WBCHSE Class 12 Physics Electromagnetism Question and Answers

Magnetic Effect Of Current And Magnetism

Electromagnetism Long Questions and Answers

Question 1. State whether the mutual distances between the circular magnetic lines of force obtained on a plane, perpendicular to a straight long current-carrying wire would be the same or not.
Answers:

According to the formula \(B=\frac{\mu_0}{4 \pi} \cdot \frac{2 I}{r}\), the greater the distance from the current carrying wire, “lesser the value of the magnetic field.

Again, according to the properties of magnetic lines of force, the greater the Strength of the magnetic field at a place, the greater the number density of magnetic lines of force, and hence their mutual distances decrease.

So, the mutual distance between the circular magnetic lilies of force gradually increases with the increase of the radius of the lines.

Question 2. A magnet and a charged particle are placed near each other. State whether a force will act on the charged particle if

1. Both the magnet and the charged particle are at rest,
2. Both travel with equal velocity,
3. The magnet is moving but the charged particle is at rest,
4. The magnet is at rest but the charged particle is in motion.

Answer:

If there is a relative motion between the charged particle and the magnet, then only a force acts on the charged particle.

1. If both the magnet and the charged particle are resting, no force acts on the charged particle because there is a relative motion between them.

2. In this case, no relative motion exists between them and hence no force acts on the charged particle

3. In both cases there is a relative motion between the magnet and the charged particle. So, a force action the charged particle. But there is a special case. If the magnetic field stated above is uniform and the direction of the relative velocity between the magnet and the charged particle is parallel to the field, then no force acts on the charged particle.

Question 3. Current flowing in a long, straight conductor passes through the axis of a circular coil carrying current. What will be the mutual force acting between them?
Answer:

In the case of the straight conductor, if we apply Maxwell’s corkscrew rule, we see that at each point on the circular coil, the direction of the magnetic field is parallel to the direction of the current. We know that no force acts when the direction of the current and that of the magnetic field are parallel to each other. Hence, the mutual force between them will be zero.

WBBSE Class 12 Electromagnetism Q&A

Question 4. An electron and a proton are revolving along circular paths of equal radii in equal magnetic fields. Compare their kinetic energies.
Answer:

The magnitude of charge (e) of an electron and a proton are equal. The force acting on a charge e traveling with a velocity v normal to a magnetic field of strength B is Bev. This force supplies the necessary centripetal force to revolve the charge along a circular path.

So, for the electron \(\frac{m v^2}{r}\) = Bev

or, mv = Ber

Similarly, for the proton, MV = Ber

∴ mv = MV (i.e., the momenta of both of them are equal)

∴ \(\frac{kinetic energy of the electron}{kinetic energy of the proton}\)

= \(\frac{\frac{1}{2} m v^2}{\frac{1}{2} M V^2}\)

= \(\frac{(m \nu)^2}{m} \cdot \frac{M}{(M V)^2}=\frac{M}{m}\)

Since the mass of a proton (M) is about 1836 times that of an electron (m), the kinetic energy of the electron will be about 1836 times that of the proton.

Question 5. Four wires of infinite lengths are placed on a plane. The same current I is flowing through each of the wires. Determine the resultant magnetic field at the center O of the square ABCD.
Answer:

Since point O is at the center of the square ABCD, it is equidistant from the conductor’s AB, BC, CD, and DA. Since, each of the conductors carries equal current I, the magnetic field produced by each of them at the point O will be the same. But the magnetic fields at point O due to the conductors AB and CD are mutually opposite and similarly for the conductors AD and BC, magnetic fields at point O are also mutually opposite. Hence, at the point O, the resultant magnetic field is zero.

Short Answer Questions on Faraday’s Law

Question 6. When a charged particle moves through a particular region it is not deflected. From this, can it be inferred that no magnetic field is present in that region?
Answer:

We know that, when a charge q passes through a magnetic field B with a velocity v, the force acting on that charged particle is, \(\vec{F}=q(\vec{v} \times \vec{B})\)

The cross product \((\vec{v} \times \vec{B})\) signifies that \(\vec{F}\) = 0 when \(\vec{\nu} \| \vec{B}\). So, if the charged particle moves along or opposite to the direction of the magnetic field, no force acts on it. Hence, when a charged particle moves without suffering any deflection, it cannot be inferred definitely that no magnetic field is present there

Question 7. A charged particle is released from rest In a region of steady and uniform electric and magnetic fields, which are parallel to each other. What will be the nature of the path followed by the charged particle?
Answer:

The force on a particle of charge q and mass m acting along the direction of the uniform electric field E = qE, and its acceleration = \(\frac{qE}{m}\) = constant. So, this particle has a uniformly accelerated motion along \(\vec{E}\).

Now, \(\vec{E}\) and \(\vec{B}\) are parallel; thus the particle velocity is also parallel to \(\vec{B}\). Then, the magnetic force on the particle = 0, i.e., the magnetic field has no effect on the motion of the particle.

Question 8. An electron is not deflected in passing through a certain region of space. Can we be sure that there is no magnetic field in that region?
Answer:

The magnetic force on the electron would be zero if its velocity \(\vec{v}\) is along the magnetic field B, because

⇒ \(\overrightarrow{F_m}=\overrightarrow{e v} \times \vec{B}\).

Then the electron would not be deflected even if a non-zero magnetic field is present.

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Question 9. Equal currents are flowing through two infinitely long conducting wires. State whether a magnetic field will exist at a point midway between the wires if they carry current.

1. In the same direction
2. In the opposite direction?

Answer:

1. When equal current flows through two parallel wires in the same direction, at a point midway between the wires, the magnetic field will be zero because at that point magnetic fields for the two current-carrying wires are equal but opposite in direction.

2. When the direction of current in the two wires are opposite, at that point, the magnetic field will be twice the field due to any one wire because in this case the magnetic field due to the two wires will be the same and in the same direction.

Question 10. How will the magnetic field intensity, at the center of a circular coil carrying current, change if the current through the coil is doubled and the radius of the coll is halved?
Answer:

We know, \(B=\frac{\mu_0 N I}{2 r}[/laex]

∴ [latex]\frac{B_1}{B_2}=\frac{l_1}{I_2} \cdot \frac{r_2}{r_1}=\frac{1}{2} \times \frac{1}{2}=\frac{1}{4} \quad \text { or }\)

Practice Questions on Induced EMF

Question 11. A rectangular loop carrying a current is placed near a long straight wire in such a way that the wire is parallel to one of the sides of the loop and in the plane of the loop. If a steady current I is passed through the wire, then the loop

1. Will rotate about an axis parallel to the wire
2. Will move away from the wire
3. Will move towards the wire
4. Will remain stationary.

Answer:

The arm of the rectangular loop nearer to the wire will be attracted by the wire because the directions of current in them are the same and parallel to each other.

In the farther arm of the loop, the direction of the current is parallel but opposite with respect to the wire and hence they will repel each other. But the like parallel current-carrying arm being nearer to the current-carrying wire, the attraction dominates over the repulsion. Hence, the loop shifts towards the wire. Therefore, alternative 3 is correct.

Question 12. A steady current is flowing through a long wire. If it is converted into a single-turn circular coil, the magnetic field produced at its center is B. Now it is converted into a circular coil having n turns. What will be the magnetic field at the center of the coil?
Answer:

Magnetic field at the center of the single turn circular coil, \(B=\frac{\mu_0 I}{2 r}\)

If a coil of n turns is made, the radius reduces to \(\frac{r}{n}\).

Thus, the magnetic field at the center of the coil

⇒ \(\frac{\mu_0 n I}{2 \cdot \frac{r}{n}}=n^2 \cdot \frac{\mu_0 I}{2 r}=n^2 B\)

Question 13. A rectangular loop made of a very thin and flexible wire is kept on a table. The two ends of the wire are connected with two joining screws and a high direct current Is allowed to pass through the wire. What will be the shape of the wire and why?
Answer:

The rectangular loop. The directions of current in the parts AB and CD are parallel and opposite in direction and hence these two parts repel each other. If the current is too high, the loop will take almost the shape of a circle; that shape is shown by the dotted line.

Question 14. In a region, a uniform electric field and uniform magnetic field are acting in the same direction. An electron is shot along the direction of the fields. What change will be observed In the magnitude and direction of the velocity of that electron?
Answer:

Since the direction of the magnetic field and that of the velocity of the electron is the same, the magnetic force will be zero. Since the charge of an electron is negative, Its velocity will decrease gradually in the direction of the electric field.

Question 15. Two wires of.equal length or bent in the form of two loops of one turn each. One of them Is square-shaped, whereas the other is circular. Both are suspended in a uniform magnetic field. When the same current is passed through them, which one will experience greater torque?
Answer:

The torque on a current loop of current I and area \(\vec{A}\), in a magnetic field \(\vec{B}, \text { is } \vec{\tau}=I \vec{A} \times \vec{B}\). For the two given loops, current I and magnetic field \(\vec{B}\) are both the same. Now, for equal lengths of the boundary, the area of a circle is greater than that of a square. So the torque would be greater for the circular current loop.

Question 16. A circular conducting loop of radius r carrying a currently placed in a magnetic field \(\vec{B}\) in such a way that the plane of the loop is perpendicular to \(\vec{B}\). What will be the magnitude of the magnetic force exerted on the loop?
Answer:

Let us consider an infinitesimal element \(\vec{dl}\) at point P of the Circular loop. The magnetic force on this element,

⇒ \(d \vec{F}=I d \vec{l} \times \vec{B}=I d l B \sin 90^{\circ}\) [∵ \(d \vec{l} \text { is perpendicular to } \vec{B}\)]

= IdlB (directed along \(\vec{OP}\))

Similarly, if we consider an infinitesimal element \(\vec{dl}\) at point Q located on diametrically opposite end of P, then the magnetic force on this element will be IdlB (directed along \(\vec{OQ}\)).

So the net force on two infinitesimal elements located on opposite ends of a diameter is zero.

In this manner, if we divide the whole loop into such pairs of infinitesimal elements, the net force on each pair will be zero. Hence the net force on the loop will be zero

Question 17. The ratios of the masses and charges of a proton and an alpha particle are respectively 1 s 4 and 1:2. They enter a uniform magnetic field of magnitude B normally with

1. Same velocity
2. Same momentum and
3. Same kinetic energy. What will be the ratio of the radii of the circular paths described by the particles in each case?

Answer:

Radius, of the circular path, \(r=\frac{m v}{a B} ; B=\text { constant }\)

1. If the velocity is the same then, \(r \propto \frac{m}{q}\)

⇒ \(\frac{r_1}{r_2}=\frac{m_1}{m_2} \times \frac{q_2}{q_1}=\frac{1}{4} \times \frac{2}{1}=\frac{1}{2}\)

2. If the momentum is the same then, \(r \propto \frac{1}{q}\)

⇒ \(\text { i.e., } \frac{r_1}{r_2}=\frac{q_2}{q_1}=\frac{2}{1}\)

3. \(r=\frac{m v}{q B}=\frac{\sqrt{\frac{1}{2} m v^2 \cdot 2 m}}{q B}\)

If kinetic energy \(\frac{1}{2}\) mv2 is the same then, \(r \propto \frac{\sqrt{m}}{q}\)

∴ \(\frac{r_1}{r_2}=\sqrt{\frac{m_1}{m_2}} \times \frac{q_2}{q_1}=\sqrt{\frac{1}{4}} \times \frac{2}{1}=\frac{1}{1}\)

Question 18. If current I passes through an infinitely long wire PQR bent at the right angle at Q, then the magnetic field at point M is H₁. Now another wire is joined along QS in such a manner that the currents along PQ, QR, and QS are 1, \(\frac{1}{2}\) and \(\frac{1}{2}\), respectively. Now, if the magnetic field at the point M is H₂, find the value of \(\frac{H_1}{H_2}\)

Answer:

Due to the current through the wire QR, no magnetic field exists at point M. With respect to point M, the two wires PQ and QS are semi-infinite in length.

So, if MQ = r,

⇒ \(H_1=\frac{\mu_0}{4 \pi} \cdot \frac{I}{r} \text { and } H_2=\frac{\mu_0}{4 \pi} \cdot \frac{I}{r}+\frac{\mu_0}{4 \pi} \cdot \frac{\frac{I}{2}}{r}=\frac{\mu_0}{4 \pi} \cdot \frac{3 I}{2 r}\)

⇒ \(\text { Hence, } \frac{H_1}{H_2}=\frac{2}{3}\)

Conceptual Questions on Magnetic Fields and Forces

Question 19. If current I pass through a square-shaped conducting loop of side a, what is the value of the magnetic field at the point of intersection of its two diagonals?
Answer:

Distance of the point of intersection of two diagonals (O) from each arm of the square is r = \(\frac{a}{2}\) For two extreme points of each side, the angle subtended at O, θ₁ = θ₂ = 45°.

So, the magnetic field at O due to current I through each arm

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

⇒ \(\frac{\mu_0}{4 \pi} \cdot \frac{I}{\frac{a}{2}}\left(\sin 45^{\circ}+\sin 45^{\circ}\right)\)

⇒ \(\frac{\mu_0}{4 \pi} \cdot \frac{2 I}{a} \cdot 2 \times \frac{1}{\sqrt{2}}\)

= \(\frac{\mu_0}{4 \pi} \frac{2 \sqrt{2} I}{a}\)

The direction of the magnetic field at point O due to current I in each arm is the same.

Hence, the resultant magnetic field at the point O

⇒ \(4 B=\frac{\mu_0}{4 \pi} \cdot \frac{8 \sqrt{2} I}{a}\)

Question 20. If the current through a conducting loop in the shape of an equilateral triangle of side a is I, what will be the magnitude of the magnetic field at the point of intersection of its three medians?
Answer:

Length of each median of the triangular loop

⇒ \(\sqrt{a^2-\left(\frac{a}{2}\right)^2}=\frac{\sqrt{3}}{2} a\)

The distance of each side from the point of intersection O of the medians

⇒ \(r=O A=\frac{1}{3} \cdot \frac{\sqrt{3}}{2} a=\frac{a}{2 \sqrt{3}}\)

The angle subtended by the two extremities of each side at O, θ₁ = θ₂ = 60°.

So, due to current 7 through each side, the magnetic field at point O,

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

⇒ \(\frac{\mu_0}{4 \pi} \cdot \frac{I}{\frac{a}{2 \sqrt{3}}}\left(\sin 60^{\circ}+\sin 60^{\circ}\right)\)

⇒ \(\frac{\mu_0}{4 \pi} \cdot \frac{2 \sqrt{3} I}{a} \cdot 2 \times \frac{\sqrt{3}}{2}=\frac{\mu_0}{4 \pi} \cdot \frac{6 I}{a}\)

For current I flowing through each side, the direction of the magnetic field at the point O is identical, and hence, the resultant magnetic field at the point \(O=3 B=\frac{\mu_0}{4 \pi} \cdot \frac{18 I}{a}\)

Question 21. Due to the flow of current I through a square-shaped conducting loop, the magnetic field generated at its center is B. The magnetic field generated at the center of a circular conducting loop having the same perimeter as that of the square and for the flow of the same current is B’. Determine the ratio of B to B’.
Answer:

The magnetic field at the center of the circular loop of radius r,

⇒ \(B^{\prime}=\frac{\mu_0 I}{2 r} ; \text { circumference of the circle }=2 \pi r\)

So, the perimeter of the given square loop = 2πr, and hence the length of each side,

⇒ \(a=\frac{2 \pi r}{4}=\frac{\pi r}{2}\)

The magnetic field at the center of the square loop,

⇒ \(B=\frac{\mu_0}{4 \pi} \cdot \frac{8 \sqrt{2} I}{a}\)

⇒ \(\frac{\mu_0}{4 \pi} \cdot \frac{8 \sqrt{2} I}{\frac{\pi r}{2}}=\frac{4 \sqrt{2} \mu_0 I}{\pi^2 r}=\frac{8 \sqrt{2}}{\pi^2} \cdot \frac{\mu_0 I}{2 r}=\frac{8 \sqrt{2}}{\pi^2} B^{\prime}\)

So, \(\frac{B}{B^{\prime}}=\frac{8 \sqrt{2}}{\pi^2}\)

Question 22. The radius of the circular path of a revolving electron (charge = e) around the nucleus is r. Due to this revolution, the magnetic field generated at the nucleus is B. What is the angular velocity of the electron?
Answer:

If the angular velocity of the electron is ω, its time period of revolution,

⇒ \(T=\frac{2 \pi}{\omega}\)

Through any point on the orbit, tire amount of charge flowing per second = effective current (I)

So, \(I=\frac{e}{T}=\frac{e Q}{2 \pi}\)

The magnetic field at the center of the circular path,

⇒ \(B=\frac{\mu_0 I}{2 r}=\frac{\mu_0}{2 r} \cdot \frac{e \omega}{2 \pi}=\frac{\mu_0 e \omega}{4 \pi r}\)

So, \(\omega=\frac{4 \pi r B}{e \mu_0}\)

Question 23. An electron (mass m, charge e), accelerated through a potential difference V, enters normally a uniform magnetic field B. What will be the radius of the circular motion of the electron?
Answer:

If the electron is accelerated through a potential, difference V; then kinetic energy gained by it,

⇒ \(e V=\frac{1}{2} m v^2\)

or, \(v=\sqrt{\frac{2 e V}{m}}\)

Magnetic force = evB and centripetal force = \(\frac{m v^2}{r}\)

So, \(e v B=\frac{m v^2}{r}\)

or, \(r=\frac{m v}{e B}=\frac{m}{e B} \sqrt{\frac{2 e V}{m}}=\left(\frac{2 m V}{e B^2}\right)^{1 / 2}\)

Question 24. To detect whether current is flowing in a wire or not, the wire is brought near a magnetic needle but the needle shows no deflection. Hut when the wire is immersed in water kept In a calorimeter, the water gets heated.
Answer:

A current-carrying wire does not produce any deflection on a magnetic needle placed in its vicinity in the following two cases:

1. If the die wire and the needle are placed perpendicular to each other in the same plane.

2. If the needle Is along the same line as the wire. So, as per the problem, the wire-carrying current may not produce any deflection on the magnetic needle. But a current in a wire always produces heat due to the Joule heating effect. This heat produced is Independent of the direction of the rent and hence the temperature of water Increases when the wire is immersed in water kept in a calorimeter.

Question 25. An α-particle and a proton are moving in the plane of a paper in a region where there is a uniform magnetic field directed normally to the plane. If two particles have equal linear momenta, what will be the ratio of the radii of their trajectories in the field?
Answer:

The magnetic force acts as the centripetal force in the circular orbit, i.e.,

⇒ \(B q v=\frac{m v^2}{r} \quad \text { or, } r=\frac{m v}{B q}\)

For the same linear momentum my, in the same magnetic field,

⇒ \(r \propto \frac{1}{q} \quad ∴ \frac{r_1}{r_2}=\frac{q_2}{q_1}=\frac{e}{2 e}=\frac{1}{2}\)

Question 26. If a particle of charge q is moving with velocity v along the axis and a magnetic field B acts along the axis, find the force acting on it. What happens to its kinetic energy?
Answer:

⇒ \(\vec{v}=\hat{j} v, \vec{B}=\hat{k} B\)

∴ \(\vec{F}=q \vec{v} \times \vec{B}=q(\hat{j} \times \hat{k}) v B=\hat{i} q v B\)

\(\vec{v} \text { and } \vec{F}\) are mutually perpendicular. So the magnetic force does not change the magnitude of the particle velocity, i.e., the kinetic energy of the particle would remain unchanged.

Question 27. A circular loop of radius r is formed by bending some portion of an infinitely long wire. If current i flows through the wire, what will be the magnetic field at the center of the circle?
Answer:

The magnetic field at the center of the circle due to the current in the infinitely long wire,

⇒ \(B_1=\frac{\mu_0}{4 \pi} \cdot \frac{2 i}{r}\) [acts vertically downwards to the plane of the paper]

Magnetic field intensity at the center due to current in the circular loop,

⇒ \(B_2=\frac{\mu_0}{4 \pi} \cdot \frac{2 \pi i}{r}\) [acts vertically upwards to the plane of the paper]

So, the resultant magnetic field at the center of the circle,

⇒ \(B=B_2-B_1=\frac{\mu_0}{4 \pi} \cdot \frac{2 \pi i}{r}-\frac{\mu_0}{4 \pi} \cdot \frac{2 i}{r}=\frac{\mu_0}{4 \pi} \cdot \frac{2 i}{r}(\pi-1)\)

Which is vertically upwards to the plane of the paper.

Question 28. State the nature of the graph showing the change of magnetic field with the perpendicular distance from an infinitely long wire carrying a steady current.
Answer:

If current I flow through an infinitely long current-carrying wire, the magnetic field at a perpendicular distance r from the wire will be,

⇒ \(B=\frac{\mu_0 I}{2 \pi r}\)

∴ \(B \propto \frac{1}{r}\left[∵ \mu_0 \text { and } I \text { are constants }\right]\)

So, the B-r graph will be a rectangular hyperbola

Question 29. Determine the force between two parallel circular coaxial coils of radius R each, which are a small distance d(d<<R) apart in free space and carry identical currents I. Assume that each of the coils has a sign.

Answer:

Let the two coils be denoted by C₁ and C₂. As d is much smaller than the radius (R) of each coil, the force per unit length on one of the coils (say C₁ ) due to the other coil (C₂) can be approximated as

⇒ \(F_0=\frac{\mu_0}{2 \pi d}(I)(I)\)

The net force on \(C_1 \text { is, } F=2 \pi R F_0=\frac{\mu_0(I)(I) R}{d}=\frac{\mu_0 I^2 R}{d}\)

The two coils will attract each other with this force if I both are in the same direction.

Examples of Electromagnetic Applications

Question 30. A particle of mass m and charge q moves with a constant velocity v along the positive x-direction. It enters a region of uniform magnetic field B directed along the negative z-direction and extending from x = a to x = b. Find the minimum value of v so that the particle can just enter the region x > b.
Answer:

When a charged particle enters a uniform magnetic field directed perpendicular to its velocity, it moves in a circular path. Let r be the radius of the circular path.

If ( b-a) >= r then the particle cannot enter the region x > b.

So to enter the region x > b, r should be greater than (b-a)

Now we know, \(r=\frac{m v}{B q}\)

So, \(\frac{m v}{B q}>(b-a) \quad .. \quad v>\frac{q(b-a) B}{m}\)

So the minimum value of v is \(\frac{q(b-a) B}{m}\)

Question 31. Two Insulated Infinitely long wires are lying mutually perpendicular to each other. If the two wires carry currents I₁ and I₂ find the locus of the point, where the magnetic field due to the two wires Is zero.

Answer:

Let the point be P(x,y) where the magnetic field Is zero.

Let B₁ and B₂ be the magnetic fields at point P due to the wires carrying currents I₁ and I₂ respectively

Then, \(B_1=\frac{\mu_0}{4 \pi} \cdot \frac{2 I_1}{x}\) (In Inward direction and perpendicular to the plane of the paper)

and, \(B_2=\frac{\mu_0}{4 \pi} \cdot \frac{2 I_2}{y}\) (in an outward direction and perpendicular to the plane of the paper)

If the magnetic field is zero at that point, then

B₁ = B₂ (in magnitude)

⇒ \(\frac{\mu_0}{4 \pi} \cdot \frac{2 I_1}{x}=\frac{\mu_0}{4 \pi} \cdot \frac{2 I_2}{y} \quad ∴ y=\frac{I_2}{I_1} x\)

This is a straight line passing through the point of intersection.

Question 32. Any two points on the circumference of a uniform circular conductor are connected to the terminals of the u cell. Show that, the resultant magnetic field at the center of the circle is zero.
Answer:

In the circular conductor of radius, a is shown. There are two points X and, Y on the circumference and O is the center of the circle.

Let I₁ and I₂ be the currents flowing through the parts XMY and XNY respectively of the circular conductor. Let XMY and XNY subtend angles θ₁ and θ₂ respectively at the center.

Now magnetic field intensity at O due to part XMY,

⇒ \(B_{X M Y}=\frac{\mu_0}{4 \pi} \cdot \frac{T_1}{a} \theta_1\)…(1)

and similarly for the part XNY,

⇒ \(B_{X N Y}=\frac{\mu_0}{4 \pi} \cdot \frac{I_2}{a} \theta_2\)…(2)

BXMY is directed into the plane of the conductor, while BXNY Is directed out of the plane of the conductor.

If r Is resistance per unit length of the conductor then the resistance of the parts XMY and XNY are given by, R₁ = (αθ₁)r and R₂ = (αθ₂)r respectively.

As potential differences across R₁ and R₂ are equal,

∴ V_XMY = V_XNY

or, \(I_1 R_1=I_2 R_2 \quad \text { or, } I_1\left(a \theta_1\right) r=I_2\left(a \theta_2\right) r\)

∵ \(I_1 \theta_1=I_2 \theta_2\)

Prom equations (1) and (2) follow that magnetic fields B1 and B2 due to the two parts of the circular conductor are equal and opposite. Hence the magnetic field at the centre of the circular conductor is zero.

Question 33. Write down the differences between electric lines of force and magnetic lines of force.
Answer:

Important Definitions in Electromagnetism

Question 34. “Increasing the current sensitivity of a galvanometer may not necessarily Increase Its voltage sensitivity”.
Answer:

The current sensitivity (Si) of a galvanometer increases with the increase in the number of its coils. But this increases the galvanometer resistance R_g. If the angle of deflection of the galvanometer, its voltage sensitivity

⇒ \(S_V=\frac{d \theta}{d V}=\frac{d i}{d V} \cdot \frac{d \theta}{d i}=\frac{1}{R_g} S_i\)

This decreases with the increase in R_g

Question 35. The range of an ammeter is Increased by n times. What is the change in its resistance?
Answer:

The shunt that is required to convert a galvanometer of range IG and resistance G into an ammeter of range I is,

⇒ \(S=\frac{I_G}{I-I_G} \cdot G=\frac{1}{\frac{I}{I_G}-1} \cdot G=\frac{G_{-}}{n-1}\) [ given \(\frac{I}{I_G}=n\)]

Now, if R is the equivalent resistance of the parallel combination of G and S, then

⇒ \(\frac{1}{R}=\frac{1}{G}+\frac{1}{S}=\frac{1}{G}+\frac{n-1}{G}={ }^{\prime} \frac{n}{G}\)

∴ \(R=\frac{G}{n}\)

So, if the range of the ammeter is n times, then its resistance decreases from G to \(\frac{G}{n}\)

Question 36. What is the change in resistance of a Voltmeter if its range is increased by n times?
Answer:

A galvanometer has a range of Ig and resistance G. When used as a voltmeter its range, = I_G.G.

Resistance to be connected in series with G to convert it into a voltmeter of range V is,

⇒ \(R=\frac{V}{I_G}-G=G\left(\frac{V}{I_G \cdot G}-1\right)=G\left(\frac{V}{V_G}-1\right)\)

⇒ \(=G(n-1)\left[∵ \frac{V}{V_G}=n\right]\)

Now, the equivalent resistance of the series combination of G and R,

G + R= G + G(n – 1)

= nG

So, when the range of a voltmeter is n times its resistance becomes n times

Question 37. If the distance X of a point on the axis of a circular loop carrying current I is much larger than the radius of the loop, show that the magnetic field at that point is proportional to \(\frac{1}{x^3}\)
Answer:

The magnetic field, \(B=\frac{\mu_0 I}{2} \cdot \frac{r^2}{\left(r^2+x^2\right)^{3 / 2}}\)

Now, for x >> r,

⇒ \(\left(r^2+x^2\right)^{\frac{3}{2}}=\left[x^2\left(\frac{r^2}{x^2}+1\right)\right]^{\frac{3}{2}} \approx\left[x^2(0+1)\right]^{\frac{3}{2}}=x^3\)

∴ \(B \propto \frac{1}{x^3}\)

Question 38. Two wires of equal length arc bent in the form of two loops. One of the loops is a square whereas the other loop Is circular. These are suspended In a uniform magnetic field and the same current is passed through them. Which loop will experience greater torque?
Answer:

Torque on a current carrying loop is given by,

⇒ \(\vec{\tau}=I \vec{A} \times \vec{B}\)

Where, \(\vec{A}\) is the area of the loop, \(\vec{B}\) is the magnetic field and I is the current. According to the question, I and \(\vec{B}\) are same for both the loops

Now, for loops having the same perimeter, the loop has a greater area than the square loop. So, the circular loop will experience greater torque.

Question 39. A rectangular coil carrying current I is placed in a uniform magnetic field B such that the direction of B is perpendicular to the plane of the coll. Calculate the torque experienced by the coil.
Answer:

Torque acting on the coil,

⇒ \(\tau=B I A \sin \theta\)

Where I = current in the coil, A = area of the coil and 0 is the angle between the magnetic field and the normal to the surface of the coil.

As the coil is placed perpendicular to the direction of B, the normal to its surface is parallel to B, i.e., θ = 0. Hence sinθ = 0.

∴ \(\tau=B I A \cdot 0=0\)

Therefore no torque acts on the coil.

Question 40. Find the magnetic field at the point of the intersection of the diagonals of a square having sides a and carrying current I.
Answer:

Distance of the point of intersection (O) of the diagonals from every side,

⇒ \(r=\frac{a}{2}\)

From the angles at 0, θ₁ = θ₂ = 45°

Hence, for each side carrying current I, the magnetic field at O,

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

⇒ \(=\frac{\mu_0}{4 \pi} \cdot \frac{I}{\frac{a}{2}}\left(\sin 45^{\circ}+\sin 45^{\circ}\right)\)

⇒ \(=\frac{\mu_0}{4 \pi} \cdot \frac{2 I}{a} \cdot 2 \times \frac{1}{\sqrt{2}}=\frac{\mu_0}{4 \pi} \frac{2 \sqrt{2}}{a}\)

Now, the directions of the magnetic field at O due to all four sides are the same.

So, the net magnetic field at \(O, 4 B=\frac{\mu_0}{4 \pi} \frac{8 \sqrt{2} I}{a}\)

Question 41. A charged particle enters a uniform magnetic field perpendicularly and experiences a force F. If the kinetic energy of the particle is doubled, then what will be the force on that particle?
Answer:

Force on a charged particle entering a magnetic field B perpendicularly with speed v is, F = qvB

or, \(F \propto v\)

Now, kinetic energy, \(E \propto v^2\)

or, \(F \propto \sqrt{E}[\text { since } F \propto \nu]\)

or, \(\frac{F}{F^{\prime}}=\sqrt{\frac{E}{E^{\prime}}}\)

or, \(F^{\prime}=F \sqrt{\frac{E^{\prime}}{E}}=F \sqrt{2}\)

Question 42. Two equally charged positive ions of Ne²⁰ and Ne²² atom enters a uniform magnetic field perpendicular to the lines of force. Which trajectory will have a larger radius of curvatures?
Answer:

Ne²² ion is heavier than Ne²⁰ ion

we know, qvB = \(q v B=\frac{m v^2}{r}\)

or, \(r=\frac{m v}{q B}\)

then, \(r \propto m\)

So, Ne²² will have a greater radius of curvature