 cm of the bridge wire. But it is better to avoid the null point at 50 cm because in that case no change is observed in the reading of the null point on interchanging the positions of R and S.

Kirchhoff’s Laws And Electrical Measurement Numerical Examples

Example 1. In a meter bridge experiment, a null point Is obtained at a length of 39.8 cm when a 2Ω resistance Is placed In the left gap and a 3Ω resistance in the right gap. If the two resistances are interchanged, the null point is obtained at 60.8 cm. Calculate the end errors of the bridge.
Solution:

Suppose, the resistance per unit length of the bridge wire = ρΩ.cm^-1. End resistance of the left end of the bridge

= resistance λ₁ of cm of the wire. End resistance of the right end of the bridge

= resistance of λ₂ cm. of the wire. These two are the end errors of the bridge.

Therefore, iΩ be the position of the null point, then according to the relation \(\frac{P}{Q}=\frac{R}{S}\) we have,

⇒ \(\frac{R}{S}=\frac{\left(l+\lambda_1\right) \rho}{\left(100-l+\lambda_2\right) \rho}=\frac{l+\lambda_1}{100-l+\lambda_2}\)

In the first case,

R = 2Ω,S = 3Ω and l = 39.8 cm

∴ \(\frac{2}{3}=\frac{39.8+\lambda_1}{60.2+\lambda_2}\)

or, 3λ₁ – 2λ₂ = 1 ….(1)

Again, in the second case,

R = 3Ω,S = 2Ω and l = 60.8 cm

∴ \(\frac{3}{2}=\frac{60.8+\lambda_1}{39.2+\lambda_2}\)

or, 2λ₁ – 3λ₂ = -4 …(2)

Solving (1) and (2) we get,

_λ1 = 2.2 and λ₂ = 2.8

So, the left-end resistance and the right-end resistance of the bridge are equal to the resistances of 2.2 cm and 2.8 cm of the bridge wire respectively

Example 2. In the left gap of a meter bridge, there is a coil of copper and in the right gap, there is a fixed resistance. If the coil of copper is dipped in ice the balance point is obtained at 41.2 cm of the bridge wire. Next, the coil is taken off from ice and placed in a container of hot water. Now the balance point is shifted by a distance of 8.1 cm towards the right. What is the temperature of hot water? (Temperature coefficient of resistance of copper = 42.5 x 10^-4 °C^-1.)
Solution:

Suppose, fixed-resistance = R, the temperature coefficient of resistance of copper = aaa, the resistance of the coil at 0°C = R₀, the position of the balance point = l₀, the resistance of the coil at t°C = Rt, and the position of the balance point =l.

∴ R_t = R₀(l + at)

According to the principle of the meter bridge,

In case of \(0^{\circ} \mathrm{C}, \frac{R_0}{R}=\frac{l_0}{100-l_0}\)….(1)

In case of \(t^{\circ} \mathrm{C}, \frac{R_0(1+\alpha t)}{R}=\frac{197}{100-l}\) …(2)

Dividing (2) by (1) we have,

⇒ \(1+\alpha t=\frac{l}{l_0} \times \frac{100-l_0}{100-l}\)

⇒ \(\frac{49.3}{41.2} \times \frac{100-41.2}{100-49.3} \cdot[l=41.2+8.1=49.3]\)

⇒ \(\frac{49.3}{41.2} \times \frac{58.8}{50.7}=1.388\)

∴ \(t^t=\frac{1.388-1}{\alpha}=\frac{0.388}{42.5 \times 10^{-4}}\)

= 91.3°C

Example 3. In a meter bridge, the balance point is found to be at 40 cm from one end when the resistor at the end is 15Ω. Find the resistance on the other side.
Solution:

If p be the resistance per 1 cm length of the metre wire, then in the present case,

P = 40ρΩ, Q = (100-40)ρ = 60ρΩ, R = 15Ω,

S =?

In balance condition \(\frac{P}{Q}=\frac{R}{S}\)

or, \(S=R \frac{Q}{P}=15 \times \frac{60 \rho}{40 \rho}\)

= 22.5Ω

Example 4. In the circuit given, E₁ = 6 V, E₂ = 2 V, E₃ = 3 V, C’ = 5μF, = 2R₂ = 6Ω, R₃ = 2R₄ = 4Ω. Find the current in R₃ and the energy stored in the capacitor

Solution:

Let the distribution of currents in the various branches be as shown below.

Considering the closed loop CDEHC

⇒ \(I_2 R_3=E_1 \quad \text { or, } 4 I_2=6 \quad \text { or, } I_2=1.5 \mathrm{~A}\)

∴ Current through the resistance = 1.5 A

Considering the closed-loop AHEFGA,

⇒ \(\left(I_1-I_2\right) \times R_2-I_2 R_3+\left(I_1-I_2\right) R_4=-E_2-E_3\)

or, \(\left(I_1-1.5\right) \times 3-1.5 \times 4+\left(I_1-1.5\right) \times 2=-2-3\)

∴ I₁ = 1.7A

Let V_A and V_C be the potentials at points A and C respectively, As the current flows from point A to C along the path AHC,

V_A = V_C + potential drop across R₂ + E₂

or, V_A-V_C= (I₁-I₂)R₂+E₂

= (1.7- 1.5) X 3+ 2

∴ V_A-V_C = 2.6 V

Through branch ABC current = 0. Hence the potential difference across the capacitor is equal to the potential difference between the points A and C.

∴ Potential differences across the capacitor,

V_A-V_C = 2.6V

∴ Energy storedin the capacitor,

⇒ \(U=\frac{1}{2} C V^2=\frac{1}{2} \times 5 \times 10^{-6} \times(2.6)^2\)

= 1.69 x 10^-5 J

Example 5. If the resistance X and Y. (X < Y) are placed in the two gaps of a meter bridge, a null point is obtained at a length of 20 cm. Keeping Y unchanged if a resistance 4X is placed in place of X what will be the position of the null point?
Solution:

In the first case, \(\frac{X}{Y}=\frac{20}{100-20}\)

or, Y = 4X

So, if resistance 4X is placed in place of X, the resistances of the two gaps will be equal, since Y = 4X.

Therefore, the null point will be in the middle of the bridge wire i.e., at 50 cm

Example 6. The distance between the positions of two null points obtained in a meter bridge wire of length 100 cm, by interchanging a known resistance of 2.5Ω and an unknown resistance in the two gaps, is 28.6 cm. Find the value of the unknown resistance.
Solution:

In the first case, if the unknown resistance r is kept at the

left gap and resistance of 2.5Ω at the right gap, the position of null point l₁.

∴ \(\frac{r}{2.5}=\frac{l_1}{100-l_1}\)…(1)

In the second case, after interchanging the resistances, the position of the null point l₂

∴ \(\frac{2.5}{r}=\frac{l_2}{100-l_2}\)…(2)

According to the question,

l₁-l₂ = 28.6…(3)

By (1) x (2) we get,

⇒ \(1=\frac{l_1 l_2}{\left(100-l_1\right)\left(100-l_2\right)}\)

or, 104 – 100l₁ – 100l₂ + l₁l₂ = l₁l₂

or, l₁ + l₂ = 100….(4)

Solving equation (3) and (4),

l₁ = 643

and l₂ = 100-l₁ = 35.7

∴ The value of the unknown resistance

⇒ \(r=2.5 \times \frac{l_1}{100-l_1}=2.5 \times \frac{64.3}{100-64.3}=4.5 \Omega\)

Kirchhoff’s Laws And Electrical Measurement Very Short Questions and Answers

Question 1. In which part of an electrical circuit can Kirchhoff’s law of
Answer: At the meeting points

Question 2. In which part of an electrical circuit can Kirchhoff’s law of voltage be applied?
Answer: At The Closed Loops

Question 3. If the potential of point C be 12 V, what will be the potential of the point D?

Answer: 3V

Question 4. At a junction of three wires, the inward currents through two of the wires are 1 A and 2 A. What is the inward current through the third wire?
Answer: -3 A

Question 5. AB = 1 m. What will be the length of the part AC for the null condition of the galvanometer?

Answer: 75cm

Question 6. Why do we prefer a potentiometer with a longer wire?
Answer: Proportional error in the recorded reading will be less

Question 7. What kind of cell should be used in a Wheatstone bridge circuit?
Answer: Leclanche cell

Question 8. Of battery and galvanometer segments which one should be closed first during the operation of a Wheatstone bridge circuit?
Answer: Battery

Question 9. The resistance of each of the four arms of a Wheatstone bridge is 10Ω and the resistance of the galvanometer is 500Ω. What is the equivalent resistance of the combination?
Answer: 10Ω

Question 10. The resistances in the left and the right gaps of a metre bridge circuit are 3Ω and 2Ω respectively. For what length of the bridge wire IN the null point found If there In no end error?
Answer: 60cm

Question 11. Keeping a resistance of 2Ω In the left gap of a metre bridge, an unknown resistance Is placed In the right gap and the null point is obtained at a distance of 40.0 cm. If no end error Is present, what will be the value of the unknown resistance?
Answer: 3Ω

Kirchhoff’s Laws And Electrical Measurement Fill In The Blanks

1. Kirchhoff’s law of current expresses the principle of conservation of charge

2. With the help of a potentiometer, the emf of a cell can be measured accurately, in this experiment lost volt of the cell becomes zero

3. A battery of steady emf 2.0 V is connected across the two ends of a potentiometer wire. With the help of this arrangement, the emf of more than 2V of a cell can not be determined.

Kirchhoff’s Laws And Electrical Measurement Assertion Reason Type

Direction: These questions have Statement 1 and Statement 2. Of the four choices given below, choose the one that best describes the two statements.

1. Statement 1 is true, Statement 2 is true; Statement 2 is a correct explanation for Statement 1.
2. Statement 1 is true, Statement 2 is true; Statement 2 is not a correct explanation for Statement 1.
3. Statement 1 is true, Statement 2 is false.
4. Statement 1 is false, and Statement 2 is true.

Question 1.

Statement 1: A balance is obtained at the position of 40 cm on the meter wire, when 2Ω and 3Ω resistances are put in the left and right gaps respectively, of a meter bridge.

Statement 2: The balanced condition of a Wheatstone bridge is P/Q = R/S.

Answer: 1. Statement 1 is true, Statement 2 is true; Statement 2 is a correct explanation for Statement 1.

Question 2.

Statement 1: An alloy like manganin or German silver is used, instead of copper, as the material of the wire of a meter bridge.

Statement 2: The temperature coefficient of resistance is very low for alloys.

Answer: 2. Statement 1 is true, Statement 2 is true; Statement 2 is not a correct explanation for Statement 1.

Question 3.

Statement 1: A potentiometer arrangement is more suitable than a voltmeter arrangement for the accurate measurement of the emf of an electric cell.

Statement 2: It is possible to connect an electric cell in a potentiometer circuit in such a way that no current passes through the cell.

Answer: 1. Statement 1 is true, Statement 2 is true; Statement 2 is a correct explanation for Statement 1.

Question 4.

Statement 1: If the resistances of the first two arms P and Q of a balanced Wheatstone bridge are exchanged, the balanced condition is not disturbed.

Statement 2: The balanced condition of a Wheatstone bridge is independent of the resistance of the galvanometer used.

Answer: 4. Statement 1 is false, Statement 2 is true.

Question 5.

Statement 1: Kirchhoff’s voltage law indicates that the static field is conservative.

Statement 2: The potential difference between two points in a circuit does not depend on the path.

Answer: 1. Statement 1 is true, Statement 2 is true; Statement 2 is a correct explanation for Statement 1.

Question 6.

Statement 1: In the balanced Wheatstone bridge,

⇒ \(R_{A C}=\frac{(P+Q)(R+S)}{(P+Q+R+S)}\)

Statement 2: This is because B and D are at the same potential.

Answer: 2. Statement 1 is true, Statement 2 is true; Statement 2 is not a correct explanation for Statement 1.

Question 7.

Statement 1: In an electrical circuit the algebraic sum of currents meeting at a point is zero.

Statement 2: In the case of the flow of current in an electrical circuit total energy is conserved.

Answer: 2. Statement 1 is true, Statement 2 is true; Statement 2 is not a correct explanation for Statement 1.

Kirchhoff’s Laws And Electrical Measurement Match The Columns

Question 1. A balance is obtained at 25 cm of a meter bridge when resistances X and Y are placed in the left and the right gaps respectively. Then, during repetitions of the experiment, the left and the right gaps are filled respectively, with the resistances of column A. Column B shows the positions of, the balance points

Answer: 1-D, 2-B, 3-A, 4-C

Question 2. Match the following two columns to specify the instrument to be used for the given electrical purposes

Answer: 1-B, 2-D, 3-A, 4-C

Capacitance And Capacitor Capacitance

Ifabodyis heated, its temperature rises. Similarly, the conductor is positively charged, and its potential increases if different materials are heated equally, their rise in temperature is not equal because the thermal capacities of the bodies are different. Similarly, even if different conductors are charged equally, their increase in potential may not be equal. The potential of a conductor depends not only on the amount of charge possessed by it but also on its shape, surface area, nature of the surrounding medium, and presence of other conductors.

For a given conductor its potential is always proportional to its charge. If a conductor is charged with Q and as a result, its potential is raised by an amount V, then

Q ∝ V or, Q = CV

The proportionality constant C is known as the capacitance or capacity of the conductor. Its value depends on shape, surface area, nature of the surrounding medium, and presence of other conductors.

From equation (1) we have,

C = \(\frac{Q}{V}\)…(2)

i.e., capacitance = \(\frac{amount of charge}{rise of potential}\)

If V = 1, then C = Q.

Read and Learn More Class 12 Physics Notes

Definition: The capacitance or capacity of a conductor is defined as the charge required to raise its potential by unity.

Units of Capacitance:

In the CGS system:

In the CGS system, a unit of capacitance is esu of capacitance. In the above equation (2), if Q = 1 esu of charge and V = 1 esu of potential, then’ C = 1 esu of capacitance, i.e., if 1 esu of charge raises the potential of a conductorby1 esu, the capacitance of the conductor is defined as 1 esu. This unit is also known as statfarad (statF).

In SI: SI unit of capacitance is farad (F). It is the practical unit of capacitance. The capacitance of a conductor is said to be 1 farad if 1 coulomb of charge is required to raise the potential of the conductor by 1 volt.

Therefore, IF = \(\frac{1 \mathrm{C}}{1 \mathrm{~V}}\)

Since farad is a very large unit, smaller units like microfarad and picofarad are most frequently used as the unit of capacitance.

1 microfarad (μF) = 10^-6 farad (F)

1 picofarad (pF) = 10^-12 farad (F)

Relation between farad and esu of capacitance:

1C = 3 X 10⁹ esu of charge

IV = esu of potential

∴ \(1 \mathrm{~F}=\frac{1 \mathrm{C}}{1 \mathrm{~V}}=\frac{3 \times 10^9 \text { esu of charge }}{\frac{1}{300} \text { esu of potential }}\)

= 9 x 10¹¹ esu of capacitance or statF

1μF = 10^-6F = 9 x 10 esu of capacitance

lpF = 10^-12F = 0.9 esu of capacitance

Dimension of capacitance

⇒ \(C=\frac{Q}{V}=\frac{Q}{W / Q}=\frac{Q^2}{W}\)

∴ \([C]=\frac{\left.T^2\right|^2}{M L^2 T^{-2}}=M^{-1} L^{-2} T^4 I^2\)

Capacitance And Capacitor Capacitance Numerical Examples

Example 1. The capacitance of a spherical conductor is 1μF and the charge on it is -10C. What is its potential in the air?
Solution:

We know, V = 2;

here C = 1μF = 10^-6F,

Q = -10C

∴ \(V=\frac{-10}{10^{-6}}=-10^7 \mathrm{~V}\)

Example 2. The potential of a conductor having 40 esu of capacitance is raised by 10 esu. What is the charge on the conductor? How much charge is to be given to another conductor, having capacitance three times that of the first conductor, to raise its potential three times that of the first one?
Solution:

Charge given to the first conductor,

Q₁ = C₁V₁

= 40 x 10

= 400 esu of charge

The capacitance of the second conductor,

C₂ = 3C₁

= 3 x 40

= 120 esu of capacitance

The potential rise of the second conductor,

V₂ = 3 x 10

= 30 esu of potential

Charge to be given to the second conductor,

Q₂ = C₂V₂

= 120 x 30

= 3600 esu of charge

Factors Affecting Capacitance of a Conductor:

A conductor, at a potential V and having a charge Q, has a capacitance, C = \(\frac{Q}{V}\).

For constant Q, \(C \propto \frac{1}{V}\) Hence, the factors affecting V also affect the value of C of a conductor. The value of the capacitance of a conductor depends on the following factors.

1. Surface area and shape of the conductor,
2. Nature of the surrounding medium
3. Presence of other conductors (especially earthed ones).

Surface area and shape of the conductor:

A conductor of greater size, i.e., a larger surface area has larger capacitance. The potential of a conductor decreases with the increase of its surface area and hence its capacitance increases.

Experiment:

A thin tin sheet is suspended from a charged ebonite rod. At the bottom of the sheet, a heavy metal rod is attached. This rod keeps the sheet stretched.

The tin sheet Is connected to the disc of a gold-leaf electroscope by a metal wire. If the sheet Is given a definite amount of charge, the leaves of the gold-leaf electroscope spread apart.

The divergence of the leaves indicates the potential of the sheet. NowIf the tin sheet is rolled up to some extent with the help of the ebonite rod, the divergence of the leaves of the electroscope will Increase.

It indicates that the potential of the sheet has increased, So, the charge of the sheet remains constant, and its capacitance decreases.

If the of the tin sheet is increased, the divergence of the leaves decreases, i.e., the potential of the sheet decreases and its capacitance increases. So capacitance of a conductor depends on its surface area.

If the experiment is performed with conductors of the same surface area but of different shapes, it will be found that the spreading of the leaves of the gold-leaf electroscope are different So capacitance of a conductor also depends on its shape.

Nature of the surrounding medium:

If the conductor is surrounded by some dielectric medium other than air, the capacitance of the conductor increases. The effect increases with the increase of the dielectric constant of the medium.

Experiment:

A charged metal sheet A placed on an insulating stand is connected to the disc of an uncharged gold-leaf electroscope. The divergence of the leaves of the electroscope is observed. The amount of deflection indicates the potential of the metal sheet.

Now a dielectric slab, say a thick glass slab, is brought slowly near sheet A. It is found that the spreading of the leaves diminishes. So, the potential of A has decreased, i.e., its capacitance has increased.

Again if the dielectric slab is removed from the vicinity of sheet A, the electroscope leaves will spread out to the same extent as earlier.

In this case, the dielectric medium is polarised from induction due to the metal sheet A. So opposite charges are developed on the two sides of the glass slab.

Induced negative charge reduces the potential of sheet A and positive charge raises the potential of A.

However, due to the close proximity of the negative charge, its effect on sheet A predominates. So as a whole, the potential of sheet A diminishes a little, and hence its capacitance increases.

Presence of other conductors:

The capacitance of a conductor depends on the presence of other conductors near it. If an uncharged conductor is present in the vicinity of the charged conductor under test, its capacitance increases. This effect becomes pronounced if the neighboring conductor is earthed.

Experiment:

If a positively charged conductor A placed on an insulating stand is connected to the disc of an uncharged gold-leaf electroscope, the leaves of the electroscope spread out. The amount of divergence indicates the potential of conductor A.

Now another uncharged conductor B placed on an insulating stand is brought near A, it will be found that the divergence of the leaves diminishes a little. So, the potential of A has decreased a little.

If conductor B is removed, the leaves spread to the same extent as earlier. From this, it is understood that if B is brought near A, the potential of A diminishes, and its capacitance increases.

The reason is that an induced negative charge is developed at the nearer end and a positive charge at the far end of B due to the inducing charge of A.

The induced negative charge reduces the potential of A and the induced positive charge enhances its potential. But due to the proximity of the negative charge its effect on A predominates. So as a whole, the potential of conductor A diminishes a little, and hence its capacitance increases.

Now if conductor B is earthed, it will be found that the divergence of the leaves decreases considerably. This proves that the potential of conductor A is highly reduced.

If conductor B is removed from the vicinity of the coil now ductor A, the leaves of the electroscope will spread out to the same extent as earlier. So it is proved that if conductor B is earthed, the potential of conductor A diminishes a lot and hence its capacitance increases to a large extent.

The reason is that, if B is connected to the earth, the induced positive charge being free charge moves to the earth. Under this condition, due to the presence of only a negative charge in B, the potential of A diminishes a lot and hence its capacitance increases to a large extent.

The capacitance of a Spherical Conductor:

Let us consider a spherical conductor of radius R charged with Q’ amount of charge. The charge Q is uniformly distributed over the surface of the sphere. Potential at the surface of the sphere, as we know, is the same as that produced by an isolated point charge Q placed at the center of the sphere.

The potential of the sphere is given by,

⇒ \(V=\frac{1}{4 \pi \epsilon_0} \cdot \frac{Q}{R}\) [∈₀ = permittivity of air or vacuum]

∴ The capacitance of the sphere,

⇒ \(C=\frac{Q}{V}=4 \pi \epsilon_0 R\)

We know, \(\frac{1}{4 \pi \epsilon_0}=9 \times 10^9 \mathrm{~N} \cdot \mathrm{m}^2 \cdot \mathrm{C}^{-2}\)

So, in vacuum or air, the capacitance of the sphere,

⇒ \(C=\frac{R}{9 \times 10^9} \text { farad }\)

In CGS system, replacing \(\epsilon_0 \text { by } \frac{1}{4 \pi}\), we have, C = R.

Hence the capacitance in the CGS unit of a spherical conductor placed in the air (or vacuum) is numerically equal to its radius in centimeters. For this reason, the capacitance CGSunit is sometimes expressed in centimeters.

Unit of ∈₀: From the relation C = 4π∈₀R,we have,

⇒ \(\epsilon_0=\frac{C}{4 \pi R}\)

So, the unit of ∈₀

⇒ \(=\frac{\text { unit of } C}{\text { unit of } R}=\frac{\mathrm{F}}{\mathrm{m}}=\text { farad } / \text { metre }\left(\mathrm{F} \cdot \mathrm{m}^{-1}\right)\)

Using this simpler unit, we may write,

= 8.854 x 10^-21 F.m^-1

That this unit F m^-1 is identical to the unit C².N^-1.m^-2 of ∈₀, used earlier, is shown here

⇒ \(\mathrm{F} \cdot \mathrm{m}^{-1}=\frac{\mathrm{F}}{\mathrm{m}}=\frac{\mathrm{C}}{\mathrm{V} \cdot \mathrm{m}}=\frac{\mathrm{C}}{\frac{\mathrm{J}}{\mathrm{C}} \cdot \mathrm{m}}=\frac{\mathrm{C}^2}{\mathrm{~J} \cdot \mathrm{m}}\)

⇒ \(\frac{\mathrm{C}^2}{(\mathrm{~N} \cdot \mathrm{m}) \cdot \mathrm{m}}=\frac{\mathrm{C}^2}{\mathrm{~N} \cdot \mathrm{m}^2}=\mathrm{C}^2 \cdot \mathrm{N}^{-1} \cdot \mathrm{m}^{-2}\)

Capacitance And Capacitor Factors Affecting Capacitance of a Conductor Numerical Examples

Example 1. The radius of the__earth is 6400km. Determine its capacitance in μF.
Solution:

The radius of the earth = 6400 km = 6400 x 10³ m.

Capacitance, \(C=4 \pi \epsilon_0 R=\frac{1}{9 \times 10^9} \times 6400 \times 10^3\)

⇒ \(=\frac{64}{9} \times 10^{-4}=\frac{6400}{9} \times 10^{-6} \mathrm{~F}\)

= 711.1μF

Example 2. A metal sphere has a diameter of 1 m. What will be the amount of charge required to raise its potential by 2.7 x 10⁶ V?
Solution:

Radius, R = \(\frac{1}{2}\) = 0.5 m;

⇒ \(C=4 \pi \epsilon_0 R=\frac{1}{9 \times 10^9} \times 0.5=\frac{0.5}{9 \times 10^9} \mathrm{~F}\)

∴ The amount of charge required,

⇒ \(Q=C V=\frac{0.5}{9 \times 10^9} \times\left(2.7 \times 10^6\right)\)

= 1.5 X 10^-4

= 150 X 10^-6 C

= 150μC

Example 3. Is it possible for a metal sphere of radius 1 cm to hold a charge of IC?
Solution:

Radius, r = 1 cm = 0.01 m

So, capacitance of the sphere

⇒ \(C=4 \pi \epsilon_0 r=\frac{1}{9 \times 10^9} \times 0.01=\frac{1}{9} \times 10^{-11} \mathrm{~F}\)

∴ The potential of the sphere,

⇒ \(V=\frac{Q}{C}=\frac{1}{\frac{1}{9} \times 10^{-11}}=9 \times 10^{11} \mathrm{~V}\)

At this very high potential, the sphere will discharge in the sureo rounding air, i.e., it will not be able to hold the charge of 1 C.

Example 4. SI The diameter of the spherical liquid drop is 2mm and its charge is 5 x 10^-6 esu.

1. What is the potential and its surface?
2. If two such liquid drops coalesce to form a bigger drop, what will be the potential on its surface?

Solution:

1. In the CGS system, the radius of the spherical conductor = its capacitance (numerically).

∴  The capacitance of the spherical liquid drop,

C = 0.1 statF [∵ Radius = 1mm = 0.1 cm ]

∴  Potential on the surface of the liquid drop,

⇒ \(V=\frac{Q}{C}=\frac{5 \times 10^{-6}}{0.1}=5 \times 10^{-5} \mathrm{statV}\)

= 5 x 10^-5 x 300 V

= 0.015 V

2. Let the radius of the bigger drop be R.

According to the question,

⇒\(\frac{4}{3} \pi R^3=2 \times \frac{4}{3} \pi(0.1)^3\)

or, \(R^3=2 \times(0.1)^3\)

or, \(R=0.1 \times 2^{\frac{1}{3}}\)

= 0.1 x 1.26

= 0.126 cm

Total charge,

Q = 2 x 5 x 10^-6

= 10~5 esu of charge.

∴  Potential on the surface of the bigger liquid drop,

⇒ \(V=\frac{Q}{C}=\frac{10^{-5}}{0.126}\)

= 7.94 x 10^-5 x 300 statV

= 7.94 x 10^-5 x 300 V

= 0.0238 V

Capacitance And Capacitor Potential Energy Of A Charged Conductor

A certain amount of work has to be done in order to charge a conductor. The energy spent for doing that work remains stored in the charged conductor as potential energy. Essentially, the electric field of the conductor stores this energy.

Calculation:

Let a conductor be charged with Q and let its capacitance be C. The potential of the conductor is V. During charging we assume that the whole amount of charge is not given to the conductor at a time, rather it is charged gradually.

At first, the charge of the conductor is zero so its potential is also zero. Gradually its potential increases due to the accumulation of charges. So at the time of charging the conductor has no particular potential. Its potential becomes V when its charge is Q.

Initial potential = 0; final potential = V

Average of these potentials = \(\frac{0+V}{2}=\frac{V}{2}\)

Therefore, work done = average potential x charge

⇒ \(\frac{V}{2} \times Q=\frac{1}{2} Q V=\frac{1}{2} C V \cdot V\) [∵ Q = CV]

⇒ \(\frac{1}{2} C V^2\)

= \(\frac{1}{2} C \times\left(\frac{Q}{C}\right)^2\)

= \(\frac{1}{2} \cdot \frac{Q^2}{C}\)

This work is stored in the charged conductor as potential energy.

∴ The potential energy of a charged conductor

⇒ \(\frac{1}{2} Q V\)

= \(\frac{1}{2} C V^2\)

= \(\frac{1}{2} \frac{Q^2}{C}\)

If C, V, and Q are expressed in esu, the unit of potential energy will be erg. Again C, V, and Q are expressed in farad, volt, and coulomb, respectively, the unit of potential energy will be joule.

Derivation using calculus: Let at any moment the charge on the conductor be q and its potential be v.

Evidently, q = Cv

When a charge dq is given to the conductor, the work done against the repulsive force due to potential v is given by,

⇒ \(d W=v d q=\frac{q}{C} d q\)

Hence the total work done to impart Q amount of charge is,

⇒ \(W=\int d W=\int_0^Q \frac{q}{C} d q=\frac{1}{2} \frac{Q^2}{C}=\frac{1}{2} C V^2=\frac{1}{2} Q V\)

This work is stored as potential energy in the charged conductor.

Capacitance And Capacitor Distribution Of Charge Between Two Conductors

Two conductors at the same potential:

Let us consider two insulated uncharged conductors A and B of capacitances C₁ and C₂, respectively. They are connected by a fine metal wire. Under this condition, if a charge Q is given to this combination, it will be distributed between the two conductors.

Let us consider that conductor A has obtained a charge Q₁ and conductor B a charge Q₂ As the two conductors are connected to each other, they have the same potential. Let their common potential be V.

Q = Q₁ + Q₂

and \(V=\frac{Q_1}{C_1}\)

= \(\frac{Q_2}{C_2}\)

= \(\frac{Q_1+Q_2}{C_1+C_2}\)

= \(\frac{Q}{C_1+C_2}\)

⇒ \(\left.\begin{array}{ll}
∴ & Q_1=C_1 V=Q \cdot \frac{C_1}{C_1+C_2} \\
\text { and } & Q_2=C_2 V=Q \cdot \frac{C_2}{C_1+C_2}
\end{array}\right\}\)…(1)

Again, \(\frac{Q_1}{Q_2}=\frac{C_1}{C_2}\)….(2)

Therefore, a charge on each conductor is proportional to its capacitance. If the two conductors have the same capacitance, the given charge will be shared equally between them. If they have radii r₁ cm and r₂ cm, then C₁ = r₁ and C₂ = r₂ (in the CGS system).

In that case,

⇒ \(\left.\begin{array}{l}
Q_1=Q \cdot \frac{r_1}{r_1+r_2} \\
Q_2=Q \cdot \frac{r_2}{r_1+r_2}
\end{array}\right\}\)….(3)

∴ \(\frac{Q_1}{Q_2}=\frac{r_1}{r_2}\)…(4)

Therefore, the charge on each spherical conductor is proportional to its radius.

We know that the surface density of charge on the surface of a charge is the same everywhere.

If σ₁ and σ₂ be the surface densities of charge of the two conductors, then

⇒ \(\sigma_1=\frac{Q_1}{4 \pi r_1^2} \text { and } \sigma_2=\frac{Q_2}{4 \pi r_2^2}\)

∴  \(\frac{\sigma_1}{\sigma_2}=\frac{Q_1}{Q_2} \cdot \frac{r_2^2}{r_1^2}=\frac{r_1}{r_2} \times \frac{r_2^2}{r_1^2}=\frac{r_2}{r_1}\)…(5)

Therefore, the surface density of the charge of a spherical conductor is inversely proportional to its radius.

Two conductors initially at different potentials:

Let us consider two insulated conductors A and B. They have capacitances C₁ and C₂ and they are given charges Q₁ and Q₂ separately.

So under this condition,

potential of the conductor A, \(V_1=\frac{Q_1}{C_1} \text { or, } Q_1=C_1 V_1\)

and potential of the conductor B, \(V_2=\frac{Q_2}{C_2} \text { or, } Q_2=C_2 V_2\)

∴  The total charge of the conductors,

Q = Q₁ + Q₂

= C₁V₁ + C₂V₂…(6)

If the two conductors are connected by a thin metal wire, a positive charge will flow from the conductor at a higher potential to that at a lower potential and this flow of charge will continue till their potentials become equal.

Suppose, V₁ > V₂; then a charge will flow from A to B. Let V be the common potential after connection. During this flow of charge, the total charge of the system remains constant.

So, total charge before connection = total charge after connection

i.e„ Q = C₁V₁ + C₂V₂

= C₁V + C₂V

= (C₁ + C₂)V

or, \(V=\frac{C_1 V_1+C_2 V_2}{C_1+C_2}\)….(7)

After connection, if A and B contain charges q x and q2 respectively, then

⇒ \(\left.\begin{array}{l}
q_1=C_1 V=C_1 \cdot \frac{C_1 V_1+C_2 V_2}{C_1+C_2}=\frac{C_1}{C_1+C_2} \cdot Q \\
q_2=C_2 V=C_2 \cdot \frac{C_1 V_1+C_2 V_2}{C_1+C_2}=\frac{C_2}{C_1+C_2} \cdot Q
\end{array}\right\}\)….(8)

As V₁ > V₂, the charge lost by A,

q’₁ =C₁V₁-C₁V = C₁(V₁-V)

⇒ \(C_1\left(V_1-\frac{C_1 V_1+C_2 V_2}{C_1+C_2}\right)\)

or, \(q_1^{\prime}=\frac{C_1 C_2\left(V_1-V_2\right)}{C_1+C_2}\)….(9)

Again, the charge gained by B,

⇒ \(q_2^{\prime}=C_2 V-C_2 V_2=C_2\left(\frac{C_1 V_1+C_2 V_2}{C_1+C_2}-V_2\right)\)

or, \(q_2^{\prime}=\frac{C_1 C_2\left(V_1-V_2\right)}{C_1+C_2}\)…..(10)

From equations (9) and (10) we get, q₁ = q₂, i.e., the charge gained by conductor B is equal to that lost by conductor A.

Loss of energy due to sharing of charge:

Before connection, the total energy of the two conductors

⇒ \(\frac{1}{2} C_1 V_1^2+\frac{1}{2} C_2 V_2^2\)

After connection, the total energy of them

⇒ \(\frac{1}{2} C_1 V^2+\frac{1}{2} C_2 V^2=\frac{1}{2}\left(C_1+C_2\right) V^2\)

⇒ \(\frac{1}{2}\left(C_1+C_2\right) \cdot\left(\frac{C_1 V_1+C_2 V_2}{C_1+C_2}\right)^2\)

⇒ \(\frac{1}{2} \cdot \frac{\left(C_1 V_1+C_2 V_2\right)^2}{C_1+C_2}\)

Therefore, loss of energy due to the sharing of charge

⇒ \(\frac{1}{2} C_1 V_1^2+\frac{1}{2} C_2 V_2^2-\frac{1}{2} \cdot \frac{\left(C_1 V_1+C_2 V_2\right)^2}{C_1+C_2}\)

⇒ \(\frac{1}{2} C_1 V_1^2+\frac{1}{2} C_2 V_2^2-\frac{1}{2} \cdot \frac{\left(C_1 V_1+C_2 V_2\right)^2}{C_1+C_2}\)

⇒ \(\frac{1}{2} \cdot \frac{C_1 C_2}{C_1+C_2} \times\left(V_1-V_2\right)^2\)….(11)

Now, C₁ and C₂ are both positive quantities, and (V₁ – V₂)², being a perfect square, is also positive. So, the relation (11) is positive.

Hence, there is always a loss of energy in the electric field of the conductors due to the sharing of charges.

According to the law of conservation of energy, this loss of energy must be converted to some other form, usually as heat in the connecting wire. This Joss is partly converted into light and sound in addition to heat if sparkling occurs.

Capacitance And Capacitor Distribution Of Charge Between Two Conductor Numerical Examples

Example 1. A conductor of capacity 4 units, charged with 100 units of positive charge is connected to another conductor of capacity 2 units, charged with 20 units of negative charge. What is the change In the potential of each conductor? What will be the charges for each of them after the connection?
Solution:

Capacity of the first conductor; C₁ = 4 units; charge, Q₁ = 100 units.

∴ \(\text { Potential, } V_1=\frac{Q_1}{C_1}\)

= \(\frac{100}{4}\)

= 25units

Capacity of the second conductor C₂ = 2 units; charge, Q₂ = -20 units.

∴ \(\text { Potential, } V_2=\frac{Q_2}{C_2}\)

= \(\frac{-20}{2}\)

= -10 units

After connection, suppose, the common potential of the two conductors becomes equal to V.

∴ \(V=\frac{Q_1+Q_2}{C_1+C_2}\)

= \(\frac{100-20}{4+2}\)

= \(\frac{80}{6}\)

= \(\frac{40}{3}\)

= 13.33 units

So, the change of potential of the first conductor

= 25-13.33

= 11.67 units

The change of potential of the second conductor

= 13.33- (-10)

= 2333 units

Residual charge in the first conductor after connection,

q₁ = C₁V

= 4 x \(\frac{40}{3}\)

= 5333 units

Residual charge in the second conductor after connection,

q₂ = C₂V

= 2 x \(\frac{40}{3}\)

= 26.67 units

Example 2. An insulated metallic vessel full of water Is charged with a potential of 3V. Drops of water are trickling from an orifice at the bottom of the vessel What is the amount of charge contained In each spherical drop of radius 1mm?
Solution:

Potential of the metal vessel full of water, V = 3 V Radius of a water drop, R = 1 mm = 10^-3m

∴ The capacitance of the water drop,

⇒ \(C=4 \pi \epsilon_0 \mathrm{R}=\frac{10^{-3}}{9 \times 10^9}=\frac{1}{9} \times 10^{-12} \mathrm{~F}\)

∴ Charge of each water drop,

⇒ \(Q=C V=\frac{1}{9} \times 10^{-12} \times 3=3.3 \times 10^{-13} \mathrm{C}\)

Example 3. The radii of two insulated metal spheres are 5 cm and 10 cm. They are charged up to potentials of 10 esu and 15 esu, respectively. If the two spheres are connected with one another, what will be the loss of energy?
Solution:

The radius of the first sphere = 5 cm

Capacitance, C₁ = 5 statF

and potential, V₁ = 10 statV

Charge, Q₁ = C₁V₁

= 5 x 10

= 50 state

The radius of the second sphere = 10 cm

∴ Capacitance, C₂ = 10 statF

and potential, V₂ = 15 statV

Charge, Q₂ = C₂V₂

= 10 x 15

= 150 statC

The total charge of the two spheres,

Q = Q₁ + Q₂

= 50 + 150

= 200 statC

Equivalent capacitance of the combination of two spheres,

C = C₁ + C₂

= 5 + 10

= 15 statF

If V is the common potential of the two spheres after connection, then

⇒ \(V=\frac{\text { total charge of the two spheres }}{\text { total capacitance of the two spheres }}\)

⇒ \(\frac{200}{15}\)

= \(\frac{40}{3} \mathrm{statV}\)

The total energy of the two spheres before connection,

⇒ \(E_1=\frac{1}{2} C_1 V_1^2+\frac{1}{2} C_2 V_2^2\)

⇒ \(\frac{1}{2}\left[5 \times(10)^2+10 \times(15)^2\right]\)

= 1375 erg.

The total energy of the two spheres after connection,

⇒ \(E_2=\frac{1}{2} C V^2=\frac{1}{2} \times 15 \times\left(\frac{40}{3}\right)^2\)

= 1333.33 erg

∴ Loss of energy due to connection

= 1375-1333.33

= 41.67 erg

Example 4. A metal sphere of radius 10 cm is charged up to a potential of 80 esu. After sharing its charge with another sphere, their common potential becomes 20 esu. What is the radius of the second sphere?
Solution:

The radius of the first sphere = 10 cm

∴ Capacitance C₁ = 10 statF

and potential, = 80 statV

∴ Charge, Q₁ = C₁V₁

= 10 x 80

= 800 state

After connection, the common potential, V = 20 statV

∴ \(V=\frac{\text { total charge of the two spheres }}{\text { equivalent capacitance of the two spheres }}\)

or, \(20=\frac{800}{10+C_2}\left[C_2=\text { capacitance of the second sphere }\right]\)

or, 10+ C₂ = 40

or, C₂ = 30 statF

Example 5. One thousand similar electrified raindrops merge into a single one so that their total charge remains unchanged. Find the change in the total electrostatic energy of the drops, assuming that all the drops are spherical and the small drops were initially at large distances from one another
Solution:

Suppose, each drop of radius r contains a charge Q.

∴ Capacitance, C = r

and energy, \(E_1=\frac{1}{2} \cdot \frac{Q^2}{C}=\frac{Q^2}{2 r}\)

The total energy of 1000 drops,

⇒ \(E=1000 E_1=\frac{500 Q^2}{r}\)

If R is the radius of the large drop then,

⇒ \(\frac{4}{3} \pi R^3=1000 \times \frac{4}{3} \pi r^3\)

or, R = 10r

Charge of the large drop = 1000 Q.

∴ The energy of the large drop,

⇒ \(E_2=\frac{1}{2} \cdot \frac{(1000 Q)^2}{R}\)

= \(\frac{5 \times 10^5 Q^2}{10 r}\)

= \(\frac{5 \times 10^4 Q^2}{r}\)

∴ \(\frac{E_2}{E}=\frac{5 \times 10^4}{500}=100\)

or, E₂ = 100E

∴ Change in electrostatic energy of the drops

E₂-E = 100E-E

= 99E

= 99 x initial energy

Example 6. Two equally charged soap bubbles of equal volume join together to form a large bubble. If each small bubble had a potential V, find the potential of the result
Solution:

Let the radius of each small bubble be r, the charge be Q, and the radius of the large bubble be R.

∴ \(\frac{4}{3} \pi R^3=2 \times \frac{4}{3} \pi r^3 \text { or, } R=2^{1 / 3} \cdot r\)

The potential of each small bubble,

V = \(\frac{Q}{r}\)

or, Q = Vr

∴ The potential of the large bubble

⇒ \(\frac{\text { total charge }}{\text { radius }}\)

⇒ \(\frac{2 Q}{R}\)

= \(\frac{2 V r}{2^{1 / 3} r}\)

= \(2^{2 / 3} V\)

Example 7. Eight spherical liquid drops join to form a large drop. The diameter of each drop is 2 mm and the charge 5μ statC. What is the potential on the surface of the large drop?
Solution:

Radius of a small drop, r = 1 mm =0.1 cm

Suppose, the radius of the large drop is R.

∴ \(\frac{4}{3} \pi R^3=8 \times \frac{4}{3} \pi r^3\)

or, R = 2r

= 2 x 0.1

= 0.2 cm

∴ The capacitance of large drop, C = R = 0.2 statF

The total charge of the small drops,

Q = 8 X 5

= 40μstatC

= 40 X 10^-6 statC

∴ Potential on the surface of the large drop,

⇒ \(V=\frac{Q}{C}=\frac{40 \times 10^{-6}}{0.2} \text { statV }\)

= 2 X 10^-4 x 300 V

= 0.06 V

Example 8. The ratio of the capacitances of two conductors A and B is 2: 3. The conductor A gains a certain amount of charge and shares it with B. Compare the initial energy of A with the total energy of A and B.
Solution:

Let the capacitance of the conductor A be 2C and that of the conductor B be 3C.

The amount of charge gained by A is Q.

Let the common potential of A and B after sharing of charge be V

∴ \(V=\frac{Q}{2 C+3 C}\)

= \(\frac{Q}{5 C}\)

The energy of the conductor A before sharing of charge,

⇒ \(E_A=\frac{1}{2} \cdot \frac{Q^2}{2 C}\)

= \(\frac{Q^2}{4 C}\)

Total energy of the conductors A and B after sharing the charge,

⇒ \(E_A=\frac{1}{2} \cdot \frac{Q^2}{2 C}\)

= \(\frac{Q^2}{4 C}\)

∴ The required ratio = \(\frac{E_A}{E}\)

= \(\frac{\frac{Q^2}{4 C}}{\frac{Q^2}{10 C}}\)

= \(\frac{5}{2}\)

Example 9. Each of the 27 identical mercury drops is charged to a potential of 10V. If the drops coalesce to form a big drop, what will be its potential? Calculate the ratio of the energy of the big drop to that of a small drop.
Solution:

Let the radius of each small drop be r.

∴ Capacitance, C₁ = 4π∈₀r

∴ Charge of each small drop, q = C₁V₁

= 10C₁

∴ Total charge, Q = 27q = 27 x 10C₁

= 270C₁

If R is the radius of the big drop, then according to the question,

⇒ \(\frac{4}{3} \pi R^3=27 \times \frac{4}{3} \pi r^3 \text { or, } R=3 r\)

∴ The capacitance of the big drop,

C₂ = 4π∈₀.3r = 3C₁

∴ The potential of the big drop,

⇒ \(V_2=\frac{Q}{C_2}=\frac{270 C_1}{3 C_1}=90 \mathrm{~V}\)

The energy of a small drop,

⇒ \(E_1=\frac{1}{2} C_1 V_1^2=\frac{1}{2} \times C_1 \times(10)^2\)

The energy of the big drop,

⇒ \(E_2=\frac{1}{2} C_2 V_2^2=\frac{1}{2} \times 3 C_1 \times(90)^2\)

∴ \(\frac{E_2}{E_1}=\frac{3 C_1}{C_1} \times \frac{(90)^2}{(10)^2}=\frac{243}{1}\)

∴ E₂: E₁ = 243: 1

Example 10. Charges of 10^-2 C and 5 x 10^-2.C are put on two metal spheres of radii 1 cm and 2 cm respectively. If they are connected with a metal wire, what will be the final charge on the smaller sphere?
Solution:

Here, the radius of the first sphere, R% = 1 cm = 0.01 m, and the radius of the second sphere, R₂ = 2 cm = 0.02 m.

∴ The capacitance of the first sphere, C₁ = 4π∈₀R₁

and capacitance of the second sphere, C₂ = 4π∈₀R₂

The total amount of charge before and after connection,

Q₁ + Q₂ = Q

= (C₁ + C₂) V [V = common potential]

∴ \(\dot{V}=\frac{Q_1+Q_2}{C_1+C_2}\)

The final charge on the smaller (first) sphere,

⇒ \(q_1=C_1 V=C_1 \frac{Q_1+Q_2}{C_1+C_2}\) [Given, Q₁ = 10^-2C and Q₂ = 5 x 10^-2C]

⇒ \(4 \pi \epsilon_0 R_1 \times \frac{10^{-2}+5 \times 10^{-2}}{4 \pi \epsilon_0 R_1+4 \pi \epsilon_0 R_2}\)

⇒ \(R_1 \times \frac{6 \times 10^{-2}}{R_1+R_2}\)

= \(0.01 \times \frac{6 \times 10^{-2}}{0.01+0.02}\)

= 0.02C

Example 11. The capacitance and potential, respectively, of conductor A are 10 units and 50 units; those of conductor B are, respectively, 5 units and 65 units. Find out the charges on the two conductors after they
Solution:

Initially, the charge on conductor A,

Q₂ = C₁V₁

= 10 x 50

= 500 unit

and charge on conductor B,

Q₂ = C₂V₂

= 5 x 65

= 325 unit

The common potential after the two conductors are connected,

⇒ \(V=\frac{Q_1+Q_2}{C_1+C_2}=\frac{500+325}{10+5}=\frac{825}{15}\)

= 55 unit

Now, charge on conductor A,

q₁ = C₁V

= 10 X 55

= 550 unit

and charge on conductor B,

q₂ = C₂V

= 5 X 55

= 275 unit

Example 12. A spherical liquid drop of capacitance I/<F breaks Into drops of the same radius. What Is the capacitance of each of these smaller drops?
Solution:

Let R = radius of the Initial drop; r = radius of each of 8 smaller drops.

∴ \(\frac{4}{3} \pi R^3=8 \times \frac{4}{3} \pi r^3 \text { or, } r=\frac{R}{2}\)

The capacitance of the bigger drop = 4π∈₀R

= 1μF

∴ The capacitance of each small drop

=4π∈₀r

⇒ \(4 \pi \epsilon_0 r=4 \pi \epsilon_0 \frac{R}{2}=\frac{1}{2} \cdot 4 \pi \epsilon_0 R=\frac{1}{2} \times 1 \mu \mathrm{F}\)

= 0.5μF

Example 13. Two isolated metallic solid spheres of radii R and 2R are charged in such a way that both of these have the same charge density. The spheres are placed far away from each other and are connected by a thin conducting wire. Find the new charge density on the bigger sphere
Solution:

Let or be the charge density of the two spheres.

So, charge of the first sphere = q₁ = 4πR₂σ and charge of the second sphere = q₂ = 4π(2R)²σ = 16πR²σ

When they are connected with a wire, let q₁ and q₂ be the new charges.

Then we may write

q’₁ +q’₂ = q₁ + q₂ ….(1)

Since the two spheres are at the same potential,

⇒ \(\frac{1}{4 \pi \epsilon_0} \frac{q_1^{\prime}}{R}=\frac{1}{4 \pi \epsilon_0} \frac{q_2^{\prime}}{2 R} \quad \text { or, } q_1^{\prime}=\frac{q_2^{\prime}}{2}\)

In the equation (1), by substituting q₁, we have

⇒ \(\frac{q_2^{\prime}}{2}+q_2^{\prime}=q_1+q_2\)

or, \(q_2^{\prime}=\frac{2}{3}\left(q_1+q_2\right) \text { or, } q_2^{\prime}=\frac{2}{3}\left(4 \pi R^2 \sigma+16 \pi R^2 \sigma\right)\)

∴ \(q_2^{\prime}=\frac{40 \pi R^2 \sigma}{3}\)

Therefore, the new charge density of the bigger sphere,

⇒ \(\sigma^{\prime}=\frac{q_2}{4 \pi(2 R)^2}=\frac{40 \pi R^2 \sigma}{3 \times 16 \times \pi R^2}=\frac{5}{6} \sigma\)

Capacitance And Capacitor Capacitor And Its Principle

Capacitor:

A capacitor (originally known as con denser) is an arrangement by which the capacitance of a conductor can be increased.

It is used for storage of charge. Hence, a capacitor can alternatively be defined as an arrangement that can store a certain amount of charge.

Usually, a counselor uses the Ilia principle of artificially looking at Ilia capacitance of mi Insulnlod clinical conductor by bringing another method of conductor noor It.

Construction of capacitor:

A capacitor Is basically an arrangement of an Insulated conductor and an earthed conductor held close to each other and separated by air or a non-conducting (dielectric) medium. The shape of the two conductors Is usually the same, e.g.,

In the case of a parallel plate capacitor, parallel metal plates are placed close to each other. Again, a spherical capacitor consists of two concentric spheres and a cylindrical capacitor of two co-axial cylinders.

Working principle of capacitor: An insulated metal plate A is connected to an electrical machine, Suppose, the potential of the plate is +V when It Is fully charged, ff C be the capacitance of the plate, the charge on the plate will be,

Q = CV

Now if a similar plate B is placed in front of plate A, then due to induction, a negative charge is induced on the inner surface of B and a positive charge on its outer surface.

The induced negative charge, being nearer, lowers the positive potential of plate A. Thus, the capacitance of plate A increases a little (since C = \(\frac{Q}{V}\)). Hence plate A takes a slight additional charge from the electrical machine and raises its potential again to V.

Now if B is earthed, the positive charge on the far side of it moves to the earth and the influence of positive charges is absent, the potential of A falls further. So the capacitance of A increases further and consequently, it will now be able to receive a greater amount of charge from the machine.

So in this way, the capacitance of an insulated charged conductor can be increased with the help of another earthed conductor, placed in its vicinity.

Factors affecting the capacitance of a capacitor:

1. Overlapping nrcu of the plates: The Greater the surface area of the plates, the greater its capacitance. The capacitance decreases with the decrease of the overlapping area.

2. Distance between the two conducting plates: Capacitance increases with the decrease of this distance and vice versa.

3. Nature of the intervening medium between the two ducting plates: Instead of air, if the intervening space of the two plates is filled up with an insulator, e.g., paraffin, glass, paper, etc., the capacitance of a capacitor increases.

Uses of capacitors: Extensive uses of tire capacitors are found in different electrical circuits. In the case of different circuits in AC, the capacitor is almost an indispensable part. Capacitors are used in electronic instruments, radio, television, telephone, the flash circuit of a camera, etc.

Discussions:

1. Charge of a capacitor: Charge of a capacitor means die magnitude of charge on any one of its plates. If one plate possesses a charge + Q, then the other plate will contain a charge -Q. So total charge = +Q + (-Q) = 0. Here, Q is called the charge of a capacitor, it is not the total charge.

2. Ideal capacitor:

If a capacitor is connected to a source of high potential, it is charged to that potential. The capacitor is called an ideal one if it is not discharged automatically even if the source of potential is removed. It preserves its acquired charge without any leakage.

3. Maximum limit of the potential of a capacitor:

A capacitor cannot be charged to any high potential at will. If the value of the potential exceeds a certain maximum limit, the intervening medium loses its insulating properties. Consequently, electric discharge takes place between the capacitor and the intervening medium.

4. Any charged conductor is a capacitor:

Any charged conductor may be considered as a capacitor. The floor or the walls of the room act as the earthed conductor in this case.

5. Circuit symbol of capacitor:

Two parallel lines of the same size in an electrical circuit diagram, represent a capacitor Symbol of a variable capacitor.

Charging and discharging of a capacitor:

When a battery Is connected to a series resistor and capacitor, charges begin to accumulate on the capacitor.

This Is called the charging of a capacitor. After removal of the battery, the capacitor loses Its accumulated charge through the resistor gradually. Tills Are called discharging of a capacitor.

The two plates A and B of a capacitor are connected to a buttery of electromotive force E through a resistor. Electrons from the negative pole of (lie battery move to plate H.

Simultaneously, a How of electrons starts from plate A to the positive pole of the battery. This produces a charging current.

As negative charges on plate B and positive charges on plate A keep on accumulating, (ho potential difference between die two plates increases.

So the plates act as a cell and consequently, a tendency of electron flow Is established In the direction opposite to that of the initial electron flow. As a result, the die charging current decreases.

When the potential difference between the two plates A and II becomes equal to the end of the battery, the charging current ceases to flow.

Then it is said that the capacitor has become fully charged. So at the start, the charging current is maximum and afterward, It gradually decreases. When the capacitor is fully charged, the charging current becomes zero.

After removal of the battery from the circuit, l.e„ during discharging electrons from plate 2 flow to plate A and begin to neutralize the positive charge of plate A.

Thus again a current flows in the circuit. This is called discharging and its direction is opposite to that of the charging current- After a while all the electrons of plate B neutralize all the positive charges of plate A.

Then the discharging current becomes zero and the capacitor is said to be completely discharged. So at the start, discharging current is. maximum and It decreases gradually and becomes zero when the capacitor is completely discharged.

In fact, no capacitor is an ideal one. A fully charged capacitor loses its charge in the course of time even if the two plates of it are not connected by a conducting wire.

Finally, the tire capacitor becomes completely discharged. Of course, in this case, the discharging action continues for a long time.

Potential and Capacitance of a Capacitor:

Potential of a capacitor: The potential difference between the two conducting plates of a capacitor is called the potential of a capacitor. Generally, the potential of a capacitor means the potential attained by the insulated plate of the capacitor due to the charge given to it, the grounded plate of the capacitor being at zero potential.

The capacitance of a capacitor:

The capacitance of a capacitor means the capacitance of the insulated conducting plate of the capacitor. So it may be defined as the amount of charge that must be given to the insulated plate to raise its potential by unity. If a charge Q raises its potential by Vi its capacitance, C = \(\frac{Q}{V}\).

Therefore, the capacitance of a capacitor may be defined as the ratio of the magnitude of the charge on any one of the two plates to die difference of potential between them, i.e., the capacitance of a capacitor

= \(\frac{\text { charge on a conducting plate of the capacitor }}{\text { difference of potential between the two plates }}\)

The capacitance of a capacitor is assigned a value of 1 faradic 1 coulomb of charge is required to maintain a potential difference of 1 V between the two conductors or plates of the capacitor.

Here capacitance is always a positive quantity and it does not depend on the nature of charge and potential. The capacitance of a conductor and that of a capacitor are expressed in the same unit.

Usually, a capacitor is rated, bearing the mark of the magnitude of its capacitance and the maximum potential difference that can be applied safely between its two plates.

A capacitor rated 0.04μF 220V means that its capacitance is 0.04μF and the maximum potential difference to be applied between its two plates is 220 V. If it is used in a potential difference higher than 220 V, it may get damaged.

Capacitance And Capacitor Dielectrics

Now we discuss the materials that do not conduct electricity and can be inserted between the plates of a capacitor.

Substances that have no free electrons cannot conduct electricity. They are called insulators or dielectrics. When they are placed in an electric field, charges are induced on their surfaces.

Classification of Dielectrics:

Dielectrics are classified into two groups according to the position of charge within their molecules: O non-polar substance and 0 polar substance.

Non-polar substance: A substance in which the center of negative charges (electrons) coincides with that of positive charges (protons) in each of its molecules, is called a non-polar substance.

In the absence of an external electric field, these molecules do not possess any permanent electric dipole moment Thus they are called non-polar molecules.

In the presence of an external electric field, a relative displacement occurs between the centers of positive and negative charge distributions. Thus a non-polar molecule when subjected to an electric field, acquires an electric dipole moment These types of dipoles are called induced dipoles.

Polar substance: A substance in which the center of negative charges (electrons) does not coincide with that of positive charges (protons) in each of its molecules, is called a polar substance.

So, even in the absence of an external electric field, each of these molecules possesses a permanent electric dipole moment So They are called polar molecules.

Example: water (H₂O), ammonia (NH₃).

Generally, the dipole moments of different molecules of a polar substance are randomly directed. So the resultant dipole moment of a polar substance is zero.

But when subjected to an electric field, each molecule of a polar substance experiences a torque and tends to fall in line with the direction of field lines of the external electric field. As a result, the sample of the polar substance acquires a resultant electric dipole moment.

So polar and non-polar molecules behave in a similar manner when subjected to an external electric field.

Polarisation of a Dielectric:

A conductor In an external electric field: If a conductor is placed in an external electric field the free electrons of the conductor orient themselves in a direction opposite to the electric field. This transfer continues until finally, the induced electric field balances the external electric field. In that case, no further displacement of charges takes place i.e., an equilibrium has been reached. So the resultant electric field \(\vec{E}\) inside a conductor is zero.

A dielectric in an electric field:

If a dielectric is placed in an external electric field \(\left(\vec{E}_0\right)\), the dipoles align themselves along the lines of force. So an electric field \(\left(\vec{E}_p\right)\) is generated inside the dielectric whose direction is opposite to that of the applied external field \(\left(\vec{E}_0\right) \cdot \vec{E}_P\) is less than \(\left(\vec{E}_0\right)\).

As a dielectric has no free electrons, the external field \(\left(\vec{E}_0\right)\) is not completely balanced by the internal field \(\left(\vec{E}_p\right)\) set up inside the dielectric. So at any point inside a dielectric the resultant intensity \((\vec{E})\) is less than the external field intensity \(\left(\vec{E}_0\right) \cdot \vec{E}_P\) but it does not become zero as in a conductor.

The alignment of the molecules of a dielectric, which behave like electric dipoles under the influence of an external field, is known as electric polarization.

It is observed that one face of the dielectric acquires a net positive charge and the other, a negative. This is because the charges between the two dotted lines neutralize each other’s effect, thus leaving an unbalanced negative charge on the left face and a positive charge on the right face of the dielectric,

The random arrangement of the molecules of a dielectric has been shown in the absence of any external electric field. Alignment of the molecules along the field lines under the influence of the external field has been shown.

Shows that inside a dielectric, electric field intensity reduces due to electric polarization. The resultant intensity of the electric field inside the dielectric is given by

⇒ \(\vec{E}=\vec{E}_0-\vec{E}_p\)

The electric polarization is directly proportional to die resultant electric field In the dielectric

The capability of storing the charge of a capacitor, i.e., its capacitance can be increased by using a suitable dielectric substance between its two plates.

For Example, air, paraffin, glass, sulfur, mica, paper, etc. are the dielectric substances used as intervention medium in a parallel plate capacitor. The increase in the capacitance of a capacitor depends on a property of dielectric materials, termed as dielectric constant (k).

Definition: The dielectric constant of a material is the ratio of the capacitance of a capacitor filled with the given dielectric material to the capacitance of a similar capacitor without any medium.

So, dielectric constant,

k = \(\frac{capacitance of capacitor with the dielectric as the intervening medium}{capacitance of the same capacitor without anymedium}\)

The dielectric constant is also known as specific inductive capacity (SIC).

The capacitor without any medium between its two plates has only a vacuum between the plates.

By the statement that the dielectric constant of glass is 8.5, we mean that the capacitance of a capacitor will increase 8.5 times if glass is used as a dielectric instead of a vacuum. Naturally, the dielectric constant of a vacuum is 1. The dielectric constant of dry air is 1.000586(≈1).

Capacitance And Capacitor Capacitance Of Some Standard Capacitors

Parallel Plate Capacitor:

It consists of two similar metal plates held parallel to each other, separated by a certain distance. The space in between the two plates contains air or any dielectric, e.g., glass, mica, etc.

Consider two parallel plates A and B separated by a distance d. The area of each plate is a. Plate A is charged with a charge +Q while plate B is grounded. The dielectric constant of the medium between the plates is K.

Now the surface density of charge on plate A will be

⇒ \(\sigma=\frac{Q}{\alpha}\)

The inner face of plate B is charged to -Q due to induction. If the area of the plates is large compared to the die distance between them, the electric lines of force between the plates are straight and parallel except near their ends.

Consequently, the intensity of the electric field between the plates may be taken to be uniform. The slight deviation from this uniformity near the edges may be neglected.

The plates A and B can be considered to be infinite plates with respect to any point in between the two plates. So, the electric field at that point due to the positive charge on plate A is,

⇒ \(E_1=\frac{\sigma}{2 \kappa \epsilon_0} \text {, along } A B\)

The electric field at that internal point will be in the same direction as AB, also due to the negative charge on plate B. Its magnitude is

⇒ \(E_2=\frac{\sigma}{2 \kappa \epsilon_0}\)

Therefore, at all points between the plates A and B, the electric field is

⇒ \(E=E_1+E_2=\frac{\sigma}{2 \kappa \epsilon_0}+\frac{\sigma}{2 \kappa \epsilon_0}=\frac{\sigma}{\kappa \epsilon_0}\)

If V is the potential difference between the plates, then

V = work done to bring a unit positive charge from plate B to plate A

= force acting on the unit charge x distance

= intensity of the electric field x distance

⇒ \(E \cdot d=\frac{\sigma}{\kappa \epsilon_0} \cdot d\)

If C is the capacitance of the capacitor, then

⇒ \(C=\frac{Q}{V}=\frac{\sigma \alpha}{\frac{\sigma}{\kappa \epsilon_0^V} \times d}=\frac{\kappa \epsilon_0 \alpha}{d}\)…(3)

⇒ \(\text { [In CGS units, } C=\frac{K \alpha}{4 \pi d} \text { ] }\)

For air, K = 1

Hence, \(C=\frac{\epsilon_0 \alpha}{d}=\frac{8.854 \times 10^{-12} \times \alpha}{d}\)…(4)

Dependence of the capacitance of a parallel plate capacitor on various factors:

Area of the plates: The capacitance is directly proportional to the area of the plates, i.e., C ∝ a.

1. Distance between the plates: The capacitance of a parallel plate capacitor is inversely proportional to the distance between the plates, i.e., C oc \(\frac{1}{d}\)

2. The nature of the medium between the plates: The capacitance is directly proportional to the permittivity or dielectric constant of the medium between the plates, i.e., C ∝ K.

3. The relation (3) clearly indicates that the capacitance does not depend on the charge Q or the potential V of the capacitor. Only the shape and the intervening medium determine its capacitance.

4. If the common overlapping area between the two plates can be changed by using a hinge arrangement, then changes, and as a result, the capacitance C changes. This technique may be used to design a capacitor, of variable capacitance.

Special case: If n number of parallel plates are alternately connected to form a multi-plate capacitor, the capacitance will be,

⇒ \(C=\frac{(n-1) \kappa \epsilon_0 \alpha}{d}\)

where a = area of each plate,

d = distance between two consecutive plates

The capacitance of a parallel plate capacitor with compound diet dice:

A and B are two parallel plates of which A is charged and B is grounded. Let a be the area of each plate, d the distance between them, and +Q the charge on plate A.

So the surface density of charge on each plate is \(\sigma=\frac{Q}{\alpha}\)

The space between the plates is now filled with two media of permittivites e₁ and e₂. The thickness of the two layers are (d-1) and t respectively. The intensity of the electric field between the plates may be taken as uniform. If the dielectric constants of the two media are k₁ and k₂, the intensity of the electric field will be given by,

⇒ \(E_1=\frac{\sigma}{\kappa_1 \epsilon_0}\)

and \(E_2=\frac{\sigma}{\kappa_2 \epsilon_0} ; \epsilon_0\) = permittivity of air or vacuum

∴ The potential difference between the plates Is,

V = E₁(d-t) + E₂t

⇒ \(\frac{\sigma}{\kappa_1 \epsilon_0}(d-t)+\frac{\sigma}{\kappa_2 \epsilon_0} t=\frac{\sigma}{\epsilon_0}\left[\frac{d-t}{\kappa_1}+\frac{t}{\kappa_2}\right]\)

So, the capacitance of the capacitor,

⇒ \(\frac{Q}{V}\)

⇒ \(\frac{\sigma \alpha}{\frac{\sigma}{\epsilon_0}\left[\frac{d-t}{\kappa_1}+\frac{t}{\kappa_2}\right]}=\frac{\epsilon_0 \alpha}{\left(\frac{d-t}{\kappa_1}+\frac{t}{\kappa_2}\right)}\)

If k₁ = 1 (for air) and k₂ = k (say),

⇒ \(C=\frac{\epsilon_0 \alpha}{d-t+\frac{t}{\kappa}}=\frac{\epsilon_0 \alpha}{d-\left(t-\frac{t}{\kappa}\right)}\)…(6)

1. Since \(\left(t-\frac{t}{K}\right)\) is a positive quantity, equation (6) shows that the capacitance of a parallel plate capacitor increases with the insertion of any dielectric medium between the plates.

2. If a dielectric of thickness t is inserted in between the two plates of a parallel plate air capacitor, the capacitance of the capacitor becomes equal to that of a capacitor having separation between the plates reduced by \(\left(t-\frac{t}{K}\right)\), i.e., the separation between the two plates effectively decreases by the amount \(\left(t-\frac{t}{K}\right)\).

If we want to get the previous value of the capacitance, the distance between the plates is to be increased. If this increase is x, then

⇒ \(x=t-\frac{t}{K}=t\left(1-\frac{1}{K}\right)\)

3. If n number of dielectric slabs of dielectric constants k₁,k₂,…,K_n of thicknesses t₁, t₂…., t_n be inserted between the two parallel plates, the capacitance of the capacitor so formed is given by,

⇒ \(C=\frac{\epsilon_0 \alpha}{\frac{t_1}{\kappa_1}+\frac{t_2}{\kappa_2}+\cdots+\frac{t_n}{\kappa_n}}=\frac{\epsilon_0 \alpha}{\sum_1^n \frac{t}{\kappa}}\)

Energy Stored in a Charged Capacitor:

During the charging of an uncharged capacitor, electrons are removed from one plate and transferred to the other gradually. Initially, the charge of the capacitor is zero and so the potential difference between the plates is also zero. As soon as an electron is transferred from one plate to the other, an electric field builds up in the space between the capacitor plates.

This field opposes further transfer. Thus as the charge accumulates on the capacitor plates, increasingly larger amounts of work is to be done to transfer more electrons.

Hence the potential difference increases due to the accumulation of charges. The energy spent for doing that work remains stored as potential energy in the electric field between the two plates of the capacitor.

Calculation: Suppose, at any moment, the charge of a capacitor be q and the potential difference between the two plates is v.

The capacitance of the capacitor, C = \(\frac{q}{v}\)

Further, when some charge of amount dq is given to the capacitor, the work done against the repulsive force due to the existing charge on the capacitor plate,

⇒ \(d W=v d q=\frac{q}{C} d q\)

∴ To give Q the amount of charge, the total work done

⇒ \(W=\int d W\)

= \(\int_0^Q \frac{q}{C} \cdot d q\)

⇒ \(=\frac{1}{2} \cdot \frac{Q^2}{C}\)

= \(\frac{1}{2} C V^2\)

= \(\frac{1}{2} Q V\)

where V = final potential difference between the plates.

This work is stored as potential energy in the capacitor

Energy stored in a charged parallel plate capacitor:

Let us consider a charged parallel plate capacitor.

Here,

a = area of each plate,

d = separation between the plates,

K = dielectric constant of the material between the plates,

σ = surface density of charge on each plate

So, volume between the plates = αd and amount of charge on each plate (Q) = σa.

The capacitance of this parallel plate capacitor,

⇒ \(C=\frac{K \epsilon_0 \alpha}{d}\)

where, ∈₀ = permittivity of air or vacuum

Therefore, the energy stored in this charged capacitor,

⇒ \(U=\frac{1}{2} \frac{Q^2}{C}\)

= \(\frac{1}{2}\left(\sigma^2 \alpha^2\right) \frac{d}{\kappa \epsilon_0 \alpha}\)

= \(\frac{\sigma^2 \alpha d}{2 \kappa \epsilon_0}\)

The unit of U is joule (J). This energy is stored in the electric field between the plates of the capacitor.

[In the CGS system, the expression for U is obtained by replacing

⇒ \(\epsilon_0 \text { by } \frac{1}{4 \pi} . \text { So, } U=\frac{2 \pi \sigma^2 a d}{\kappa}\); its unit is erg.]

Energy stored per unit volume or energy density between the plates:

Energy stored per unit volume,

⇒ \(u=\frac{U}{\alpha d}=\frac{\sigma^2}{2 \kappa \epsilon_0}\)

This is called the energy density in the electric field of the
capacitor.

Now, the electric field is uniform, except at the ends, inside a parallel plate capacitor, provided the plate area is very large compared to the separation between the plates.

From Gauss’ theorem, we already know that the uniform electric field between the two plates of a parallel plate capacitor, neglecting end effects, is

⇒ \(E=\frac{\sigma}{\kappa \epsilon_0} ; \text { then } \sigma=\kappa \epsilon_0 E\)

So, the energy density between the two plates is,

⇒ \(u=\frac{\left(\kappa \epsilon_0 E\right)^2}{2 \kappa \epsilon_0}=\frac{1}{2} \kappa \epsilon_0 E^2\)…(1)

For vacuum or air, K = 1 . Then equation (1) becomes,

⇒ \(u=\frac{1}{2} \epsilon_0 E^2\)…(2)

[In the CGS system, the equation (1) and (2) become, due to the replacement of
\(\epsilon_0 \text { by } \frac{1}{4 \pi}, u=\frac{1}{8 \pi} \kappa E^2 \text { and } u=\frac{1}{8 \pi} E^2\)]

In SI, the unit of u is J.m^-3 [In the CGS system, it is erg.cm^-3 ]

Dimension of \(u=\frac{\text { dimension of energy }}{\text { dimension of volume }}\)

⇒ \(\frac{M L^2 T^{-2}}{L^3}\)

= \(M L^{-1} T^{-2}\)

The expressions (1) and (2) of energy density in an electric field have been derived by considering the special case of a parallel plate capacitor. However, it can be proved that these two expressions are quite general expressions – true not only for parallel plate capacitors but also for electric fields of any other type.

These expressions give the energy density, i.e. energy in a unit volume around any, point in an electric field of any type. The rigorous proofs of the expressions are beyond the scope of our present discussions.

Energy density around a point In an electric field.’ Suppose, a point charge q is placed at a point 0. Another point P in air, is at a distance r from 0. Then the electric field at P,

⇒ \(E=\frac{1}{4 \pi \epsilon_0} \frac{q}{r^2}\)

So, the energy density of the tire electric field around point P is,

⇒ \(u=\frac{1}{2} \epsilon_0 E^2=\frac{1}{2} \epsilon_0 \frac{q^2}{16 \pi^2 \epsilon_0^2 r^4}=\frac{q^2}{32 \pi^2 \epsilon_0 r^4}\)

Then, in a small volume v around the point P, the energy stored is,

⇒ \(U=u v=\frac{q^2 v}{32 \pi^2 \epsilon_0 r^4}\)

Capacitance And Capacitor Capacitance Of Some Standard Capacitors Numerical Examples

Example 1. The area of each plate of a parallel plate glass capacitor is 314 cm2. Its plates are separated distance 1cm. What will be the radius of a sphere having a capacitance equal to that of this capacitor? [k of glass = 8 ].
Solution:

The capacitance of the sphere

⇒ \(C=\frac{\kappa \alpha}{4 \pi d}=\frac{8 \times 314}{4 \times \pi \times 1} \approx 200 \mathrm{statF}\)

So, the radius of the sphere = 200 cm.

Example 2. The area of each plate of a parallel plate capacitor is 22 cm² and the plates are kept separated by a paraffin paper of thickness 1 mm. The specific inductive capacity (SIC) of paraffin paper is 2. What are the capacitance of the capacitor and the surface density of charge under a potential difference of 330 V?
Solution:

The capacitance of the capacitor

⇒ \(C=\frac{\kappa \alpha}{4 \pi d} \quad\left[\text { Here, } \kappa=2 ; \alpha=22 \mathrm{~cm}^2 ; d=0.1 \mathrm{~cm}\right]\)

⇒ \(\frac{2 \times 22}{4 \pi \times 0.1}=35 \mathrm{statF}\)

∴ \(Q=C V=35 \times \frac{330}{300} {statC}\left[∵ V=330 \mathrm{~V}=\frac{330}{300} \text { statV }\right]\)

∴ Surface density of charge,

⇒ \(\sigma=\frac{Q}{\alpha}=\frac{35 \times 330}{300 \times 22}\)

= 1.75 statC.cm^-2

Example 3. The distance between the two plates of a parallel plate air capacitor is d. A piece of metal of thickness \(\frac{d}{2}\) and of area equal to that of the plates is inserted between the plates. Compare the capacitances in the two cases.
Solution:

In the first case, the capacitance of the parallel plate capacitor,

⇒ \(C_1=\frac{\epsilon_0 \alpha}{d}\)

We know that the intensity of the electric field inside a metal is zero. So, if a piece of metal of thickness \(\frac{d}{2}\) is inserted between the plates now becomes \(\left(d-\frac{d}{2}\right)=\frac{d}{2}\). Therefore, the capacitance of the capacitor becomes

⇒ \(C_2=\frac{\epsilon_0 \alpha}{d / 2}=\frac{2 \epsilon_0 \alpha}{d}\)

∴ \(\frac{C_1}{C_2}=\frac{\left(\epsilon_0 \alpha\right) / d}{\left(2 \epsilon_0 \alpha\right) / d}\)

= \(\frac{1}{2}\)

Example 4. The conducting plates of a parallel plate capacitor are separated by 2 cm from each other. A dielectric slab (K = 5) of thickness 1 cm is inserted between the two plates. The distance between the plates is now so changed that the capacitance of the capacitor remains the same. What will be the new distance between the plates?
Solution:

Let the present distance between the plates be d. According to the question,

⇒ \(\frac{\epsilon_0 \alpha}{2}=\frac{\epsilon_0 \alpha}{\left[d-t+\frac{t}{\kappa}\right]}\) [t = thickness of the dielectric slab = 1 cm; K = 5 ]

or, \(\frac{1}{2}=\frac{1}{\left[d-1+\frac{1}{5}\right]}\)

or, \(2=\left[d-1+\frac{1}{5}\right]=d-\frac{4}{5}\)

or, d = \(\frac{14}{5}\)

= 2.8 cm

Example 5. The surface area of each plate of a parallel plate capacitor is 50 cm². They are separated by 2mm in air. It is connected with a 100V power supply. Now a dielectric (K = 5) is inserted between its two plates. What will happen

1. If the voltage source remains connected and
2. If the voltage sources are absent during this insertion?

Solution:

We know, that the capacitance of a parallel plate capacitor is

⇒ \(C=\frac{k \epsilon_0 \alpha}{d}\)

For air, the capacitance,

⇒ \(C_1=\frac{\epsilon_0 \alpha}{d}\) [∈₀ = 8.854 x 10^-12 F.m^-1, a = 50cm² = 5 X 10^-2m², d = 2 mm = 2 X 10^-3 m]

⇒ \(\frac{8.854 \times 10^{-12} \times 5 \times 10^{-3}}{2 \times 10^{-3}}\)

= 2.21 x 10^-11 F

For the dielectric (k = 5), the capacitance,

⇒ \(C_2=\frac{\kappa \epsilon_0 \alpha}{d}\)

= 5 x 2.21 x 10^-11

= 1.11 x 10^-10 F

1. If a dielectric is inserted, the capacitance of a parallel plate capacitor increases. Since the capacitor is still connected to the power supply, its potential will remain constant.

When the intervening medium is air, charge on the capacitor,

Q₀ = capacitance x potential

= 2.21 X 10^-11 X 100

= 2.21 X 10^-9 C

When the intervening medium is the dielectric (k = 5), charge on the capacitor,

Q = 1.11 x 10^-10 x 100

= 1.11 x 10^-8C

∴ Change in charge of the capacitor

= Q – Q₀

= 1.11 x 10^-8 – 2.21 x 10^-9

= 8.89 x 10^-9 C

Change in potential difference = 0.

2. If the battery is removed, the charge stored remains the same.

So change in charge of the capacitor = 0.

According to the question, the potential difference between the plates of the capacitor before the insertion of the dielectric = 100 V.

After the insertion of the dielectric, the potential difference between the plates is,

⇒ \(V=\frac{Q_0}{C_2}=\frac{2.21 \times 10^{-9}}{1.11 \times 10^{-10}}\)

= 19.91 V

So, the potential difference decreases.

Change in potential difference

= 100 – 19.91

= 80.09 V

Example 6. Each of the two square plates of a capacitor has sides of length l. The angle d between the two plates is very small, If the medium between the plates is air and the minimum distance between them is t, determine the capacitance of the capacitor

Solution:

The average distance between the plates,

⇒ \(d=\frac{t+(t+l \sin \theta)}{2}=t+\frac{l}{2} \theta\) [∵ θ is very small, sinθ ≈ 0]

∴ The capacitance of the capacitor,

⇒ \(C=\frac{\epsilon_0 \alpha}{d}=\frac{\epsilon_0 l^2}{t+\frac{l}{2} \theta}=\frac{\epsilon_0 l^2}{t\left(1+\frac{l \theta}{2 t}\right)}=\frac{\epsilon_0 l^2}{t}\left(1+\frac{l \theta}{2 t}\right)^{-1}\)

⇒ \(\approx \frac{\epsilon_0 l^2}{t}\left(1-\frac{l \theta}{2 t}\right)\)

Example 7. A parallel plate air capacitor has a capacitance of 2pF. Now, the separation between the plates is doubled, and the space is filled with wax. If the capacitance rises to 6 pF, what is the dielectric constant of wax?
Solution:

Initial separation between the plates = d; area of each plate = α.

∴ Capacitance in the 1st case,

⇒ \(C_1=\frac{\epsilon_0 \alpha}{d}\)

The final separation between the plates = 2d; dielectric constant of wax = k.

∴ Capacitance in the 2nd case,

⇒ \(C_2=\frac{\kappa \epsilon_0 \alpha}{2 d}\)

∴ \(\frac{C_1}{C_2}=\frac{2}{\kappa} \text { or, } \kappa=2 \times \frac{C_2}{C_1}=2 \times \frac{6 \mathrm{pF}}{2 \mathrm{pF}}=6\)

Example 8. The area of each plate of a parallel plate capacitor is A – 600 cm² and their separation is d = 2.0 mm. The capacitor is connected to a 200 V DC source. Find out

1. The uniform electric field between the plates In the SI unit and
2. The surface density, of charge on any plate. Given, ∈₀ = 8.85 x 10^-12F m^-1

Solution:

Here, A = 600 cm²

= 600 x 10^-4m²

= 6 x 10^-2m²;

d = 2.0 mm

= 2 x 10^-3 m.

1. The uniform electric field between the plates,

⇒ \(E=\frac{V}{d}=\frac{200}{2 \times 10^{-3}}=10^5 \mathrm{~V} \cdot \mathrm{m}^{-1}\)

2. If cr = surface density of charge on any plate, then

⇒ \(E=\frac{\sigma}{\epsilon_0}\)

∴ \(\sigma=\epsilon_0 E=\left(8.85 \times 10^{-12}\right) \times 10^5\)

= 8.85 x 10^-7 C m^-2

Example 9. The potential of a capacitor increases from zero to 150 V when a charge of 10 esu Is imparted to it. What will be the energy stored in the capacitor?
Solution:

Energy stored within the capacitor

⇒ \(\frac{1}{2} Q V=\frac{1}{2} \times 10 \times \frac{150}{300}\)

= 2.5 erg.

Example 10. Find out the energy content in a volume of 1 cm³ around a point, situated in the electric field of a point charge of 10 C, at a distance of 2 m in air from the position of the point charge. Given,∈₀ = 8.85 x 10^-12 F.m^-1
Solution:

The electric field at the referred point due to the point charge,

⇒ \(E=\frac{1}{4 \pi \epsilon_0} \frac{q}{r^2}[\text { Here, } q=10 \mathrm{C}, r=2 \mathrm{~m}]\)

∴ Energy density at the point

⇒ \(u=\frac{1}{2} \epsilon_0 E^2=\frac{1}{2} \epsilon_0 \cdot \frac{1}{16 \pi^2 \epsilon_0^2} \frac{q^2}{r^4}=\frac{q^2}{32 \pi^2 \epsilon_0 r^4}\)

∴ The energy content in a volume of 1 cm³, i.e., 10^-6 m³,

U = u x 10^-6

⇒ \(\frac{q^2}{32 \pi^2 \epsilon_0 r^4} \times 10^{-6}\)

⇒ \(\frac{10^2 \times 10^{-6}}{32 \times(3.14)^2 \times\left(8.85 \times 10^{-12}\right) \times 2^4} \approx 2240 \mathrm{~J}\)

Example 11. Estimate the percentage change of the energy stored in a parallel plate capacitor, if the separation between its plates is reduced by 10%, keeping the voltage of the charging source unchanged.
Solution:

Let d = initial separation between the plates.

∴ Final separation,

d’ = d – (10% of d)

= d – \(\frac{d}{10}\)

= 0.9d

The initial and final values of the energy stored in the capacitor are,

⇒ \(U=\frac{1}{2} C V^2=\frac{1}{2} \frac{\epsilon_0 \alpha}{d} V^2 \text { and } U^{\prime}=\frac{1}{2} \cdot \frac{\epsilon_0 \alpha}{0.9 d} V^2\)

∴ Percentage increase in energy

⇒ \(\frac{U^{\prime}-U}{U} \times 100=\left(\frac{U^{\prime}}{U}-1\right) \times 100=\left(\frac{1}{0.9}-1\right) \times 100\)

⇒ \(\frac{0.1}{0.9} \times 100=\frac{100}{9}\)

= 11.1 %

Example 12. A 900 pF capacitor is charged to 100 V by a battery. How much energy is stored in the capacitor?
Solution:

Here, C = 900 pF

= 900 x 10^-12F

=9 x 10^-10F;

Energy stored within the capacitor,

E = \(\frac{1}{2}\)CV²

= \(\frac{1}{2}\) x (9 X 1010) X (100)²

= 4.5 x 10^-6J

Example 13. The capacitance of a parallel plate air capacitor is C. The capacitor is immersed halfway into an oil of dielectric constant 1.6 with the plates perpendicular to the surface of the oil. What will be the capacitance of this capacitor?
Solution:

The capacitance of the half of the capacitor immersed in oil,

⇒ \(C_1=\frac{\kappa \epsilon_0}{d} \cdot \frac{\alpha}{2}=\frac{1.6 \epsilon_0}{d} \cdot \frac{\alpha}{2}\)

The capacitance of the other half of the capacitor in air,

⇒ \(C_2=\frac{\epsilon_0}{d} \cdot \frac{\alpha}{2}\)

Net capacitance = \(C_1+C_2=\frac{1.6 \epsilon_0}{d} \cdot \frac{\alpha}{2}+\frac{\epsilon_0}{d} \cdot \frac{\alpha}{2}=\frac{2.6 \epsilon_0}{d} \cdot \frac{\alpha}{2}\)

⇒ \(\frac{1.3 \epsilon_0}{d} \cdot \alpha=1.3 C\left[∵C=\frac{\epsilon_0 \alpha^2}{d}\right]\)

Capacitance And Capacitor Combination Of Capacitors

Series combination: in this type of combination of capacitors, the first plate of the first capacitor is joined to the electric source, its second plate is joined to the first plate of the second capacitor, the second plate of the second capacitor is joined to tire first plate of tire third capacitor and so on. The second plate of the last capacitor is grounded, the rest of the system being kept insulated.

Calculation of equivalent capacitance: Let three capacitors of capacitances C₁, C₂, and C₃ be connected in series. Now a charge +Q be given from a source to the first plate A of the first capacitor, this will induce a charge -Q on the other plate B of this capacitor and a charge +Q on the first plate C of the second capacitor, and so on. All the capacitors will have the same charge Q. The final free positive charge from the last plate of the system moves to the earth.

If V₁, V₂, and V₃ are the potential differences across the capacitors C₃, C₂, and C₃ and V is the potential difference between the first plate A and the last plate F of the combination, then

V= V₁+ V₂ + V₃ …(1)

For the first capacitor, \(V_1=\frac{Q}{C_1}\)

For the second capacitor, \(V_2=\frac{Q}{C_2}\)

For the third capacitor, \(V_3=\frac{Q}{C_3}\)

∴ From equation (1) we have,

⇒ \(V=\frac{Q}{C_1}+\frac{Q}{C_2}+\frac{Q}{C_3}\)

or, \(V=Q\left(\frac{1}{C_1}+\frac{1}{C_2}+\frac{1}{C_3}\right)\)…..(2)

Now suppose that instead of this combination, a single capacitor is used such that the same charge (Q) given to this new capacitor produces the same potential difference ( V) between its two plates. This single capacitor is known as the equivalent capacitor of the combination and its capacitance is known as the equivalent capacitance

If C is the equivalent capacitance of the series combination of the capacitors C₁, C₂, and C₃, then

V = \(\frac{Q}{C}\)…(3)

From equations (2) and (3) we get,

⇒ \(\frac{Q}{C}=Q\left(\frac{1}{C_1}+\frac{1}{C_2}+\frac{1}{C_3}\right)\)

or, \(\frac{1}{C}=\frac{1}{C_1}+\frac{1}{C_2}+\frac{1}{C_3}\)…(4)

For n number of capacitors connected in series, the equivalent capacitance C is given by,

⇒ \(\frac{1}{C}=\frac{1}{C_1}+\frac{1}{C_2}+\cdots+\frac{1}{C_n}\)…(5)

Thus if several capacitors are connected in series, the reciprocal of the capacitance of the equivalent capacitor is equal to the sum of the reciprocals of the capacitances of the individual capacitors.

Now, \(\frac{1}{C}=\frac{1}{C_1}+\frac{1}{C_2}+\frac{1}{C_3}+\cdots+\frac{1}{C_n}\)

∴ \(\frac{1}{C}>\frac{1}{C_1}, \frac{1}{C_2}, \frac{1}{C_3}, \cdots, \frac{1}{C_n}\)

or, \(C<C_1, C_2, C_3, \cdots, C_n\)

Clearly, for series combinations, the equivalent capacitance is always less than any individual capacitance in the series.

Parallel combination:

In this type, of combination the first plates, i.e., the insulated plates of all the capacitors are connected to a common point A, and the second plates, i.e., the grounded plates to another common point B. Point A is connected to an electric source and point B is earthed

Calculation of equivalent capacitance: Three capacitors of capacitances C₁, C₂, and C₃ connected in parallel. The insulated plates of the three capacitors are connected to an electric source having potential V and other plates are earthed. So the potential difference between the two plates of each capacitor is V. A charge +Q drawn from the supply divides into Q₁, Q₂, and Q₃ according to the capacity of the different capacitors. So,

Q = Q₁ + Q₂+ Q₃….(6)

For the first capacitor,

Q₁ = C₁V

For the second capacitor,

Q₂ = C₂V

For the third capacitor,

Q₃ = C₃V

∴ From the equation (6) we get,

Q = C₁V + C₂V + C₃V

or, Q = V(C₁ + C₂ + C₃)….(7)

If the capacitors connected in parallel are replaced by a single capacitor so that the same potential difference V is produced if charge +Q is given to its insulated plate, the single capacitor is known as the equivalent capacitor of the combination and its capacitance is known as the equivalent capacitance. If C is the equivalent capacitance of the parallel combination of the capacitors C₁, C₂, and C₃, then

Q = CV…..(8)

From equations (7) anil1(8) we have

CV = V(C₁ + C₂ + C₃)

or, C = C₁ + C₂ + C₃….(9)

For n number of capacitors connected in parallel, the equivalent capacitance C is

C = C₁ + C₂ + …. + C_n ,

Thus the equivalent capacitance of the capacitors joined in parallel is equal to the sum of their individual capacitances.

Clearly, the equivalent capacitance of a number of capacitors in parallel is greater than any individual capacitance in the combination.

Capacitors are connected in parallel when a large capacitance for a small potential is required.

Unit 1 Electrostatics Chapter 4 Capacitance And Capacitor Combination Of Capacitors Numerical Examples

Example 1. H A condenser Is composed of 21 circular plates placed one after the other. The diameter of each plate is 10 cm. The consecutive plates are separated by 0.2 mm thick mica sheets of dielectric constant 6. If the alternate circular plates are connected, calculate the capacitance of the condenser μF.
Solution:

The condenser is composed of 21 circular plates and alternate plates are connected. So here we get 20 identical capacitors connected in parallel whose capacitance is

⇒ \(C=\frac{20 \kappa \epsilon_0 \alpha}{d}\)

⇒ \(\frac{20 \times 6 \times 8.854 \times 10^{-12} \times \pi \times 2.5 \times 10^{-3}}{2 \times 10^{-4}}\)

= 4.17 x 10^-8

F = 0.0417μF

[Here, k = 6, a = n x (5)² cm² = n x 2.5 x 10^-3 m², d = 2 x 10^-4m, ∈₀ = 8.854 x 10^-12 C².N^-1.m² ]

Example 2. A condenser is composed of 200 circular tin plates placed one after the other. The consecutive plates are separated by 0.5 mm thick mica sheets of dielectric constant 6. If the alternate tin plates are connected and the capacitance of the entire condenser is 0.4μF, what is the radius of each tin plate?
Solution:

The condenser is composed of 200 circular plates and alternate plates are connected. So, here we get 199 identical capacitors connected in parallel. Now capacitance of each capacitor is,

⇒ \(C=\frac{\kappa \alpha}{4 \pi d}=\frac{\kappa \pi r^2}{4 \pi d}=\frac{\kappa r^2}{4 d}\)

∴ \(\frac{\kappa r^2}{4 d} \times 199=0.4 \times 10^{-6} \times 9 \times 10^{11}\) [∵ 0.4μF =0.4 x 10^-6F and IF = 9 x 10¹¹ statF]

or, \(\frac{6 r^2}{4 \times 0.05} \times 199=3.6 \times 10^5\)

or, \(r^2=\frac{3.6 \times 10^5 \times 4 \times 0.05}{6 \times 199}\)

or, r^-2 = 60.3

or, r = 7.76

So, the radius of each tin plate = 7.76 cm.

Example 3. The equivalent capacitances of the parallel and the series combinations of two capacitors are 5μF and 1.2μF, respectively. Calculate the capacitances of each capacitor
Solution:

Let the capacitances of the two capacitors be C₁μF and C₂μF. According to the question,

C₁ + C₂ = 5…(1)

and \(\frac{C_1 C_2}{C_1+C_2}=1.2\)

or, C₁C₂ = 1.2 x 5

= 6 ….(2)

or, C₁(5-C₁)-6 = 0 [with the help of equation (1)]

or, \(C_1^2-5 C_1+6=0\)

or, (C₁-3)(C₁-2) = 0

∴ C₁ = 3 or, 2

If C₁ = 3;

C₂ = 5-3

= 2

and if C₁ = 2;

C₂ = 5-2

= 3

So, the capacitances of the two capacitors are 3μF and 2μF.

Example 4. Two capacitors of capacitances 20μF and 60μF are connected in series. If the potential difference between the two ends of the combination is 40 V, calculate the terminal potential differences of each capacitor
Solution:

If C is the equivalent capacitance of the combination,
then

⇒ \(C=\frac{20 \times 60}{20+60}=15 \mu \mathrm{F}=15 \times 10^{-6} \mathrm{~F}\)

∴ The total charge of the combination,

Q = CV

= 15 x 10^-6 x 40

= 6 x 10^-4C

Since the two capacitors are connected in series, the charge on each capacitor is equal to the total charge of the combination, i.e., 6 x 10^-4C.

∴ The potential difference between the two plates of the capacitor having capacitance C₁ is,

⇒ \(V_1=\frac{Q}{C_1}=\frac{6 \times 10^{-4}}{20 \times 10^{-6}}\)

= 30 V

Again, the potential difference between the two plates of the capacitor having capacitance C₂ is,

⇒ \(V_2=\frac{Q}{C_2}=\frac{6 \times 10^{-4}}{60 \times 10^{-6}}\)

= 10V

Example 5. A charged condenser is made to share its charge with an uncharged condenser of twice its capacitance. Find the sum of the energy of the two condensers.
Solution:

Let the capacitance of the charged condenser be C and let its charge be Q.

Before sharing of charge, the energy of the charged condenser is,

⇒ \(E_1=\frac{Q^2}{2 C}\)

The capacitance of the other condenser = 2C; since the charged condenser is made to share its charge with the second condenser, it is clear that the condensers are connected in parallel with each other.

∴ Equivalent capacitance of them

= C + 2C

= 3C

∴ Energy of the combination = \(=\frac{1}{2} \cdot \frac{Q^2}{3 C}=\frac{1}{3} \cdot \frac{1}{2} \frac{Q^2}{C}=\frac{1}{3} E_1\)

Example 6. A spherical drop of water carries a charge of 10 X 10^-12 C and has a potential of 100 V at its surface.

1. What is the radius of the drop?
2. If eight such charged drops combine to form a single drop, what will be the potential at the surface of the new drop?

Solution:

1. Charge of a spherical water drop,

Q = 10 x 10^-12C

= 10 x 10^-12 x 3 x 10⁹ esu of charge

= 3 x 10^-2 esu of charge

Potential of the drop, \(V=\frac{100}{300}=\frac{1}{3} \text { stat } \mathrm{V}\)

The capacitance of the drop,

⇒ \(C=\frac{Q}{V}=\frac{3 \times 10^{-2}}{\frac{1}{3}}\)

= 0.09 statF

In CGS units, the radius of a spherical conductor = its capacitance.

The radius of the drop, r = 0.09 cm.

2. If R is the i&dius of the large drop, then

⇒ \(\frac{4}{3} \pi R^3=8 \times \frac{4}{3} \pi r^3\)

or, R = 2r

= 2x 0.09

= 0.18 cm

Total charge, Q = 8 x 3 x 10^-2

= 24 x 10^-2 statC

∴ Potential at the surface of the new drop,

⇒ \(V=\frac{Q}{C}=\frac{Q}{R}=\frac{24 \times 10^{-2}}{0.18}=\frac{4}{3}\)

= 1.33 statV

= 1.33 X 300 V

= 400 V

Example 7. Three plates of the same size form a system of capacitors. Each plate has an area a. The intermediate differences between the plates are d₁ and d₂, respectively. The space between the first two plates is occupied by a dielectric of constant and that between the second and third plates by a dielectric of constant K₂. Calculate the capacitance of the system.
Solution:

The three plates form two capacitors connected in series.

The capacitance of the first capacitor,

⇒ \(C_1=\frac{\kappa_1 \epsilon_0 \alpha}{d_1}\)

The capacitance of the second capacitor

⇒ \(C_2=\frac{\kappa_2 \epsilon_0 \alpha}{d_2}\)

If C is the equivalent capacitance of the whole system, then

⇒ \(\frac{1}{C}=\frac{1}{C_1}+\frac{1}{C_2}=\frac{d_1}{\kappa_1 \epsilon_0 \alpha}+\frac{d_2}{\kappa_2 \epsilon_0 \alpha}=\frac{1}{\epsilon_0 \alpha}\left(\frac{d_1}{\kappa_1}+\frac{d_2}{\kappa_2}\right)\)

∴ \(C=\epsilon_0 \alpha \cdot \frac{1}{\frac{d_1}{\kappa_1}+\frac{d_2}{\kappa_2}}\)

Example 8. The capacitance of a parallel plate air capacitor is 9 pF. The separation between the plates is d. The intermediate space is filled up by two dielectric media. The widths of them are,\(\frac{d}{3} \text { and } \frac{2 d}{3}\), and their dielectric constants are 3 and 6, respectively. Find the capacitance of the parallel plate capacitor.
Solution:

As the capacitors are connected in series the equivalent capacitance,

⇒ \(C_{\mathrm{eq}}=\frac{C_1 C_2}{C_1+C_2}\)

We know that,

⇒ \(C_1=\kappa_1 \frac{\epsilon_0 A}{\frac{d}{3}} \text { and } C_2=\kappa_2 \frac{\epsilon_0 A}{\frac{2 d}{3}}\) [where A = area of each plate]

Putting the values of C₁ and C₂ in equation (1),

⇒ \(C_{\mathrm{eq}}=\frac{\frac{\epsilon_0 A}{\frac{d}{3}} \times \frac{\epsilon_0 A}{\frac{2 d}{3}} \times \kappa_1 \kappa_2}{\frac{\epsilon_0 A}{\frac{d}{3}}\left(\kappa_1+\frac{\kappa_2}{2}\right)}\)

∵ \(\frac{\epsilon_0 A}{d}=9 \mathrm{pF} \text { and } \kappa_1=3, \kappa_2=6\)

So, \(C_{\text {eq }}=\frac{3 \times 9 \times 3 \times \frac{9}{2} \times 3 \times 6}{3 \times 9\left(3+\frac{6}{2}\right)}\)

= 40.5 pF

Example 9. Three capacitors having capacitances 1μF, 2μF, and 3μF are Joined in series. A potential difference of 1100 V is applied to the combination. Find the charge and potential difference across each capacitor.
Solution:

If C is the equivalent capacitance of the combination, then,

⇒ \(\frac{1}{C}=\frac{1}{1}+\frac{1}{2}+\frac{1}{3}=\frac{11}{6} \text { or, } C=\frac{6}{11} \mu \mathrm{F}=\frac{6}{11} \times 10^{-6} \mathrm{~F}\)

∴ The total charge of the combination,

Q = CV

= \(\frac{6}{11}\) x 10^-6 x 1100

= 6 x 10^-4 c

Since the capacitors are connected in series, the charge on each capacitor is equal to the total charge, i.e., 6 x 10^-4 C. Potential difference across the plates of the first capacitor,

⇒ \(V_1=\frac{Q}{C_1}=\frac{6 \times 10^{-4}}{1 \times 10^{-6}}=600 \mathrm{~V}\)

Similarly, for the other two capacitors, respectively,

⇒ \(V_2=\frac{Q}{C_2}=\frac{6 \times 10^{-4}}{2 \times 10^{-6}}=300 \mathrm{~V}\)

and, \(V_3=\frac{Q}{C_3}=\frac{6 \times 10^{-4}}{3 \times 10^{-6}}=200 \mathrm{~V}\)

Example 10. Two capacitors having capacitances 0.1μV and 0.01μF are joined in series. A potential difference of 22 V is applied to the combination. If the capacitors are now joined in parallel, what will be the change in stored energy?
Solution:

If the capacitors are joined in series, their equivalent capacitance,

⇒ \(C_s=\frac{C_1 \times C_2}{C_1+C_2}=\frac{0.1 \times 0.01}{0.1+0.01}=\frac{1}{110} \mu \mathrm{F}=\frac{10^{-6}}{110} \mathrm{~F}\)

∴ Energy stored in the combination,

⇒ \(E_1=\frac{1}{2} C_s V^2=\frac{1}{2} \times \frac{10^{-6}}{110} \times(22)^2 \mathrm{~J}\)

= 22 erg

If the capacitors are joined in parallel, their equivalent capacitance,

Cp = C₁ + C₂

= 0.1 + 0.01

= 0.11μF

= 0,11 x 10^-6F

∴ Energy stored in this combination,

⇒ \(E_2=\frac{1}{2} C_p V^2=\frac{1}{2} \times 0.11 \times 10^{-6} \times(22)^2 \mathrm{~J}\)

= 266.2 erg

The change in stored energy = E₂-E₁

= 266.2- 22

= 244.2 erg

∴ Stored energy increases by 244.2 erg.

Example 11. Five capacitors have been arranged in a circuit. The capacitance of each capacitor is C. Determine the effective capacitance between points A and B.

Solution:

The equivalent circuit is like that,

From symmetries see that P and Q are equipotential points. So no current will pass through C₃.

So, the effective capacitance between A and B \(=\frac{C}{2}+\frac{C}{2}=C\)

Example 12. Show that the equivalent capacitance of an infinite circuit formed by the repetition of a similar loop made of two similar capacitors, each of capacitance C = (√5 + l) μF, is 2μF.

Solution:

Let the capacitance of the infinite circuit be C’.

So the circuit on the right-hand side starting from the second loop is also infinite. Therefore, its capacitance will also be C’.

In this case, the equivalent circuit will be that

∴ \(\frac{1}{C^{\prime}}=\frac{1}{C}+\frac{1}{C^{\prime}+C}\)

or, \(\frac{1}{C^{\prime}}-\frac{1}{C^{\prime}+C}=\frac{1}{C}\)

or, \(\frac{C^{\prime}+C-C^{\prime}}{C^{\prime}\left(C^{\prime}+C\right)}=\frac{1}{C}\)

or, \(C^{\prime 2}+C^{\prime} C-C^2=0\)

or, \(C^{\prime}=\frac{-C \pm \sqrt{C^2+4 C^2}}{2}=\frac{(-1 \pm \sqrt{5}) C}{2}\)

Neglecting the negative value of C’ we have

⇒ \(C^{\prime}=\frac{(\sqrt{5}-1) C}{2}=\frac{(\sqrt{5}-1)}{2} \cdot(\sqrt{5}+1)\)

= 2μF

Example 13. Twelve capacitors, each of capacitance 10μF, are inserted at the sides of a cube made of conducting wires. Determine the equivalent capacitance between A and B.

Solution:

Let a charge Q be given to the circuit from a source. The distribution of charge in the different capacitors has been shown

Let C’ be the equivalent capacitance between the points A and B.

Capacitance of each capacitor = C = 10μF

∴ \(V_{\text {A}}-V_B=\frac{Q}{C^{\prime}}\)….(1)

Again considering the path AMNB we have,

⇒ \(V_A-V_B=\frac{Q}{\frac{3}{C}}+\frac{Q}{C}+\frac{Q}{C}=\frac{Q}{3 C}+\frac{Q}{6 C}+\frac{Q}{3 C}=\frac{5}{6} \cdot \frac{Q}{C}\)…(2)

∴ From equations (1) and (2) we have,

⇒ \(\frac{Q}{C^{\prime}}=\frac{5}{6} \cdot \frac{Q}{C}\)

or, \(C^{\prime}=\frac{6 C}{5}=\frac{6}{5} \times 10\)

= 12μF

Example 14. Two identical parallel plate air capacitors are connected to a battery. At first, the key S is closed and then It is opened. The spaces between the two capacitors are now filled up with a dielectric trie, having dielectric constant 3. Determine the ratio of the energy stored in the two capacitors before and after insertion of the dielectric.

Solution:

When the key was closed, the potential of each capacitor was equal.

Let this value ofpotential be V.

Energy of the capacitor A = \(\frac{1}{2}\) CV²

and energy of the capacitor B = \(\frac{1}{2}\) CV²

∴ Total energy = \(\frac{1}{2} C V^2+\frac{1}{2} C V^2=C V^2\)

When the key is opened and the intervening space is filled up with the dielectric, the capacitance of each capacitor becomes 3C. Since the capacitor A is still connected to the battery, its potential remains at V.

∴ The energy of the capacitor A

⇒ \(\frac{1}{2} \times 3 C \times V^2\)

= \(\frac{3}{2} C V^2\)

The total charge of the capacitor B, Q = CV

This charge of B remains unaltered even after opening the key and inserting the dielectric.

Energy of \(B=\frac{1}{2} \cdot \frac{Q^2}{3 C}\)

= \(\frac{1}{2} \frac{C^2 V^2}{3 C}\)

= \(\frac{C V^2}{6}\)

∴ Energy of the two capacitors \(\frac{3}{2} C V^2+\frac{1}{6} C V^2\)

= \(\frac{5}{3} C V^2\)

∴ The required ratio = \(\frac{C V^2}{\frac{5}{3} C V^2}\)

= \(\frac{3}{5}\)

Example 15. Determine the capacitance A of the capacitor C when the equivalent capacitance between A and B is 1μF. The unit of all the capacitances is μF.

Solution:

The two capacitors of capacitance 6μF and 12μF are connected in series. If C₁ is their equivalent capacitance, then

⇒ \(C_1=\frac{6 \times 12}{6+12}=4 \mu \mathrm{F}\)

This C₁ and the capacitor of capacitance 4μF are connected in parallel. If C₂ is their equivalent capacitance, then

C₂ = 4 + 4

= 8μF

Again C₂ and the capacitor of capacitance 1μF are connected in series. If C₃ be their equivalent capacitance, then

⇒ \(C_3=\frac{8 \times 1}{8+1}=\frac{8}{9} \mu \mathrm{F}\)

On the other side, the capacitors of capacitance 2μF each arein parallel. If C₄ is their equivalent capacitance, then

C₄ = 2 + 2

= 4μF

Again C4 and the capacitor of capacitance 8mmF are in series.If C5 be their equivalent capacitance, then

⇒ \(C_5=\frac{4 \times 8}{4+8}=\frac{8}{3} \mu \mathrm{F}^{\prime}\)

Now C₃ and C₅ are in parallel combination. Their equivalent capacitance is given by,

⇒ \(C_6=\frac{8}{9}+\frac{8}{3}=\frac{32}{9} \mu \mathrm{F}\)

This C6 and C are in series and their equivalent capacitance = 1μF.

∴ \(\frac{\frac{32}{9} \times C}{\frac{32}{9}+C}=1\)

or, \(\frac{32}{9} \times C=\frac{32}{9}+C\)

or, \(\frac{23}{9} C=\frac{32}{9}\)

or, \(C=\frac{32}{23}=1.39 \mu \mathrm{F}\)

Example 16. Three capacitors A, B, and C are connected in such a way that their equivalent capacitance Is equal to the capacitance of B. The capacitances of A and B are 10μF and 30μF respectively and C ≠ 0. Determine three possible values of C and also show how the capacitors are to be connected in the three cases.
Solution:

The capacitor B cannot be joined in parallel to any combination of the capacitors A and C because in that case, the equivalent capacitance will be greater than the capacitance of B. The following three combinations are possible.

1. A and B are connected in series. Now C is connected in parallel with the series combination of A and B.

According to the question,

⇒ \(\frac{10 \times 30}{10+30}+C=30\)

or, \(\frac{300}{40}+C=30\)

or, C = 22.5μF

2. B and C are connected in series. Now A is connected in parallel with the series combination of B and C.

According to the question,

⇒ \(\frac{30 \times C}{30+C}+10=30\)

or, \(\frac{30 \times C}{30+C}=20\)

or, C = 60μF

3. A and B are connected in parallel. Now C is connected in series with the parallel combination of A and B.

According to the question,

⇒ \(\frac{40 \times C}{40+C}=30\)

or, 3(40 + C) = 4C

or, C = 120μF

Example 17. Two parallel plate capacitors of capacitances C and 2C are connected parallel and charged to a potential difference V. The battery then disconnected and the region between the plates of the capacitor C is completely filled with a material of dielectric constant k. What is the potential difference across the capacitors now?
Solution:

If charges Q₁ and Q₂ are given to the capacitors, then

Q₁ = CV; Q₂ = 2CV

∴ Total charge = Q₁ + Q₂

= CV+ 2CV

= 3CV

If the region between the plates of capacitor C is filled with a material of dielectric constant K, its capacitance will be kC.

∴ Total capacitance = kC+2C = (k + 2)C

∴ Potential difference across the capacitors = \(\frac{3 C V}{(\kappa+2) C}=\frac{3 V}{\kappa+2}\)

Example 18. A potential difference of 20 V is applied across a parallel combination of three Identical capacitors. If the total charge in the combination is 30 C, determine the capacitance of each capacitor. What will be the charge of the series combination of these three capacitors with the same potential difference?
Solution:

Let the capacitance of each capacitor be C.

The potential difference across each capacitor when they are connected in parallel combination, V = 20 V.

∴ Charge on the three capacitors,

Q₁ = Q₂ = Q₃ = CV

∴ Q₁ + Q₂ + Q₃ = 3CV

or, 30 = 3 X C X 20

= 60C

or, C = \(\frac{1}{2}\)

= 0.5F

∴ The capacitance of each capacitor = 0.5 F

In the second case when the capacitors are connected in series, let us suppose that the equivalent capacitance is C’

∴ \(\frac{1}{C^{\prime}}=\frac{1}{C}+\frac{1}{C}+\frac{1}{C}=\frac{3}{C}\)

or, \(C^{\prime}=\frac{C}{3}=\frac{0.5}{3}=\frac{1}{6} \mathrm{~F}\)

In this case, the total charge of the combination,

⇒ \(Q_i^{\prime}=C^{\prime} V\)

= \(\frac{1}{6} \times 20\)

= \(\frac{10}{3}\)

= 3.33 C

Example 19. The capacitance of a parallel plate air capacitor is C. Now, half the areas of its plates are vertically dipped in oil of dielectric constant 1.6. What would be Its capacitance?
Solution:

The upper half would still be an air capacitor. As the plate areas are halved, its capacitance, C₁ = \(\frac{C}{2}\). The lower half, immersed in oil, would have a capacitance, C₂ = \(\frac{kC}{2}\). C₁ and C₂ would form a parallel combination; the equivalent capacitance,

\(C^{\prime}=C_1+C_2=\frac{C}{2}+\frac{\kappa C}{2}=\frac{C}{2}(1+\kappa)\)

= \(\frac{C}{2}\)(1 + 1.6)

= 1.3C

Example 20. Each of the plates of a parallel plate capacitor is a circular disc of radius 5 cm. Find out its capacitance if the separation between the plates is 1 mm.
Solution:

The radius of each plate, r = 5 cm = 0.05 m.

Separation between the plates, d = 1 mm = 0.001 m.

∴ Capacitance, \(C=\frac{\epsilon_0 \alpha}{d}=\frac{\epsilon_0 \pi r^2}{d}\)

= \(\frac{\left(8.85 \times 10^{-12}\right) \times 3.14 \times(0.05)^2}{0.001}\)

= 6.95 X 100^-11 F

= 69.5 pF

Example 21. A 2μF capacitor is charged to a potential of 20 V. Another 3μF uncharged capacitor is connected in parallel with the first capacitor. What would be the minimum potential difference of the combination? Find out the charges on the two capacitors
Solution:

Initially, the charge on the first capacitor,

Q₁ = C₁V₁

= (2 x 10^-6) x 20

= 40 x 10^-6 C

and charge on the second capacitor, as it was uncharged, Q₂ = 0.

∴ Net charge, Q = Q₁ + Q₂

= 40 x 10^-6 C

Equivalent capacitance of the parallel combination,

C = C₁ + C₂

= (2 + 3)μF

= 5 X 10^-6 F

∴ Terminal potential difference of the combination,

⇒ \(V=\frac{Q}{C}=\frac{40 \times 10^{-6}}{5 \times 10^6}=8 \mathrm{~V}\)

Charge on the first capacitor,

Q₁ = C₁V

= (2 x 10^-6)x 8

= 16 x 10^-6 C

= 16μC

and charge on the second capacitor,

Q₂ = C₂V

= (3 X 10^-6) X 8

= 24 X 10^-6 C

= 24pC

Example 22. A 20μF capacitor is charged to a potential of 20 V and is then connected to an uncharged 10μF capacitor. Find out the common potential and the ratio of the energies stored in the two capacitors.
Solution:

Here, the capacitance of the first capacitor,

C₁ = 20μF

= 20 x 10^-6 F

The capacitance of the second capacitor,

C₂ = 10μF

= 10 x 10^-6 F

Initially, the charge on the first capacitor,

Q₁ = C₁V₁

= (20 X 10^-6) X 20

= 400 X 10^-6 C

and the charge on the second capacitor, as it was uncharged, Q₂ = 0.

∴ Net charge, Q = Q₁ + Q₂

= 400 x 10^-6 C

A parallel combination of the capacitors attains a common potential, say V.

Equivalent capacitance,

C = C₁ + C₂

= (20 + 10) x 10^-6 F

= 30 x 10^-6 F

⇒ \(V=\frac{Q}{C}=\frac{400 \times 10^{-6}}{30 \times 10^{-6}}=\frac{40}{3}\)

= 13.33V

The ratio of the energies stored in the two capacitors is,

⇒ \(\frac{E_1}{E_2}=\frac{\frac{1}{2} C_1 V^2}{\frac{1}{2} C_2 V^2}=\frac{C_1}{C_2}=\frac{20}{10}=\frac{2}{1}\)

Example 23. How should three capacitors of capacitances 3μF, 3μF and 6μF, be connected to get an equivalent capacitance of 5μF?
Solution:

For a parallel connection of all three capacitors, the 1 equivalent capacitance = 3 + 3 + 6

= 12μF; it is greater than 5μF.

Again, for a series connection, if the equivalent capacitance is C, then

⇒ \(\frac{1}{C_s}=\frac{1}{3}+\frac{1}{3}+\frac{1}{6}=\frac{5}{6}\)

or, \(C_s=\frac{6}{5}=1.2 \mu \mathrm{F}\) ; it is less than 5μF.

So, the above two connections are not applicable. Then, the following four arrangements are possible.

Now, for arrangement (1), the equivalent capacitance is clearly greater than 6μF; so it cannot be 5μF.

For the arrangement (2), equivalent capacitance

= \(\frac{(6+3) \times 3}{(6+3)+3}=2.25 \mu \mathrm{F}\)

For the arrangement (3), equivalent capacitance

= \(\frac{(3+3) \times 6}{(3+3)+6}=3 \mu \mathrm{F}\)

For the arrangement (4), equivalent capacitance

= \(3+\frac{3 \times 6}{3+6}=5 \mu \mathrm{F}\)

So, the arrangement (4) represents the desired connection.

Example 24. The capacitance of a parallel plate capacitor with plate area A and separation s is C. The space between the plates is filled with two wedges of dielectric constants k₁ and k₂ Find the capacitance of the resultant capacitor.

Solution:

Let the length of the capacitor plates be l and width b.

Area of each plate, A = lb

Now consider a small element of the capacitor of width dx at a distance x from the end. Its area, dA = bdx

The capacitance of this small element

⇒ \(d C=\frac{\epsilon_0 d A}{\frac{y}{\kappa_1}+\frac{s-y}{\kappa_2}}\)

According,

⇒ \(\frac{y}{x}=\frac{s}{l}=\tan \theta\)

∴ \(d C=\frac{\epsilon_0 b d x}{\frac{x \tan \theta}{\kappa_1}+\frac{s-x \tan \theta}{\kappa_2}}\)

∴ \(C=\int_0^l d C=\int_0^l \frac{\epsilon_0 b(d x) \kappa_1 \kappa_2}{\kappa_2 x \tan \theta+\kappa_1(s-x \tan \theta)}\)

Putting tanθ = \(\frac{s}{l}\)

⇒ \(C=\int_0^l \frac{\epsilon_0 A \kappa_1 \kappa_2 d x}{s \kappa_2 x+(l-x) s \kappa_1}=\frac{\epsilon_0 A \kappa_1 \kappa_2}{s\left(\kappa_2-\kappa_1\right)} \log _e \frac{\kappa_2}{\kappa_1}\)

Example 25. Consider a parallel plate capacitor of plate separation d. Each plate has the length l and the width a. A dielectric slab of permittivity e and thickness d is partially inserted between the plates. The plates are kept at a constant potential difference V. If x is the length of the dielectric slab within the plates, determine the force exerted on the slab.

Solution:

As per the diagram, the system may be regarded as the parallel combination of two capacitances C₁ and C₂. The part containing the dielectric of permittivity e has capacitance q and the other part (through free space) has capacitance C₂.

Hence, \(C_1=\frac{\epsilon a x}{d}\)

and \(C_2=\frac{\epsilon_0 a(l-x)}{d}\)

∴ Equivalent capacitance is,

⇒ \(C=C_1+C_2\)

= \(\frac{a}{d}\left[\left(\epsilon-\epsilon_0\right) x+\epsilon_0 l\right]\)

Now electrostatic energy of the system is

⇒ \(U=\frac{1}{2} C V^2\)

= \(\frac{1}{2} \frac{a}{d}\left[\left(\epsilon-\epsilon_0\right) x+\epsilon_0 l\right] V^2\)

∴ V is constant, the force experienced by the slab is,

⇒ \(F_x=\frac{d U}{d x}\)

= \(\frac{a}{2 d}\left(\epsilon-\epsilon_0\right) V^2\)

Example 26. A parallel plate capacitor of capacitance C Is connected to a battery to charge to a potential V. Similarly, another capacitor of capacitance 2C is charged to a potential 2V. Now the batteries are removed, and the two capacitors are connected in parallel by joining the positive plate of one with the negative plate of the other. Find out the final energy of the system.
Solution:

Charge on the first capacitor of capacitance C and connected to the battery of voltage V is,

Q₁ = CV

Similarly charge on the second capacitor,

Q₂ = (2C) x (2 V)

= 4CV

Since the plates of opposite polarity are connected together, the common potential is,

⇒ \(V^{\prime}=\frac{Q_2-Q_1}{C_1+C_2}=\frac{4 C V-C V}{C+2 C}=V\)

Now equivalent capacitance,

C’ = C+2C

= 3C

So final energy of the configuration,

⇒ \(U_f=\frac{1}{2} C^{\prime}\left(V^{\prime}\right)^2\)

= \(\frac{1}{2} \times 3 C \times V^2\)

= \(\frac{3}{2} C V^2\)

Example 27.  In the circuit, the values of the capacitances of the four capacitors are C₁ = C, C₂ = 2C, C₃ = 3C and C₄ = 4C. Find out the ratio between the charges on C₂ and C₄

Solution:

Here C₁ = C, C₂ = 2C, C₃ = 3C and C₄ = 4C.

C₁, C_2, and C₃ are connected in series, so equivalent capacitance is,

⇒ \(\frac{1}{C_{123}}=\frac{1}{C}+\frac{1}{2 C}+\frac{1}{3 C}=\frac{11}{6 C}\)

∴ \(C_{123}=\frac{6 C}{11}\)

Let q₁, q₂, q₃ and q₄ be the charges of the capacitors C₁, C₂, C₃ and C₄ respectively.

As C₁, C₂, and C₃ are in series, the charge on them is the same.

∴ \(q_1=q_2=q_3=C_{123} \times V=\frac{6 C V}{11}\)

Again charge on capacitor C4 is,

Q₄ = C₄ x V

= 4CV

So, \(\frac{\text { charge on } C_2}{\text { charge on } C_4}=\frac{\frac{6 C V}{11}}{4 C V}\)

∴ \(\frac{q_2}{q_4}=\frac{6}{4 \times 11}=\frac{3}{22}\)

∴ q₂: q₄ = 3: 22

Example 28. Five capacitors, capacitance 10μF, form a network. The network is connected to a 100 V dc supply. Calculate the equivalent capacitance between A and B, and the charge accumulated in the network.

Solution:

The given network is a balanced bridge so the middle part (CD) does not play any role.

So equivalent capacitance of branch ACB is,

⇒ \(C_{A C B}=\frac{10 \times 10}{10+10}=5 \mu \mathrm{F}\) [∴ capacitors in branches AC and CB are in series]

Similarly equivalent capacitance of branch ADB is,

⇒ \(C_{A D B}=\frac{10 \times 10}{10+10}=5 \mu \mathrm{F}\)

∴ Equivalent capacitance for the whole network is,

Cwhole = (5 + 5)μF

= 10μF

Now charge accumulated in branch ACB is,

q_ACB = C_ACB x V

= 5 x 10-6 x 100C

and charge accumulated in branch ADB,

q_ADB = C_ADB x V

= 5 x 10^-6 x 100 C

∴ Total charge accumulated in the whole network,

q = q_ACB + q_ADB

= 10 x 10^-6 x 100 C

= 10^-3 C

Example 29. In the network of the capacitance of each capacitor is 1μF. Determine the equivalent capacitance between P and Q.

Solution:

Let C₁, C₂, C₃, C₄, …∞ be the capacitances of the capacitors contained in the first row, second row, third row, fourth row, and so on.

Since the capacitors are in series in each row, the equivalent capacitances for the rows are given by,

C₁ = 1μF

⇒ \(C_2=\frac{1 \times 1}{1+1}=\frac{1}{2} \mu \mathrm{F}\)

⇒ \(\frac{1}{C_3}=\frac{1}{1}+\frac{1}{1}+\frac{1}{1}+\frac{1}{1}=4\)

∴ \(C_3=\frac{1}{4} \mu \mathrm{F}\)

Similarly,

⇒ \(\frac{1}{C_4}=\frac{1}{1}+\frac{1}{1}+\frac{1}{1}+\frac{1}{1}+\frac{1}{1}+\frac{1}{1}+\frac{1}{1}+\frac{1}{1}=8\)

∴ \(C_4=\frac{1}{8} \mu \mathrm{F}\)

Hence, the equivalent capacitance of the infinite number of rows,

⇒ \(C_{\mathrm{eq}}=C_1+C_2+C_3+C_4+\cdots+\infty\) [∵ they are connected in parallel]

∴ \(C_{\mathrm{eq}}=1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\cdots \infty\)

∴ Sum of the infinite geometric progression is given by,

⇒ \(C_{\mathrm{eq}}=\frac{1}{1-\frac{1}{2}}\)

∴ \(C_{\mathrm{eq}}=2 \mu \mathrm{F}\)

This is equivalent capacitance.

Example 30. A 0.1 F capacitor is charged by a 10 V battery. After disconnecting the battery, this charged capacitor is connected with an uncharged capacitor. If the charge is equally shared between the two, then what will be the energy stored in the two capacitors? Compare this energy with the energy stored initially in the first capacitor.
Solution:

Let C₁ and C₂ be the capacitances of the capacitors and their voltages be V₁ and V₂ respectively.

Given C₁ = 0.1 F,

initially V₁ = 10 V

and V₂ = 0.

∴ The initial energy stored in the two capacitors is,

⇒ \(U_i=\frac{1}{2} C_1 V_1^2+\frac{1}{2} C_2 V_2^2\)

= \(\frac{1}{2}\) x 0.1 x (10)² + 0

= 5J

It Q is the initial charge on capacitor C₁, its initial energy is,

⇒ \(U_i=\frac{Q^2}{2 C_1}\)

When the two capacitors are connected together, the charge is distributed equally, so the charge on each capacitor is \(\frac{Q}{2}\). Since the potential difference (in a parallel connection) across the two capacitors is also the same, it follows that their capacitances are equal.

Thus, C₁ = C₂ = C (say)

Also, Q₁ = Q₂ = \(\frac{Q}{2}\)

∴ The final energy stored in the two capacitors is

⇒ \(U_f=\frac{Q_1^2}{2 C_1}+\frac{Q_2^2}{2 C_2}=\frac{\left(\frac{Q}{2}\right)^2}{2 C}+\frac{\left(\frac{Q}{2}\right)^2}{2 C}=\frac{Q^2}{4 C}\)

But, \(U_i=\frac{Q^2}{2 C}, \text { hence } U_f=\frac{U_i}{2}=2.5 \mathrm{~J}\)

Thus, \(\frac{U_f}{U_i}=\frac{2.5}{5}=\frac{1}{2}\)

Example 31. The equivalent capacitance of two capacitors connected in series and in parallel are C_s and C_p respectively. Determine the capacitance of each capacitor.
Solution:

Let C₁ and C₂ be the values of the two capacitances.

According to the question,

C₁ + C₂ = Cp…(1)

⇒ \(\frac{C_1 C_2}{C_1+C_2}=C_s\)…(2)

From equations (1) and (2),

\(\frac{C_1 C_2}{C_p}=C_s\)

∴ \(C_2=\frac{C_p C_s}{C_1}\)

Using this value in equation (1)

⇒ \(C_1+\frac{C_p C_s}{C_1}=C_p \quad\)

or, \(C_1^2-C_p C_1+C_p C_s=0\)

∴ \(C_1=\frac{C_p \pm \sqrt{C_p^2-4 C_p C_s}}{2}\)

Using the positive sign (+) for C₁, we have,

So, the two capacitances are,

⇒ \(\frac{1}{2}\left[C_p+\sqrt{C_p^2-4 C_p C_s}\right] \text { and } \frac{1}{2}\left[C_p-\sqrt{C_p^2-4 C_p C_s}\right]\)

Unit 1 Electrostatics Chapter 4 Capacitance And Capacitor Different Types Of Capacitors

Nowadays, different types of capacitors are used for practical purposes. Some of them are discussed below.

Mica capacitor or block capacitor:

This is actually a parallel combination of a few parallel plate capacitors having a fixed capacitance. It consists of a number of metal sheets with mica as a dielectric between them. The alternate sheets are connected to A while the others are to B. If a potential difference is applied between A and B, the capacitor becomes charged.

If α be the area of each metal plate, d is the distance between two plates, K is the dielectric constant of mica and n is the number of metal plates, then the capacitance of such a capacitor in SI is given by

⇒ \(C=\frac{(n-1) \kappa \epsilon_0 \alpha}{d}\)

This capacitor looks like a block. So it is often called a block capacitor. This type of capacitor is mainly used in wireless receiving sets.

Paper capacitor: Paper coated with paraffin is a suitable dielectric. In between two aluminum or tin foils, paraffin-waxed paper sheets are inserted. The foils are then rolled into a cylindrical shape for the economy of space. This type is essentially a single parallel plate capacitor. It is cheaper than a mica capacitor. Of course one of its disadvantages is that it cannot work at high potential difference.

In low-voltage ac circuits, having semiconductor diodes and transistors in particular, paper capacitors are widely used.

Variable air capacitor: It is also one type of parallel plate capacitor whose capacitance can be varied at will. It consists of two sets (F and M) of parallel plates made of aluminum or brass. The set is fixed while the other set M can be turned with the help of a knob K.

As the knob turns, the set M either moves into the spacings of the set F or comes out of them, thereby changing the area of overlap of tire plates. Since the capacitance of a capacitor depends on the area of overlap of the plates, the capacitance also changes.

It is used in radio sets or wireless receiving sets and other electronic instruments where variation of capacitance is required. Electrolytic capacitor: It consists of a pair of aluminum plates partly immersed into a solution of ammonium borate [NH₄B₅O₆].

The two aluminum plates are connected to the positive and negative terminals of a source of steady current. Due to electrolysis, a fine layer (10^-6 cm) of aluminum oxide [Al₂O₃] is formed on the positive plate.

This layer acts as a dielectric medium. The solution of ammonium borate acts as an extension of the other plate. Since the capacitance of the capacitor is inversely proportional to the thickness of the dielectric and the thickness of the aluminum oxide layer is very small, the capacitor has a large capacitance, often a few thousand microfarads.

The plate on which aluminum oxide is deposited must be connected to the positive potential. Otherwise, the capacitor would be damaged. For this reason, the anode plate is marked with a + sign or a red dot.

Further, the maximum voltage applicable is restricted by the thickness of the oxide layer. This is also marked on the capacitor. Higher voltage would break the layer and destroy the device. Electrolytic capacitors are largely used as filters in rectifier circuits.

Electrostatic Machines

Classification of Electrostatic Machines:

Electrostatic machines can generate large quantities of electric charges rapidly. These machines are also used to set up high-potential differences. Electrostatic machines are of two types

1. Frictional machine
2. Induction machine.

Frictional machines were not very effective. So after the invention of induction machines, frictional machines became obsolete. The principle of action of induction machines depends on the principle of electrostatic induction. In this chapter, we shall discuss a familiar electrostatic machine known as the Van de Graaff generator.

Van de Graaff Generator:

In 1931, Van de Graaff invented this machine. With the help of this machine, very high potential difference (up to a few million volts) can be produced. The principle of action of this machine is based on the discharging action of points and on the property of a collection of charges of a hollow conductor. This machine is very useful at atomic research centers. At present, many changes and modifications of this machine have been introduced.

Description: The sketch of a Van de Graaff generator. A and B are two hollow spherical conductors.

These are placed on two big insulating stands (X, X). P₁ and P₂ form two pairs of pulleys. The pulleys are situated at the centers of the spherical conductors and are connected by two electric motors.

Silk or rubber belt moves in the path shown by arrows along the body of each pair of pulleys. The belt enters the spherical conductor through the hole Sx and comes out through the hole S₂. C, D, F, and G are four pointed conductors. The pointed ends are directed towards the belt.

The positive charge is supplied to the small sphere (N) placed in front of the pointed conductor D and the negative charge to the small sphere (M) placed in front of the pointed conductor C with the help of a dc generator. The pointed conductors F and G are connected with the spheres A and B respectively.

Working principle:

A positive charge on the small sphere (N) in front of the pointed conductor D induces a negative charge on D and the induced positive charge, being free, moves to the earth. The pointed conductor D discharges the negative charge to the belt in front of it.

The belt, being a non-conductor, does not distribute the charge all over its body; it remains connected to one place. The belt carries the charge upwards and when the charge comes near the pointed conductor G, it induces a positive charge on G and a negative charge on sphere B. Very soon the positive charge on G gets neutralised by the negative charge on the belt. So sphere B is charged with negative electricity.

Similarly, due to the negative charge on the small sphere (M) in front of the pointed conductor C, sphere A is charged with positive electricity.

Discharge of electricity and its remedy:

Due to the continuous movement of the belt by the electric motor, a large quantity of charge accumulates on the two spheres A and B and the potential difference between them increases quickly.

Due to continuous increases in the potential difference between the two spheres, an electric discharge may start in the neighboring air, as air cannot bear high potential differences under normal pressure.

To avoid pointed-end discharge, the spheres and the belts are made very smooth. The whole instrument is installed in a large metallic case connected to the earth and air inside the case is pumped out with the help of an exhaust pump.

Next, the entire case is filled up with nitrogen or freon gas under high pressure. Because, even under high potential difference, the tendency of nitrogen or freon molecules to be ionized, is low.

Uses:

1. Production of high energy charged particles, in nuclear research.

2. Production of hard X-rays. In Science City, Kolkata there is a Van de Graaff generator for display

 

Unit 1 Electrostatics Chapter 4 Capacitance And Capacitor Synopsis

The capacitance of a conductor is equal to the charge necessary to increase its potential by unity.

Units of capacitance:

• If 1 esu of charge raises the potential of a conductor by 1 esu, the capacitance of the conductor is defined as 1 esu of capacitance or 1 stat farad (statF).
• If 1 coulomb of charge is required to raise the potential of a conductor by 1 volt, the capacitance of the conductor is defined as 1 farad.
• 1 C = 3 x 10⁹ esu of charge; 1 V = \(\frac{1}{300}\) esu ofpotential difference; so,
• IF = \(\frac{1C}{1V}\) = 9 x 10¹¹ esu of capacitance or statF
• An arrangement by which the capacitance of an insulated charged conductor placed in the vicinity of another conductor (usually earthed) is increased artificially, is called a capacitor.
• When a capacitor is connected to a battery, charges begin to accumulate on it. This is called the charging of a capacitor. After disconnection of the battery, if any conduction occurs between the two plates of the capacitor, it begins to lose charge. This is called the discharging of a capacitor.
• The potential difference between the two conducting plates of a capacitor is called the potential of a capacitor.
• The capacitance of a capacitor is equal to the charge given to the insulated conductor of the capacitor to raise its potential by unity.
• Dielectric is an insulator. Examples—are air, paraffin, glass, sulphur, mica, ebonite, etc.
• The dielectric constant of an insulator,
• k = \(\frac{capacitance of a capacitor with the dielectric ns the intervening medium}{capacitance of the same capacitor with vacuum ns the intervening medium}\)
• Effectively, the dielectric constant of air or vacuum is 1.
• To charge a capacitor, a certain amount of work is to be done. The energy spent for doing that work remains stored as potential energy in the electric field between the two plates of the capacitor.
• For a series combination, the equivalent capacitance is always less than any of the capacitances in the series.
• The equivalent capacitance of capacitors joined in parallel is greater than any of the capacitances in the combination.
• Electrostatic machines can generate large quantities of electric charge rapidly. These machines are also used to set up high-potential differences.
• Electrostatic machines arc of two types:
  • Frictional machine and
  • Induction machine.
• The working principle of an induction machine is based on electrostatic induction. Van de Graaff generator is an electrostatic machine of this type.
• With the help of a Van de Graaff generator, a very high potential difference (up to a few million volts) can be produced.
• The working principle of a Van de Graaff generator depends on the discharging action of points and on the property of collection of charge by hollow conductors.
• Van de Graaff generator is used
  • To produce high energy charged particles which are required for nuclear research
  • To produce hard X-rays.
• If charge Q given to a conductor raises its potential by V,then Q ∝ For, Q = CVwhere C is the capacitance of the conductor.
• The capacitance of a spherical conductor,
  C = 4π∈₀kR (in SI)
• [Here, R = radius of spherical conductor, K = dielectric constant of the surrounding medium]
• The capacitance (in CGS) of a spherical conductor air or vacuum is numerically equal to its radius in centimetres.
• The energy of a charged conductor,

⇒ \(U=\frac{1}{2} Q V\)

= \(\frac{1}{2} C V^2\)

= \(\frac{1}{2} \frac{Q^2}{C}\)

where Q = charge, V = potential.

• If two conducting spheres of radii rx and r2, connected by a conducting wire, are given a charge Q, then charges on the spheres are,

⇒ \(Q_1=Q \cdot \frac{r_1}{r_1+r_2} ; Q_2=Q \cdot \frac{r_2}{r_1+r_2}\)

• Two conductors having capacitances C₁ and C₂, are charged to potentials V₁ and V₂. If they are connected, the common potential becomes,

⇒ \(V=\frac{C_1 V_1+C_2 V_2}{C_1+C_2}\)

= \(\frac{Q_1+Q_2}{C_1+C_2}\)

Q₁ and Q₂ are their charges before connection. Loss of energy due to sharing of charge

⇒ \(\frac{1}{2} \frac{C_1 C_2}{C_1+C_2}\left(V_1-V_2\right)^2\)

• The capacitance of a parallel plate capacitor,

⇒ \(\left.C=\frac{\kappa \epsilon_0 \alpha}{d} \text { (in } \mathrm{SI}\right)\)

⇒ \(C=\frac{\kappa \alpha}{4 \pi d} \text { (in CGS system) }\)

where α = area of each plate; d = distance between the plates.

If a combination is formed by connecting alternately n number of equispaced parallel plates, the capacitance of such a block capacitor

⇒ \(C=\frac{(n-1) \kappa \epsilon_0 \alpha}{d}(\text { in SI) }\)

⇒ \(C=\frac{(n-1) k \alpha}{4 \pi d} \text { (in CGS system) }\)

The capacitance of a parallel plate capacitor with compound dielectric,

⇒ \(C=\frac{\epsilon_0 \alpha}{\frac{d_1}{\kappa_1}+\frac{d_2}{\kappa_2}} \text { (in SI) }\)

where k₁ and k₂ are the dielectric constants of the dielectrics of thickness d₁ and d₂, respectively.

Energy stored in a charged capacitor,

⇒ \(U=\frac{1}{2} \cdot \frac{Q^2}{C}=\frac{1}{2} C V^2=\frac{1}{2} Q V\)

[where Q = charge of the capacitor and V = potential difference between the capacitorplates]

The potential energy of a parallel plate capacitor

⇒ \(U=\frac{1}{2} \cdot \frac{\alpha \sigma^2 d}{\kappa \epsilon_0} \text { (in SI); }\)

⇒ \(U=\frac{2 \pi \sigma^2 \alpha d}{\kappa} \text { (in CGS system) }\)

Energy density between the capacitor plates,

⇒ \(u=\frac{1}{2} \kappa \epsilon_0 E^2\)

[where E = electric field between the two plates of the capacitor, k = dielectric constant of the medium between the plates.]

For vacuum or air, K = 1.

Then u = \(\frac{1}{2} \epsilon_0 E^2\)

• For a series combination of capacitors, the equivalent capacitance is given by the following equation:

⇒ \(\frac{1}{C}=\frac{1}{C_1}+\frac{1}{C_2}+\cdots+\frac{1}{C_n}=\sum_1^n \frac{1}{C_n}\) [n = number of capacitors in series]

• The equivalent capacitance of the capacitors joined in parallel,

⇒ \(C=C_1+C_2+\cdots+C_n=\sum_1^n C_n\) [n = number of capacitors in parallel]

• Charge in different quantities due to the Introduction of a dielectric slab of dielectric constant K between the plates of a parallel plate capacitor after the capacitor has been chanted with the help of a battery.

• In n number of similar electrically charged drops coalesce to form a larger drop,
  • Charge of the large drop =nx charge of each small drop
  • Capacitance of eh large drop = n^1/3 x capacitance of each small drop
  • The surface potential of the large drop = n^2/3 x surface potential of each small drop
  • The energy of the large drop = n^5/3 x energy of each small drop
  • Surface density of charge of the large drop = n^1/3 x surface density of each small drop
• A capacitor of capacitance C can be charged by connecting
  it is in series with a battery of emf E and a resistance R. The charge on the capacitor after a time t,

⇒ \(q=C E\left(1-e^{-\frac{t}{R C}}\right)=q_0\left(1-e^{-t / \tau}\right)\)

Maximum charge on the capacitor, q₀ = CE

The time constant of the charging circuit, \(\tau\) = RC

• In the circuit, when the key Ky is pressed, the battery of emf E is connected in series with the capacitor C and resistor R, and the capacitor starts charging. After some time, the key Ky is switched off and the key K2 is pressed, then the capacitor starts discharging

The charge on the capacitor after a time t,

⇒ \(q=C E e^{-\frac{t}{C R}}=q_0 e^{-t / \tau}\)

The time constant of the discharging circuit, \(\tau\) = RC

Unit 1 Electrostatics Chapter 4 Capacitance And Capacitor Very Short Questions and Answers

Question 1. What is the name of the physical quantity whose unit is coulomb volt-1?
Answer: Capacitance

Question 2. What is the radius of a conducting sphere of capacitance 10μF?
Answer: 9 cm

Question 3. Two copper spheres of the same radius, one of them being hollow and the other solid, are charged to the same potential. Which of them does contain a greater amount of charge?
Answer: Both contain the same amount of charge

Question 4. If a charged soap bubble expands, what will be the change in its potential?
Answer: The potential will decrease

Question 5. What is the unit of dielectric constant?
Answer: No unit

Question 6. The surface area of a conducting sphere is 10.18 cm². If it is placed in the air, what will be its capacitance in the picofarad?
Answer: 1

Question 7. n small drops of the same size are initially at the. same potential. They coalesce to form a big drop. What is the ratio of the capacitance of this big drop to that of any of these small drops?
Answer: n^1/3

Question 8. Two conductors of capacitances C₁ and C₂ are connected by a conducting wire. Some amount of charge is given to the system. How is the ratio of charges acquired by the conductors related to their capacitances?
Answer: C₁/C₂

Question 9. Two conductors, of capacitances C₁ and C₂, initially have charges q₁ and q₂ at potentials V₁ and V₂, respectively. They are now connected by a thin conducting wire. What quantity the net energy loss of the system would be proportional to?
Answer: (V₁-V₂)²

Question 10. Two spheres of radii r and 2r have charges 2q and q on them, respectively. If they are connected by a copper wire, what will be the amount of charge flowing through the wire?
Answer: q

Question 11. A capacitor is marked as 0.05μF 200V. What is the maximum charge it can accumulate without being damaged?
Answer: 10^-3C

Question 12. What is the dielectric constant of conducting materials?
Answer: Infinite

Question 13. What happens when the space between the two plates of a capacitor is filled with a conducting material?
Answer: It will be discharged completely

Question 14. What is the permittivity ofmica if its dielectric constant is 5.4?
Answer: 4.78 x 10^-11C²N^-1m^-2

Question 15. What is the unit of the permittivity of vacuum, ∈₀?
Answer: C².N^-1 m^-2

Question 16. Write down the dimensional formula of ∈₀
Answer: M^-1L^-3T⁴l²

Question 17. s the permittivity of any medium greater or less than that of vacuum?
Answer: Greater

Question 18. If a dielectric is placed in an electric field, what change in the intensity of the electric field takes place inside the dielectric?
Answer: Electric field intensity decreases

Question 19. The space between the two plates of a parallel plate air capacitor is filled with an insulator. What will be the nature of the change in its capacitance?
Answer: Increase

Question 20. The space between the two plates of an isolated charged parallel plate air capacitor is filled with an insulator. What will be the nature of the change of the charge accumulation?
Answer: No change

Question 21. In a parallel plate capacitor, the capacitance increases from 4μF to 80μF when a dielectric medium is introduced between the plates. What is the dielectric constant of the medium?
Answer: 20

Question 22. What will be the effect on the capacity of a parallel plate capacitor when the area of each plate is doubled and the distance between them is also doubled?
Answer: No change

Question 23. The distance between the plates of a parallel plate capacitor is d. A metal plate of thickness d/2 is placed between the plates. What will be its effect on the capacitance of the system?
Answer: Capacitance Is Doubled

Question 24. Two protons A and B are placed between two parallel plates having a potential difference V. Will these protons experience equal or unequal forces?
Answer: Equal forces

Question 25. How can a capacitance of 10μF be designed from a few supplied 2μF capacitors?
Answer: Five of them are to be connected in parallel

Question 26. How can two capacitors be connected so that the charges on them are equal?
Answer: In series

Question 27. How can two capacitors be connected so that the potential differences between their plates are equal?
Answer: In parallel

Question 28. A combination is formed by connecting alternately n number of equidistant parallel plates. The capacitance between any two consecutive plates is C. What will be the equivalent capacitance of the combination?
Answer: (n-1)C

Question 29. Two charged conductors, each of which is effectively a capacitor, are connected by a conducting wire. Which type of combination of capacitors is this—series or parallel?
Answer: Parallel

Question 30. Three capacitors of equal capacitance, when connected in series, have a net capacitance C₁. When connected parallel, they have a net capacitance of C₂. What is the value of C₁/C₂?
Answer: 1/9

Question 31. Two plates of an isolated charged capacitor are connected by a copper wire. What will happen to the energy stored in the capacitor?
Answer: It will be zero

Question 32. The separation between the two plates of an isolated charged parallel plate air capacitor is d. The capacitor stores an energy U. Now a metal plate of thickness d/2 and of area equal to that of the capacitor plates is introduced in the intermediate space. What will be the energy stored in the capacitor?
Answer: U/2

Question 33. The separation between the two plates of an isolated charged parallel plate air capacitor is d. The capacitor stores an energy U. Now an insulating plate of thickness d/2, of dielectric constant K, and of area equal to that of the capacitor plates is introduced in the intermediate space. What will be energy stored in the capacitor?
Answer: \(\frac{K+1}{2 K} U\)

Question 34. n capacitors, each of capacity C, are connected in parallel and to a source of V volt. What will be the energy storedin the arrangement?
Answer: \(\frac{1}{2} n C V^2\)

Question 35. n capacitors, each of capacity C, are connected in series and to a source of V volt. What will be the energy storedin the arrangement?
Answer: \(\frac{1}{2 n} C V^2\)

Unit 1 Electrostatics Chapter 4 Capacitance And Capacitor Fill In The Blanks

1. The radius of the earth is 6400 km. Its capacitance in microfarad is 711.1

2. The intensity of the electric field in a dielectric decrease due to electric polarisation

3. A metal plate of negligible thickness is introduced between the two plates of a parallel plate air capacitor. The capacitance will remain the same

4. The distance between the two plates of an isolated charged parallel plate air capacitor is increased. The potential difference between the plates will increase

5. The space between the plates of a capacitor is filled up with a liquid of specific inductive capacity K. The capacitance will change by a factor of k

6. V a de Graaff generator is used to produce hard X-ray

Unit 1 Electrostatics Chapter 4 Capacitance And Capacitor Assertion Reason Type

Direction: These questions have Statement 1 and Statement 2. Of the four choices given below, choose the one that best describes the two statements.

1. Statement 1 is true, statement 2 is true; Statement 2 is, a correct explanation for statement 1.
2. Statement 1 is true, Statement 2 is true; Statement 2 is not a correct explanation for Statement 1.
3. Statement 1 is true, statement is false.
4. Statement 1 is false, and Statement 2 is true.

Question 1.

1. Statement 1: Two capacitors of the same capacity are first connected in parallel and then in series. The ratio of equivalent capacitances in two cases is 2: 1.
2. Statement 2: The equivalent capacitance is less than any of the capacitances in series.

Answer: 4. Statement 1 is false, Statement 2 is true.

Question 2.

1. Statement 1: If the distance between the parallel plates of a capacitor is halved and the dielectric constant is made three times then the capacitance becomes 6 times.
2. Statement 2: The capacitance of the capacitor does not depend on the nature of the material of the plates of the capacitor.

Answer: 2. Statement 1 is true, Statement 2 is true; Statement 2 is not a correct explanation for Statement 1.

Question 3.

1. Statement 1: The larger the sphere, the larger is its capacity, and the smaller the sphere, the smaller is its capacity.
2. Statement 2: The capacitance of a spherical conductor is directly proportional to its radius.

Answer: 1. Statement I is true, statement II is true; Statement 2 is, a correct explanation for statement I.

Question 4.

1. Statement 1: Dielectric has no significance in a parallel plate capacitor.
2. Statement 2: A Dielectric is an insulator that can be easily polarised on the application of an electric field.

Answer: 4. Statement 1 is false, Statement 2 is true.

Question 5.

1. Statement 1: The force with which one plate of a parallel plate capacitor is attracted towards the other plate is equal to the square of surface charge density per e per unit area.
2. Statement 2: The total amount of charge that resides, on the unit surface area is known as surface charge density.

Answer: 4. Statement 1 is false, Statement 2 is true.

Question 6.

1. Statement 1: The capacity of a conductor, under given circumstances, remains constant irrespective of the charge present on it.
2. Statement 2: Capacity depends on the size and shape of the conductor and also on the surrounding medium.

Answer: 1. Statement 1 is true, statement 2 is true; Statement 2 is, a correct explanation for statement 1.

Question 7.

1. Statement 1: A dielectric slab is inserted between the ‘ plates of an isolated charged capacitor. The charge on the capacitor will remain the same.
2. Statement 2: Charge on an isolated system is conserved.

Answer: 1. Statement 1 is true, statement 2 is true; Statement 2 is, a correct explanation for statement I.

Question 8.

1. Statement 1: The potential energy of a capacitor is obtained at the cost of chemical energy from the battery used for charging the capacitor.
2. Statement 2: In a battery potential energy is converted to chemical energy.

Answer: 3. Statement 1 is true, the statement is false.

Unit 1 Electrostatics Chapter 4 Capacitance And Capacitor Match The Columns

Question 1. A capacitor of capacitance C is charged to a potential V. Now, it is connected to a battery of emf E. Based on this information match the entries of column 1 with entries of column 2 in the following table.

Answer: 1-B, 2-C, 3-A, 4-D

Question 2. Mathematical expressions of some physical quantities and their corresponding units are given in column 1 and column 2 respectively.

Answer: 1-B, 2-A,C, 3-D, 4-A,C

Question 3. 

Answer: 1-A, 2-A,B, 3-D, 4-C

Question 4. 

Answer: 1-B,C 2-A,D 3-D, 4-C

Question 5. In the area of each plate is A. Match the following.

Answer: 1-E, 2-D, 3-B, 4-A