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Class 12 Physics Electromagnetic Induction And Alternating Current

Alternating Current Direct Current Or DC And Alternating Current Or Ac

Before discussing alternating current thoroughly, it would be helpful to recapitulate on direct current in short.

Direct Current (dc): In an electrochemical cell, the electrical nature of the positive and negative electrodes remains unchanged, i.e., it does not vary with time. The current in a circuit from this cell always remains unidirectional.

• This kind of current is known as direct current or unidirectional current However, despite being unidirectional, the magnitude of the current may decrease or increase with time.
• In the graphs, different kinds of direct current concerning time are shown. Of these, the current shown in the is steady current the magnitude of current in this case always remains unchanged.

Alternating Current (ac): An electrochemical cell does not provide large amounts of electrical energy. All electric power plants use a machine called a dynamo or generator for this purpose.

• The characteristics of this machine are—the electrical nature of its two electrodes does not remain constant, but changes from positive to negative, and from negative to positive periodically.
• As a result, current through the external circuit connected to this source gets reversed periodically, i.e., the flow of current does not remain unidirectional.
• The electromotive force obtained from this source is called alternating emf and the current in the circuit is called alternating current.
• All modern electrical appliances, simple or delicate, are enabled. Hence the importance of the study of ac. Current-time graphs of some alternating currents are shown.

Read and Learn More Class 12 Physics Notes

The following characteristics of alternating current are to be noted.

1. An alternating current is of a wave nature.
2. The magnitude of the current above the time axis is taken as positive, and that below is taken as negative. This implies that the direction of the current gets reversed periodically, and at the moment of this transition, the instantaneous magnitude of the current becomes zero.
3. The waveforms of different alternating currents may be different. The waveforms of the three different currents shown are respectively sinusoidal, square, and triangular. Of them, the discussion about the sinusoidal waveform is of particular importance; because, by appropriate mathematical analysis, the square, triangular, or any other waveform can be reduced to a combination of a large number of sinusoidal waves.

Class 12 Physics Electromagnetic Induction And Alternating Current

Alternating Current Source Of Alternating Current Ac Dynamo

Almost the entire electrical energy requirement of the present-day world is derived from the phenomenon of electromagnetic induction.

The machine employing this phenomenon is called a dynamo or generator. A conducting coil is set to rotate inside a magnetic field, as a result of which a current is induced in the coil. This is the basic mechanism.

Definition: The machine in which the mechanical energy of a rotating conducting coil placed in a magnetic field is converted into electrical energy, is called a dynamo or generator.

Description: The main parts of an alternating current dynamo are shown.

N, S: The poles of a strong horseshoe magnet which produces a uniform magnetic field directed from the north to the south pole in the gap between them. Some lines of force are shown in the figure. This magnet is called the field magnet.

ABCD: A rectangular coil called an armature is placed in a uniform magnetic field, which usually contains several turns.

The coil in this case is made to rotate about the axis normal to the magnetic field; the direction of rotation in this case is as shown in the diagram.

R₁, R₂: Two smooth brass rings called slip rings. They are connected to the coil at its open ends A and D.

T₁, T₂: Two carbon brushes fitted to the rings with the help of springs, which keep the brushes pressed against the rings R₁ and R₂.

L: An electric lamp, to indicate the presence of current.

Observation: The lamp glows as long as the coil rotates. From this, we can infer that an electric current is flowing through the external circuit attached to T₁ and T₂. As soon as the rotation of the coil ceases, the current stops and the lamp goes out.

Working Principle: when the coil ABCD arms AB and CD intersect the magnetic lines of force. As a result, electromagnetic induction takes place.

• At a certain moment, when the motion of the arm AB is downwards, according to Fleming’s right-hand rule, a current will be induced in the direction BA. At the same time due to the upward motion of the arm CD, the direction of induced current will be along DC.
• So, a current will be set up in the direction of DCBA in the coil and will flow from T₁ to T₂ in the external circuit In this situation T₁ and T₂ behave as the positive and negative poles of a battery, respectively.
• Here, arms BC and AD do not intersect the magnetic lines of force.
• When, after half-rotation, the positions of the arms AB and CD are interchanged, by applying Fleming’s right-hand rule, it is observed that current is now induced in the direction ABCD.

As a result, current flows in the second half of rotation from T₂ to T₁ in the external circuit. Thus, the polarities of T₁ and T₂ get reversed, reversing the direction of the current.

• This reversal of the direction of the current goes on periodically. Every time the coil crosses its vertical position, the direction of the current gets reversed. Hence, this dynamo is called an AC dynamo.
• Because the heating effect of electric current {αI², i.e., the same for +I and -I) does not depend on the direction of current, the lamp continues to glow as long as the coil rotates.
• If a DC galvanometer (for example., a moving coil galvanometer) is placed instead of the electric lamp, no deflection of its pointer would be observed.

Factors Affecting The Emf And Current: The electromotive force, and hence the current is directly proportional to

1. Area of the coil,
2. Number of turns of the coil,
3. speed of rotation of the coil and
4. Strength of the magnetic field.

So, if any one of them is increased, the emf will also increase. The current however will decrease with an increase in the resistance of the circuit.

Electric motor: The principle of action of an electric motor is just opposite to that of a dynamo.

• A motor is a device in which a current-carrying coil, placed suitably in a magnetic field, rotates on the principle of action of magnets on currents.
• A motor converts the electrical energy of the coil into its mechanical energy.

Differences Between Dynamo And Motor:

However, the discussions on both DC and AC motors are beyond our present syllabus.

Class 12 Physics Electromagnetic Induction And Alternating Current

Alternating Current Variation Of Alternating Current

Waveform Of Alternating Current: Let a rectangular. coil ABCD is rotating with uniform angular velocity in a uniform magnetic field. The coil is viewed in a way such that only the AD end of the coil becomes visible. The different positions of this end concerning time are shown.

Let the period of the coil be T. If the coil starts rotating from its vertical position, then at \(\frac{T}{2}\) and T, it again comes to its vertical position.

At these moments both AB and CD are moving parallel to the magnetic field. Considering the equation, e = Blvsind with θ = 0° the induced emf and hence the induced current will be zero.

On the other hand, at the times \(\frac{T}{4}\) and \(\frac{3T}{4}\), the coil lies horizontally. At those positions,r. AB and CD are directed normally to the magnetic field (θ = 90°), and the induced current becomes maximum.

However, at the time \(\frac{T}{4}\), the direction of current is along DCBA, while at the time \(\frac{3T}{4}\), it is along ABCD. So, if the current in the first case is taken as positive, then the current in the second case will be negative.

This figure clearly shows a sine-wave; this is the wave nature of current if the coil starts rotating from its vertical position.

On the other hand, if the coil, starts rotating from its horizontal position, the current Will be a cosine-wave. Note that, both sine and cosine waves are called sinusoidal waves.

The current completes a cycle of change in a time equal to the period of rotation of the coil. A complete wave is thus formed in every cycle.

Expression Of Alternating Current: Let a rectangular coil be rotating with uniform angular velocity ω in a uniform. magnetic field B. At any moment t, the angle between the normal to the coil and the direction of the magnetic field is θ, say.

If the area of the plane of the coil is A, the magnetic flux linked with the coil,

Φ = BAcosθ → (1)

Initially, the angle between the normal to the plane of the coil and the direction of the magnetic field = α (say). After rotation for time t, this angle becomes,

θ = ωt + α

So, from equation (1) we get, magnetic flux,

Φ= BA cos (ωt+α) → (2)

Hence, the magnitude of the induced emf in the coil of a single turn

⇒ \(e=\frac{d \phi}{d t}=\omega B A \sin (\omega t+\alpha)\)

For N turns,

⇒ \(e=\omega B A N \sin (\omega t+\alpha)=e_0 \sin (\omega t+\alpha)\) → (3)

[here, e₀ = ωBAN]

If the total resistance of the coil and the external circuit is R, the induced current,

⇒ \(i=\frac{e}{R}=\frac{\omega B A N}{R} \sin (\omega t+\alpha)=\frac{e_0}{R} \sin (\omega t+\alpha)\)

⇒ i₀sin(ωt + α) → (4)

Phase: The state of alternating current at any moment is expressed by its phase. In equations (3) and (4), (ωt + α) is the phase of the alternating current.

Phase Difference: If the phases of two alternating currents are δ₁ and δ₂, then(δ₁-δ₂) is the phase difference of those two currents.

If co is the same for the two currents, the phase difference
becomes (α₁– α₂)

Peak Value: As -1 ≤ sinθ ≤ 1, it can be concluded that

• The maximum and minimum values of the electromotive force are e₀ and -e₀ respectively.
• Accordingly, the maximum and minimum values of the current are i₀ and -i₀.

These magnitudes, e₀ and i₀ are called the peak values of electromotive force and current.

Dependence Of Peak Value On Different Factors: it is obvious from equations (3) and (4) that the peak values of emf as well as current depend directly on each of the quantities involved, viz., co, B, A, N. Moreover, the peak value of current also varies inversely with the resistance (R) of the circuit.

Period And Frequency: The definite time interval in which a complete cycle repeats itself, is called the period (T) of an alternating current.

If the angular velocity, of the coil, is co, then

∴ \(T=\frac{2 \pi}{\omega}\) → (5)

The number of complete waves produced in unit time is called the frequency (n) of an alternating current.

So, \(n=\frac{1}{T}=\frac{\omega}{2 \pi}\) → (6)

Frequency is the most important quantity in the expression of an ac. It is noted that the frequency of domestic electric supply in India is 50 Hz or 50 cycles per second(cps). It indicates that the direction of the current is reversed (50 x 2) or 100 times per second.

Class 12 Physics Electromagnetic Induction And Alternating Current

Alternating Current Circuit Symbol Of An AC Source

It represents the symbols of an AC source used in a circuit. Any of these symbols may be used for the purpose.

Average Values Of Alternating Voltage And Current

Alternating voltage (V) or Current (I) always increases, decreases, and changes periodically following the functions since or cos ωt. But the average values of these quantities, do riot change with time.

Calculation: Let an expression for sinusoidal alternating voltage,

V = V₀ Sin ωt

Average value of it in a full cycle (i.e., for t = T)

= \(\frac{1}{T} \int_0^T V d t=\frac{1}{T} \int_0^T V_0 \sin \omega t d t=0\)

This is because the voltage in a half cycle is positive and in the next half it is of the same magnitude, but negative.

For convenience, the average value in a half-cycle, instead of in a full-cycle, in ac is considered as the average value \((\bar{V})\) of an alternating voltage.

∴ \(\bar{V}=\frac{1}{T / 2} \int_0^{T / 2} V_0 \sin \omega t d t=\frac{2 V_0}{\omega T}[-\cos \omega t]_0^{T / 2}\)

= \(\frac{2 V_0}{\pi}=0.637 V_0\)

Similarly, the average value of alternating current,

∴ \(\bar{I}=\frac{2 I_0}{\pi}=0.637 I_0\)

RMS Values Of Alternating Voltage And Current

Alternating voltage or current cannot be measured directly with instruments like galvanometers. They can only be measured indirectly following the thermal effect of current.

We know that heat generated in a current-carrying conductor is directly proportional to V² or l², i.e., heat thus generated does not depend on the direction of the current.

• Due to the periodical changes of an alternating current, the heat in the conductor fluctuates from zero to a certain positive value. Hence the average value of heat should be proportional to the average value of V² or I².
• The value most commonly used for an ac is its effective value. The effective value of ac is the amount of ac that produces the same heating effect as an equal amount of dc.
• This is calculated by squaring all the amplitudes of the sine wave over one period, taking the average of these values, and then taking the square root.

The effective value, being the root of the average (or mean) of the square of the currents, is known as the root mean square (in short, rms) value.

Calculation: Let an expression for sinusoidal alternating voltage,

v = v₀ sinωt

So, the mean square of V,

∴ \(\vec{V}^2=\frac{1}{T} \int_0^T V^2 d t=\frac{1}{T} \int_0^T \cdot V_0^2 \sin ^2 \omega t d t\)

= \(\frac{V_0^2}{2 T} \int_0^T(1-\cos 2 \omega t) d t=\frac{V_0^2}{2 T}\left[t-\frac{\sin 2 \omega t}{2 \dot{\omega}}\right]_0^T\)

= \(\frac{V_0^2}{2 T}\left(T-\frac{\sin 4 \pi}{2 \omega}+\frac{\sin 0}{2 \omega}\right)\) [∵ ωt = 2π]

= \(\frac{v_0^2}{2}\)

∴ rms value of V,

⇒ \(V_{\mathrm{rms}}=\sqrt{\bar{V}^2}=\sqrt{\frac{V_0^2}{2}}=\frac{V_0}{\sqrt{2}}=0.707 V_0\)

Similarly, rms value of I,

⇒ \(I_{\mathrm{rms}}=\frac{I_0}{\sqrt{2}}=0.707 I_0\)

i.e., rms voltage = 70.7% of peak voltage

and rms current = 70.7% of peak current

Relation between peak value, average value, and rms value of alternating voltage and current: From the above discussion we have,

⇒ \(\bar{V}=\frac{2 V_0}{\pi}=0.637 V_0\)

and \(V_{\mathrm{rms}}=\frac{V_0}{\sqrt{2}}=0.707 V_0\)

where V₀ is the peak value of alternating voltage.

∴ \(V_{\mathrm{rms}}: \bar{V}=\frac{V_0}{\sqrt{2}}: \frac{2 V_0}{\pi}=\frac{\pi}{2 \sqrt{2}}\)

or, \(V_{\mathrm{rms}}=\frac{\pi}{2 \sqrt{2}} \bar{V}=1.11 \bar{V}\)

∴ \(V_{\text {rms }}>\bar{V}\)

Similarly, \(I_{\mathrm{rms}}>\bar{I}\)

This relation is shown graphically.

We know that the heating effect of electric current is directly proportional to I². Hence ac ammeters and ac voltmeters, are designed based on the heating effect of current. They directly measure rms values of alternating current and voltage respectively.

Form Factor: The ratio of the rms value and the average value of alternating voltage or current is known as the form factor.

Thus form factor, \(f=\frac{V_{\mathrm{rms}}}{\bar{V}}=\frac{\frac{V_0}{\sqrt{2}}}{\frac{2 V_0}{\pi}}=\frac{\pi}{2 \sqrt{2}}=1.11\).

Note that, the above value of form factor applies only to sinusoidal voltage arid current. For different waveforms, the values of the form factor are different. For example, for a square wave

∴ \(V_{\mathrm{rms}}=V_0 \text { and } \bar{V}=V_0\)

∴ Form factor, \(f=\frac{V_{\mathrm{rms}}}{\bar{V}}=1\)

From the form factor of an alternating voltage or current an effective idea about the waveform can be obtained.

Effects Of Oil An Ac Or DC Currents On The Human Body: the three main factors that determine what kind of shock one experiences are the amplitude, frequency, and duration of the current passing through the body.

Direct current has zero frequency i.e., it has a constant amplitude. On the other hand, the peak value of an ac voltage (V₀) is √2. times its rms value (Vrms), for example., a 220V ac supply is going 220√2 or 311 V (approx.) before coming down to zero.

So, one can get a 311V shock from a 220V-50Hz ac supply line for (50 x 2) hor 100 times per second. The calculation indicates the severity of electrocution from ac compared to that from the decision of average for the same duration of time.

Class 12 Physics Electromagnetic Induction And Alternating Current

Alternating Current Circuit Symbol Of An AC Source Numerical Examples

Example 1. Equation of an ac is \(I=, 10 \sin \left(200 \pi t-\frac{\pi}{15}\right)\) ampere. Determine the frequency and peak value of the current.

Solution:

Comparing with the general equation of ac, I = I₀ sin(ωt + α) we get,

the peak value of current, I₀ = 10 A

and angular frequency, ω = 200 π Hz

∴ Frequency, \(I=\frac{\omega}{2 \pi}=\frac{200 \pi}{2 \pi}=100 \mathrm{~Hz}\)

Example 2. If an ac is represented by I = 100 sin 200 πt ampere, determine the peak value of the current and period.

Solution:

Comparing with the general equation of ac, I = I₀ sin(ωt+α) we get,

the peak value of current, I₀ = 100 A

and angular frequency, ω = 200π HZ

∴ Time period, T = \(T=\frac{2 \pi}{\omega}=\frac{2 \pi}{200 \pi}=0.01 \mathrm{~s}\)

Example 3. An alternating current having a peak value of 141 A Is used to heat a metal wire. To produce the same rate of heating effect, another constant current IA is used. What is the value of I?

Solution:

∴ \(I_{\mathrm{rms}}=\frac{I_0}{\sqrt{2}}=\frac{141 \mathrm{~A}}{\sqrt{2}}=100 \mathrm{~A}\)

If a steady DC I produces the same rate of heating, then I = I_rms = 100A.

Example 4. The peak value of an alternating magnetic field B is 0.01 T and the frequency is 100 Hz. If a conducting ring of radius 1 m is held normal to the field, what emf will be induced in the ring?

Solution:

n = 100 Hz

∴ Time period, \(T=\frac{1}{n}=\frac{1}{100} \mathrm{~s}\)

The time taken by the field to increase from 0T to, 0.01 T is \(\frac{T}{4}=\frac{1}{400} \mathrm{~s}\).

Now the area, A = π. m² = π m²

Hence induced emf,

∴ \(|e|=\frac{d \phi}{d t}=\frac{d}{d t}(B A)=A \frac{d B}{d t}\)

or, \(|e|=\pi\left(\frac{0.01-0}{\frac{1}{400}}\right)=4 \pi \mathrm{V}=12.57 \mathrm{~V}\)

Class 12 Physics Electromagnetic Induction And Alternating Current

Alternating Current Power Consumed In Ac Circuits Some Important Ac Circuits

Power consumed In ac circuits: In any dc circuit, the potential difference (voltage) and the current are always in phase. But this is not so for ac circuits—in general, a phase difference is evolved between the voltage and the current Mathematically the voltage V and the current I are expressed as,

V = V₀ sinωt and I = I₀ sin(ωt-θ)

where, θ = phase difference between the voltage and the current. θ always lies between -90° and +90°, i.e., -90° ≤ θ ≤ 90°. Then, power consumed (or power dissipated) in the circuit is,

∴ \(P=V I=V_0 I_0 \sin \omega t \sin (\omega t-\theta)\)

= \(V_0 I_0 \sin \omega t[\sin \omega t \cos \theta-\cos \omega t \sin \theta]\)

=\(V_0 I_0\left[\sin ^2 \omega t \cos \theta-\sin \omega t \cos \omega t \sin \theta\right]\)

= \(V_0 I_0\left[\sin ^2 \omega t \cos \theta-\frac{1}{2} \sin 2 \omega t \sin \theta\right]\)

Over a complete cycle, the average of \(\sin ^2 \omega t=\frac{1}{2}\) and that of sin²ωt = 0. So the average power of the circuit is,

∴ \(\bar{P}=\frac{1}{2} V_0 I_0 \cos \theta=\frac{V_0}{\sqrt{2}} \frac{I_0}{\sqrt{2}} \cos \theta=V_{\mathrm{rms}} I_{\mathrm{rms}} \cos \theta\)

No experiment can measure the instantaneous power P; every measurement leads to the average power \(\bar{P}\). This \(\bar{P}\) is referred to as the effective power, and is usually denoted by the simple symbol P. So the effective power (or true power) of an ac circuit is,

∴ \(P=\frac{1}{2} V_0 I_0 \cos \theta=V_{\mathrm{rms}} I_{\mathrm{rms}} \cos \theta\) → (1)

Power Factor: The factor cosθ in equation (1) is vitally important; this factor arises, clearly, due to the voltage-current phase difference θ. This cosθ is called the power factor of an ac circuit.

• If θ ≠ 0, cos θ< 1; so the power factor, in general, reduces the power consumed in an ac circuit to some extent Also, as -90°≤ θ ≤ 90°, the power factor cosθ is never negative; as a result, the power consumed in the circuit can never be negative.
• The unit of power P is watt (W); to distinguish between P and the product V_rms I_rms, watt is never used as the unit of V_rmsI_rms — the usual unit of this product is volt. ampere or V.A.
• Incidentally, from the similarity with the dc expression P = VI, the product V_rms I_rms is often called the apparent power of an ac circuit. Clearly,

true power = apparent power x power factor.

If voltage and current are in the same phase, then θ = 0 and cos θ = 1. In this condition, an ac circuit consumes the maximum power.

If the voltage and the current are either -90° or +90° out of phase, then cos θ = 0 and P = 0. An ac circuit of this type consumes no power. The corresponding current is referred to as wattless current.

Time Interval Between The Peak Values Of Voltage And Current:

v = v₀ sinωt, I = I₀sin(ωt-θ)

So, \(V=V_0, \text { when } \omega t=\frac{\pi}{2}, \text { i.e., } t=\frac{\pi}{2 \omega}\)

Similarly, \(I=I_0 \text {, when }\left(\omega t^{\prime}-\theta\right)=\frac{\pi}{2} \text {, i.e., } t^{\prime}=\frac{\pi}{2 \omega} 4 \frac{\theta}{\omega}\)

Thus, the minimum time interval between the peak values of ac voltage and current is,

∴ \(t_0=t^{\prime}-t=\frac{\theta}{\omega}\)

Purely Resistive Circuit

shows the circuit. Alternating voltage applied in the circuit,

V = V₀sinω t → (1)

Here, the peak value of alternating voltage = v₀

rms value, v_rms = \(\frac{V_0}{\sqrt{2}} ; \text { frequency, } f=\frac{\omega}{2 \pi}\).

According to Ohm’s law,

⇒ \(I=\frac{V}{R}=\frac{V_0}{R} \sin \omega t\)

Or, \(I=I_0 \sin \omega t\)

So, the peak value of ac, \(I_0=\frac{V_0}{R}\)

⇒ rms value, \(I_{\mathrm{rms}}=\frac{I_0}{\sqrt{2}}=\frac{V_0}{\sqrt{2} R}=\frac{V_{\mathrm{rms}}}{R}\);

⇒ frequency, \(f=\frac{\omega}{2 \pi}\)

The factor sinωt in equations (1) and (2) indicates that

1. Voltage and current are in phase;
2. There is no change in frequency in this type of circuit.

Power in the circuit: Phase difference between voltage and current, θ = 0; so, power factor, cosθ = 1. For this, maximum power is consumed in a purely resistive circuit, which is

∴ \(P=\frac{1}{2} V_0 I_0 \cos \theta=\frac{1}{2}\left(I_0 R\right) I_0 \cdot 1=\frac{1}{2} I_0^2 R=\left(\frac{I_0}{\sqrt{2}}\right)^2 R\)

i.e., P = I_rms².R → (3)

Purely Inductive Circuit

shows this circuit. Let the alternating voltage applied in the circuit be,

V = V₀sin ωt → (1)

If the instantaneous change of current in the circuit is dl, the emf induced in the two ends of the inductor = \(-L \frac{d I}{d t}\)

This emf reduces the main voltage in the circuit. For this the effective voltage of the circuit = \(V_0 \sin \omega t-L \frac{d I}{d t}\).

A purely inductive circuit has no resistance i.e., R = 0. So according to Ohm’s law,

⇒ \(V_0 \sin \omega t-L \frac{d I}{d t}=0 \quad \text { or, } \frac{d I}{d t}=\frac{V_0}{L} \sin \omega t\)

or, \(\int d I=\int \frac{V_0}{L} \sin \omega t d t\)

or, \(I=-\frac{V_0}{\omega L} \cos \omega t+k\) [k = integration constant] → (2)

Dimensionally k is the same as I. Also, fc is a time-independent quantity. As the source voltage oscillates symmetrically about zero, the current also oscillates symmetrically about zero. For this, no constant or time-independent current can flow through the circuit, i.e., k = 0.

∴ \(I=-\frac{V_0}{\omega L} \cos \omega t=+\frac{V_0}{\omega L} \sin \left(\omega t-90^{\circ}\right)\)

or, I = I₀sin(ωt-90°) →(3)

where \(I_0=\frac{V_0}{\omega L}\) = peak value of current.

From the equations (1) and (3) we come to the conclusions

current lags behind the voltage by 90°.

The quantity coL plays the same role in an inductive circuit as the resistance R in a resistive circuit. This quantity (i.e., ωL) is known as the inductive reactance of an ac circuit. Inductive reactance is the opposition offered by an inductor to the flow of alternating current through it. It is denoted by the symbol X_L.

Hence for the above circuit,

X_L = ωL = inductive reactance

Similar to that of R its unit is also ohm (Ω).

Power In The Circuit: Phase difference between voltage and current, θ = 90°, so, power factor, cosθ = 0.

Therefore, power consumption, \(P=\frac{1}{2} V_0 I_0 \cos \theta=0\)

The current I in this circuit is wattless.

Purely Capacitive Circuit

shows the circuit. Alternating voltage applied in the circuit,

V = V₀sin ωt →(1)

At any moment, if Q is the charge stored in the capacitor C, then its terminal potential difference is \(\frac{Q}{C}\)

This potential difference opposes the applied instantaneous voltage in the circuit.

So, the effective voltage of the circuit, \(V_e=V_0 \sin \omega t-\frac{Q}{C}\)

As there is no resistance in the circuit, according to Ohm’s law, V_e = IR = 0

and hence, \(V_e=V_0 \sin \omega t-\frac{Q}{C}=0\)

or, Q = CV₀sinωt → (2)

Therefore alternating current,

∴ \(I=\frac{d Q}{d t}=\omega C V_0 \cos \omega t=\omega C V_0 \sin \left(\omega t+90^{\circ}\right)\)

= I₀sin(ωt+90°) → (3)

Where, \(I_0=\omega C V_0=\frac{V_0}{1 / \omega C}=\frac{V_0}{X_C}\) = peak value of current From the equations (1) and (3) we come to the conclusions

Current leads the applied voltage by 90°;

The quantity \(\frac{1}{\omega C}\)plays the same role in a capacitive circuit as the resistance R in a resistive circuit. This quantity (i.e., \(\frac{1}{\omega C}\) is known as capacitive reactance (X_C).

Capacitive reactance is the opposition offered by a capacitor to the flow of alternating current through it. Hence for the above circuit.

∴ \(x_C=\frac{1}{\omega C}\) = capacity reactance

Similar to that of R or X_L, its unit is also ohm (Ω).

Power In The Circuit: Phase’ difference between voltage and current, θ = -90°; so, power factor, cosθ = 0. Therefore, power consumption \(P=\frac{1}{2} V_0 I_0 \cos \theta=0\).

Here again, this current is wathers.

Wattless or Idle current: Power consumed in an ac circuit, \(P=\frac{1}{2} V_0 I_0 \cos \theta\). Now the phase difference between voltage and current in a pure inductive circuit, θ = 90°.

So power in the circuit is zero. On the other hand phase difference between voltage and current in a pure capacitive circuit, θ = -90°.

So, again power is equal to zero, which means, a purely inductive or capacitive circuit does not dissipate any power; current through the inductor or capacitor is hence called wattless or idle current.

LR Circuit

shows the circuit

The applied alternating voltage,

V = V₀sinωt → (1)

If the instantaneous change in current in of the circuit is dl, then the induced emf in the inductor \(L=-L \frac{d I}{d t}\);

i.e., the effective voltage in the circuit,

⇒ \(V_e=V_0 \sin \omega t-L \frac{d I}{d t}\)

According to Ohm’s law, V_e = IR, where the resistance, if any, of the inductor L is included in R.

∴ \(V_0 \sin \omega t-L \frac{d I}{d t}=I R\)

or, \(L \frac{d I}{d t}+R I=V_0 \sin \omega t\) → (2)

Let I = I₀sin(ωt+ α)

or, \(\frac{d I}{d t}=\omega I_0 \cos (\omega t+\alpha)\)

So, putting the values of I and \(\frac{d I}{d t}\) in identity (2) we get

ωLI₀ cos(ωt+α) + RI₀ sin(ωt+α) = V₀ sinωt

or, I₀{R sin((ωt+ α) + L cos(ωt+α)} = V₀ sinωt

or, \(I_0 \sqrt{R^2+(\omega L)^2}\{\frac{R}{\sqrt{R^2+(\omega L)^2}} \sin (\omega t+\alpha)+\frac{\omega L}{\sqrt{R^2+(\omega L)^2}} \cos (\omega t+\alpha)\}=V_0 \sin \omega t\)

or, \(I_0 Z\left\{\frac{R}{Z} \sin (\omega t+\alpha)+\frac{\omega L}{Z} \cos (\omega t+\alpha)\right\}=V_0 \sin \omega t \text { [where } Z=\sqrt{R^2+(\omega L)^2} \text { ] }\)

or, \(I_0 Z\{\sin (\omega t+\alpha) \cos \theta+\cos (\omega t+\alpha) \sin \theta\}=V_0 \sin \omega t \text { [where } \cos \theta=\frac{R}{Z} \text { and } \sin \theta=\frac{\omega L}{Z} \text { ] } \)

or, I₀Zsin(ωt+ α + θ) = V₀sin ωt

From this identity, comparing both sides we get,

⇒ \(I_0 Z=V_0 \quad \text { or, } I_0=\frac{V_0}{Z}=\frac{V_0}{\sqrt{R^2+(\omega L)^2}}\)

and α + θ = 0 or, α = -θ

So, the ac in the circuit,

∴ \(I=I_0 \sin (\omega t-\theta)=\frac{V_0}{Z} \sin (\omega t-\theta)\) → (3)

From equations (1) and (3) we conclude,

The current lags behind the applied voltage by a phase angle θ given by \(\tan \theta=\frac{\sin \theta}{\cos \theta}=\frac{\omega L}{Z} \times \frac{Z}{R}=\frac{\omega L}{R}\)

i.e., \(\theta=\tan ^{-1}\left(\frac{\omega L}{R}\right)\)

This phase relation is shown. We know that in a purely resistive circuit V and I are in the same phase i.e., phase difference, θ = 0. On the other hand, in a pure inductive circuit, I always lag behind V by θ = 90°. So in an LR circuit, I should lag behind V by 0 < θ < 90°.

In this circuit, Z plays the same role as R in a pure resistive circuit. Z is known as the impedance of an ac circuit.

Impedance, \(Z=\sqrt{R^2+(\omega L)^2}=\sqrt{R^2+\left(X_L\right)^2}\) → (4)

where, X_L = ωL = inductive reactance

Impedance in an LR circuit is the effective resistance of the circuit arising from the combined effects of ohmic resistance and inductive reactance.

We can express the relation among R, X_L, and Z with a suitable right-angled triangle. This triangle is called the impedance triangle.

Note that, R, ωL, and Z have, the same unit ohm(Ω).

Power In The Circuit: 

Here, the power factor of the circuit, \(\cos \theta=\frac{R}{Z}\)

∴ \(P=\frac{1}{2} V_0 I_o \cos \theta=\frac{1}{2}\left(I_0 Z\right) I_0 \cdot \frac{R}{Z}=\frac{1}{2} I_0^2 R=\left(\frac{I_0}{\sqrt{2}}\right)^2 R\)

i.e., \(P=I_{\mathrm{rms}}^2 R\)

This means that the power is dissipated only in the resistance R. Current through the inductor is wattless.

CR Circuit

It shows the circuit. Alternating voltage applied to the circuit,

V = V₀sin ωt → (1)

At any moment if Q is the charge stored in the capacitor C, then, the effective voltage in the circuit,

∴ \(V_e=V_0 \sin \omega t-\frac{Q}{C}\)

According to Ohm’s law

∴ \(V_0 \sin \omega t-\frac{Q}{C}=I R\)

or, \(R I+\frac{1}{C} Q=V_0 \sin \omega t\)

Now let us assume that Q = Q₀ sin(ωt+ α)

∴ \(I=\frac{d Q}{d t}=\omega Q_0 \cos (\omega t+\alpha)\)

So, from the equation (2) we get,

∴ \(\omega R Q_0 \cos (\omega t+\alpha)+\frac{1}{C} Q_0 \sin (\omega t+\alpha)=V_0 \sin \omega t\)

or, \(\omega Q_0\left[R \cos (\omega t+\alpha)+\frac{1}{\omega C} \sin (\omega t+\alpha)\right]=V_0 \sin \omega t\)

or, \(\omega Q_0 Z\left[\frac{R}{Z} \cos (\omega t+\alpha)+\frac{1 /(\omega C)}{Z} \sin (\omega t+\alpha)\right]=V_0 \sin \omega t\)

[where \(Z=\sqrt{R^2+\left(\frac{1}{\omega C}\right)^2}\)(say)]

or, \(\omega Q_0 Z\{\cos (\omega t+\alpha) \cdot \cos \theta+\sin (\omega t+\alpha) \cdot \sin \theta\}=V_0 \sin \omega t\)

[where \(\frac{R}{Z}=\cos \theta \text { and } \frac{1 / \omega C}{Z}=\sin \theta\)(say)]

or, \(\omega Q_0 Z \cos (\omega t+\alpha-\theta)=V_0 \sin \omega t\)

or, \(\omega Q_0 Z \sin \left(\omega t+\alpha-\theta+90^{\circ}\right)=V_0 \sin \omega t\)

Comparing both sides we get,

∴ \(\omega Q_0 Z=V_0 \quad \text { or, } Q_0=\frac{V_0}{\omega Z}\)

and α – θ + 90° = 0 or, α = θ – 90°

So, alternating current in the circuit,

I = ωQ₀cos(ωt+ α)

= \(\omega \frac{V_0}{\omega Z} \cos \left(\omega t+\theta-90^{\circ}\right)=\frac{V_0}{Z} \sin (\omega t+\theta)\)

i.e., \(I=I_0 \sin (\omega t+\theta)=\frac{V_0}{Z} \sin (\omega t+\theta)\) → (3)

Here, \(I_0=\frac{V_0}{Z}=\frac{V_0}{\sqrt{R^2+\left(\frac{1}{\omega C}\right)^2}}\)

From equations (1) and (3) we conclude the following.

1. The current I leads the voltage V by a phase angle θ, where \(\tan \theta=\frac{\sin \theta}{\cos \theta}=\frac{1 / \omega C}{Z} \times \frac{Z}{R}=\frac{1}{\omega C R}\),

i.e., \(\theta=\tan ^{-1}\left(\frac{1}{\omega C R}\right)\)

This phase relation is shown. We know that in a purely resistive circuit, V and I are in the same phase, i.e., phase difference, θ = 0. On the other hand, in a pure capacitive circuit, I always lead V by θ = 90°. So, in a CR circuit, I should lead V by 0 < θ < 90°.

2. Z plays the same role as R in a pure resistive circuit. This Z is known as the impedance of the circuit.

Impendance, \(Z=\sqrt{R^2+\left(\frac{1}{\omega C}\right)^2}=\sqrt{R^2+X_C^2}\) → (4)

where \(X_C=\frac{1}{\omega C}\) = capacitive reactance.

Impedance in a CR circuit is the effective resistance of the circuit arising from the combined effects of ohmic resistance and capacitive reactance.

shows the impedance triangle for the circuit.

Power In The Circuit: Here, the power factor of the circuit, \(\cos \theta=\frac{R}{Z}\)

so, \(P=\frac{1}{2} V_0 I_0 \cos \theta=\frac{1}{2}\left(I_0 Z\right) I_0 \cdot \frac{R}{Z}=\frac{1}{2} I_0^2 R=\left(\frac{I_0}{\sqrt{2}}\right)^2 R\)

∴ P = I²_rms.R²

Hence, power is dissipated only in the resistance R . Current through the capacitor C is wattless.

Series LCR Circuit

shows the circuit. Alternating voltage applied in the circuit,

V = V₀sin ωt → (1)

If the instantaneous change in current in the circuit is dI, then the emf induced in the inductor

= \(-L \frac{d I}{d t}\).

On the other hand, if Q is the instantaneous charge stored in the capacitor, the opposite emf thus developed in the circuit = \(-\frac{Q}{C}\).

So, the effective voltage in the circuit,

∴ \(V_e=V_0 \sin \omega t-L \frac{d I}{d t}-\frac{Q}{C}\)

According to Ohm’s law,

∴ \(V_0 \sin \omega t-L \frac{d I}{d t}-\frac{Q}{C}=I R\)

or, \(L \frac{d I}{d t}+R I+\frac{Q}{C}=V_0 \sin \omega t\)

Now, let us assume, Q = Q₀sin(ωt+ α)

So, current, \(I=\frac{d Q}{d t}=\omega Q_0 \cos (\omega t+\alpha)\)

or, \(\frac{d I}{d t}=-\omega^2 Q_0 \sin (\omega t+\alpha)\)

Putting these values in equation (2) we get,

= \(-\omega^2 L Q_0 \sin (\omega t+\alpha)+\omega R Q_0 \cos (\omega t+\alpha)+\frac{Q_0}{C} \sin (\omega t+\alpha)=V_0 \sin \omega t\)

or, \(\omega Q_0\left\{R \cos (\omega t+\alpha)-\left(\omega L-\frac{1}{\omega C}\right) \sin (\omega t+\alpha)\right\}=V_0 \sin \omega t\)

or, \(\omega Q_0 Z\left\{\frac{R}{Z} \cos (\omega t+\alpha)-\frac{\omega L-1 / \omega C}{Z} \sin (\omega t+\alpha)\right\}=V_0 \sin \omega t\)

[Where \(Z=\sqrt{R^2+\left(\omega L-\frac{1}{\omega C}\right)^2}\)]

or, \(\omega Q_0 Z\{\cos (\omega t+\alpha) \cos \theta-\sin (\omega t+\alpha) \sin \theta\}=V_0 \sin \omega t=\)

[Where \(\frac{R}{Z}=\cos \theta \text { and } \frac{\omega L-1 / \omega C}{Z}=\sin \theta\)]

or, \(\omega Q_0 Z \cos (\omega t+\alpha+\theta)=V_0 \sin \omega t\)

or, \(\omega Q_0 Z \sin \left(\omega t+\alpha+\theta+90^{\circ}\right)=V_0 \sin \omega t\)

Comparing both sides we get,

⇒ \(\omega Q_0 Z=V_0 \quad \text { or, } Q_0=\frac{V_0}{\omega Z}\)

and α + θ + 90° = 0 or, α = -θ-90°

Then, cos(ωt+α) = cos(ωt- θ – 90°) = sin(ωt – θ)

So, alternating current in the circuit,

⇒ \(I=\omega Q_0 \cos (\omega t+\theta)\)

= \(\frac{V_0}{Z} \sin (\omega t-\theta)=I_0 \sin (\omega t-\theta)\) → (3)

Here \(I_0=\frac{V_0}{Z}=\frac{V_0}{\sqrt{R^2+\left(\omega L-\frac{1}{\omega C}\right)^2}}\) → (4)

From equations (1) and (3) we conclude,

Current lags behind voltage by a phase angle θ where \(\tan \theta=\frac{\sin \theta}{\cos \theta}=\frac{\frac{(\omega L-1 / \omega C)}{Z}}{\frac{R}{\bar{Z}}}=\frac{\omega L-\frac{1}{\omega C}}{R}\)

i.e., \(\theta=\tan ^{-1}\left(\frac{\omega L-\frac{1}{\omega C}}{R}\right)\) → (5)

This phase relation is shown. Now if V_L < V_C i.e., \(\omega L<\frac{1}{\omega C}\) will be negative. In such case, current I leads voltage V by angle θ. The voltage across the resistance R i.e., V_R and current I are in the same phase. On the other hand, the current I lags behind V_L by 90° but leads V_C by a 90° phase angle.

It is clear that M V_L< V_C i.e., if \(\omega L<\frac{1}{\omega C}\), I leads V by an angle θ.

2. In an LCR circuit, Z plays the same role as R in a pure resistive circuit. So, Z is the impedance of the circuit.

Impedance, \(Z=\sqrt{R^2+\left(\omega L-\frac{1}{\omega C}\right)^2}=\sqrt{R^2+X^2}\)

where \(X=\omega L-\frac{1}{\omega C}=X_L-X_C\) = reactance of the circuit.

Impedance in an LCR circuit is the effective resistance of the circuit arising from the combined effect of ohmic resistance and reactance of the circuit.

shows the impedance triangle for the circuit.

Power In Series LCR Circuit:

Here, the power factor of the circuit, \(\cos \theta=\frac{R}{Z}\)

Therefore power consumed in the circuit

∴ \(P=\frac{1}{2} V_0 I_0 \cos \theta=\frac{1}{2}\left(I_0 Z\right) I_0 \cdot \frac{R}{Z}=\frac{1}{2} I_0^2 R=\left(\frac{I_0}{\sqrt{2}}\right)^2 R\)

i.e., P = I²_rms R

This indicates that in this circuit power is dissipated neither in the inductor nor in the capacitor, but in the resistor only.

Hence currents through L or C are wattless.

The effective resistance of an LCR alternating current circuit is essentially the impedance of that circuit.

The inverse of impedance is known as admittance, i.e., admittance = \(\frac{1}{Z}\). Its unit is ohm” 1 or Siemens.

A pure resistance R opposes the current in any circuit and electrical energy is dissipated through it.

An inductive reactance X_L and a capacitive reactance X_C also oppose this current in a circuit, but no energy is dissipated through a pure inductor or a pure capacitor.

Both X_L = coL and X_C = depend on the frequency \(X_C=\frac{1}{\omega C}\) a) of the applied voltage. Clearly, for a dc voltage, (o = 0 . so that X_L = 0, but X_C = ∞. This means that dc passes freely through a pure inductor, but is blocked by a pure capacitor, which acts like an open switch.

Units of impedance and reactance: Both quantities have the same unit ohm (Ω).

Advantages Of The Capacitor Dependent Regulator Over A Resistor Dependent Regulator: In a resistor-dependent regulator the resistance also dissipates some energy which heats the regulator.

• On the other hand, only a capacitor but no resistor is used in an electronic regulator. Current in the circuit can be changed by regulating the capacitance of the capacitor and hence the speed of an electric fan can be controlled at will.
• As the current flowing through a pure capacitive ac circuit is wattless, such a regulator almost does not dissipate any power.
• So power is saved more in a capacitor-dependent regulator than in a resistor-dependent regulator. An electric regulator is useful in many devices—running an electric fan is just one of them.

Series resonance: It is observed from equation (4) that for \(\omega L=\frac{1}{\omega C}\), the denominator becomes minimum, and hence the current I will be maximum. This phenomenon is called the series resonance of the LCR circuit. If f₀ is the frequency for which the circuit reaches the above state, the condition for resonance, then

\(\omega_0 L=\frac{1}{\omega_0 C} \quad\left[\omega_0=2 \pi f_0\right]\) → (6)

i.e., \(\omega_0=\frac{1}{\sqrt{L C}}\)

or, \(f_0=\frac{\omega_0}{2 \pi}=\frac{1}{2 \pi \sqrt{L C}}\) → (7)

This frequency f₀ is called resonant frequency. The particular frequency of current In an LCR series circuit for which inductive reactance (X_L) and capacitive reactance (X_C) become equal to each other is called resonant frequency.

Under this condition, the circuit is termed a resonant circuit. Thus, whatever the value of frequency other than f₀, current I i.e., Irms is always less than its maximum value I.

The change of ac with angular frequency in an LCR circuit is shown graphically. It is known as the resonance curve. As \(I_m=\frac{V_0}{R}, I_m\) increases with the decrease of R.

Properties:

1. In the case of resonance,

∴ \(\omega_0 L=\frac{1}{\omega_0 C} \text { or, } X_L=X_C\)

which indicates that the inductive and capacitive reactance balance each other. Hence the circuit becomes equivalent to a pure resistive circuit.

2. According to equation (5), for resonance,

∴ \(\tan \theta=\frac{0}{R}=0 \text { or, } \theta=0\)

i.e., in a resonance circuit, alternating voltage V and alternating current I are in equal phases. In this condition, according to equation (4) the maximum value of I₀ i.e., Im = V₀/R, which is equivalent to a pure resistive circuit.

3. LCR series circuit finds use in the receiver of a radio set. The resonant frequency of this LCR circuit is tuned with the frequency of the signal transmitted from a radio station.

Hence resonance occurs. As a result, the magnitude of the current increases a lot and the transmitted signal can easily be received. LCR series circuit is also known as acceptor circuit.

Sharpness Of Resonance And Q-Factor: Power consumed at LCR series circuit,

∴ \(P=\frac{1}{2} V_0 I_0 \cos \theta\)

where cosθ = power factor = \(\frac{R}{Z}\)

∴ \(P=\frac{1}{2} V_0 I_0 \cdot \frac{R}{Z}=\frac{1}{2} V_0 \cdot \frac{V_0}{Z} \cdot \frac{R}{Z}\)

= \(\frac{1}{2} V_0^2 \frac{R}{R^2+\left(\omega L-\frac{1}{\omega C}\right)^2}\) → (8)

At resonance, \(\omega=\omega_0 \text { and } \omega_0 L=\frac{1}{\omega_0 C}\) and as a result power dissipation becomes maximum. From equation (8) we come to the value of maximum power dissipation, i.e.,

∴ \(P_m=\frac{V_0^2 R}{2 R^2}=\frac{V_0^2}{2 R}\) → (9)

From (8) and (9) we get,

∴ \(P=P_m \cdot \frac{R^2}{R^2+\left(\omega L-\frac{L}{\omega C}\right)^2}\) → (10)

This shows the average power (P) versus frequency (ω) curve using the same circuit parameters.

The points A and B have some special importance. Each of these points denotes half maximum power \(\left(\frac{P_m}{2}\right)\).

The rapidity with which resonance phenomena arise and then disappear is a measurement of the sharpness of resonance. Resonance will be sharp if the value of bandwidth (Δω) is small.

This is of course possible only when the resonance curve falls steeply around ω = ω₀.

Putting \(P=\frac{P_m}{2}\) in equation (10) we get,

= \(\frac{1}{2}=\frac{R^2}{R^2+\left(\omega L-\frac{1}{\omega C}\right)^2} \quad \text { or, }\left(\omega L-\frac{1}{\omega C}\right)= \pm R\)

or, \(\omega^2-\frac{1}{L C}= \pm \frac{R}{L} \omega\) → (11)

But we have \(\omega_0^2=\frac{1}{L C}\) (from series resonance condition)

Therefore from equation (11),

= \(\omega^2-\omega_0^2= \pm \frac{R}{L} \omega\)

i.e., \(\omega^2-\frac{R}{L} \omega-\omega_0^2=0 \quad \text { and } \omega^2+\frac{R}{L} \omega-\omega_0^2=0\)

Solving these quadratic equations neglecting the negative values of ω we get,

∴ \(\omega_1=\frac{R}{2 L}+\left(\omega_0^2+\frac{R^2}{4 L^2}\right)^{1 / 2}\)

and \(\omega_2=-\frac{R}{2 L}+\left(\omega_0^2+\frac{R^2}{4 L^2}\right)^{1 / 2}\)

Hence, bandwidth \((\Delta \omega)=\omega_1-\omega_2=\frac{R}{L}\) → (12)

and Q-factor = \(\frac{\omega_0}{\Delta \omega}=\frac{\omega_0 L}{R}=\frac{1}{\sqrt{L C}} \cdot \frac{L}{R}=\frac{1}{R} \sqrt{\frac{L}{C}}\) → (13)

Q-factor measures the sharpness of resonance in an LCR circuit. Inevitably, as Acy goes lesser, the Q-factor becomes greater and so resonance becomes sharper.

From equation (13), \(Q=\frac{\omega_0 L}{R}=\frac{X_L}{R}=\frac{I X_L}{I R}=\frac{V_L}{V_R}\)

where V_L = voltage difference across inductor

and V_R = voltage difference across the resistor

At resonance, V_R = applied voltage ( V)

and V_L = V_C [V_C = voltage difference across capacitor]

∴ \(Q=\frac{V_L}{V}=\frac{V_C}{V}\)

i.e., voltage difference (also called voltage drop) across inductor or capacitor concerning applied voltage in LCR series circuit is termed as Q-factor of the circuit.

The Q-factor is often much greater than 1. This means that in a series resonant circuit, V_L and V_C are much greater than the applied voltage V. This signifies a prominent voltage amplification across the inductor L and capacitor C.

Q-factor is a dimensionless parameter, which denotes the degree of damping of a resonator or oscillator. The more the value of Q, the less the rate of dissipation of energy concerning die energy stored, and the less the damping of the oscillator.

Class 12 Physics Electromagnetic Induction And Alternating Current

Alternating Current Power Consumed In Ac Circuits Numerical Examples

Example 1. In an LCR series ac circuit R = 10Ω, L = 50mH, and C = 5μV. Find out the resonant frequency and Q-factor. Find also the bandwidth and half-power frequencies.

Solution:

Here, L = 50mH = 5 x 10^-2H;

C = 5μF = 5 X 10^-6F

∴ Resonant frequency,

⇒ \(f_0=\frac{1}{2 \pi \sqrt{L C}}=\frac{1}{2 \times 3.14 \times \sqrt{\left(5 \times 10^{-2}\right) \times\left(5 \times 10^{-6}\right)}}\)

⇒ \(\frac{10^4}{2 \times 3.14 \times 5}=318.5 \mathrm{~Hz}\)

and Q-factor = \(\frac{1}{R} \sqrt{\frac{L}{C}}=\frac{1}{10} \sqrt{\frac{5 \times 10^{-2}}{5 \times 10^{-6}}}=\frac{1}{10} \times 100=10\)

Now, \(Q=\frac{\omega_e}{\Delta \omega}=\frac{f_0}{\Delta f}\)

∴ Bandwidth, \(\Delta f=\frac{f_0}{Q}=\frac{318.5}{10}=31.85 \mathrm{~Hz}\)

Half-power frequencies are,

⇒ \(f_1=f_0-\frac{\Delta f}{2}=318.5-\frac{31.85}{2}=302.6 \mathrm{~Hz}\)

and \(f_2=f_1+\frac{\Delta f}{2}=302.6+15.92=318.52 \mathrm{~Hz}\)

Example 2. In an LCR series combination, R = 400Ω, L = JOOmH, and C = 1 μF. This combination is connected to a 25 sin 2000t volt source. Find

1. The impedance,
2. Peak value of current,
3. The phase difference of voltage and current,
4. Power factor and
5. Dissipated power in the circuit.

Solution:

Applied ac voltage V = 25 sin 2000 t volt

Peak value of voltage, V₀ = 25V;

angular frequency, ω = 2000 Hz

and L = 100 mH = 0.1H

So, ωL = 2000 x 0.1 =200Ω

Here, \(C=1 \mu \mathrm{F}=10^{-6} \mathrm{~F} \quad \text { So, } \frac{1}{\omega C}=\frac{10^6}{2000}=500 \Omega\)

1. Impedance of the circuit,

∴ \(Z=\sqrt{R^2+\left(\omega L-\frac{1}{\omega C}\right)^2}\)

= \(\sqrt{(400)^2+(200-500)^2}=500 \Omega\)

2. Peak value of current, \(I_0=\frac{V_0}{Z}=\frac{25}{500}=0.05 \mathrm{~A}\)

3. If the phase difference between voltage and current is θ then,

∴ \(\tan \theta=\frac{\omega L-\frac{1}{\omega C}}{R}=\frac{200-500}{400}=-\frac{3}{4}=\tan \left(-36.9^{\circ}\right)\)

i.e., current leads voltage by a phase angle of 36.9°.

4. Power factor of the circuit,

∴ \(\cos \theta=\frac{R}{Z}=\frac{400}{500}=0.8\)

5. Power dissipated,

∴ \(P=\frac{1}{2} V_0 I_0 \cos \theta=\frac{1}{2} \times 25 \times 0.05 \times 0.8=0.5 \mathrm{~W}\)

Example 3. The power factor of an LR circuit is \(\frac{1}{\sqrt{2}}\). If the frequency of ac is doubled, what will be the power factor?

Solution:

Power factor, \(\cos \theta=\frac{R}{Z}=\frac{1}{\sqrt{2}}\)

∴ Z² = 2R² or, R² + (ωL)² = 2R² or, (ωL)² = R²

Now if the angular frequency co is doubled,

(ω’L)² = (2ωL)² = 4(ωL)² = 4R²

So, impendance, \(Z^{\prime}=\sqrt{R^2+\left(\omega^{\prime} L\right)^2}=\sqrt{R^2+4 R^2}=R \sqrt{5}\)

Hence, power factro, \(\frac{R}{Z^{\prime}}=\frac{R}{R \sqrt{5}}=\frac{1}{\sqrt{5}}\)

Example 4. f the value of inductor L is 1 mH and the applied ac source frequency is 50 Hz, find the inductive reactance in the above case.

Solution:

Here,1 = 1 mH = 10^-3 H

and ω = 2πf= 2 x 3.14 x 50 = 314 Hz

∴ X_L = ωL = 314 x 10^-3 = 0.314 Ω

Example 5. A series LC circuit has L = 0.405-H and C = 25 μF. The resistance R is zero. Find the frequency of resonance.

Solution:

Here, L = 0.405 H, C = 25 μF = 25 x 10^-6 F

∴ \(f_r=\frac{1}{2 \pi \sqrt{L C}}=\frac{1}{2 \times 3.14 \times \sqrt{0.405 \times 25 \times 10^{-6}}}=50 \mathrm{~Hz}\)

Example 6. An inductor and a capacitor of reactances 25Ω and 75Ω, respectively, are connected across a 250 V ac source in series. Find the potential difference between the inductor and the capacitor. Establish their relationship with the main voltage.

Solution:

The impedance of the series circuit having capacitor and inductor,

Z = X_C-X_L = (75-25) Ω = 50Ω

So, current, \(I=\frac{250}{50}=5 \mathrm{~A}\)

∴ Potential differences across the inductor,

V_L = 5 x 25V = 125V

and potential differences across the capacitor,

V_C = 5 x 75 V = 375 V

Now, main voltage,

V = 250 V

Hence, the relationship between V, V_L and V_C is

V = V_C – V_L

Example 7. A capacitor, a resistor of 5Ω, and an inductor of 50 mH are in series with an ac source marked 100 V, 50Hz. It is found that the voltage is in phase with the current. Calculate the capacitance of the capacitor and the impedance of the circuit.

Solution:

Since both the voltage and current of the circuit are in the same phase, the circuit is purely resistive. So impedance of the circuit, Z = R = 5 Ω

Frequency of the resonant circuit,

∴ \(f=\frac{1}{2 \pi \sqrt{L C}}\)

∴ \(C=\frac{1}{4 \pi^2 L f^2}=\frac{49}{4 \times 484 \times 50 \times 10^{-3} \times 50 \times 50}\)

= 2.03 x 10^-4 F

Hence, the capacitance is 2.03 x 10^-4 F and the impedance is 5Ω.

Example 8. A capacitor and a resistor are connected in series with an ac source. If the potential differences across C, R are 120 V, 90 V respectively and If the rms current of the circuit; is 3 A, calculate

1. The impedance and
2. The power factor of the circuit.

Solution:

Alternating voltage in the circuit,

∴ \(V=\sqrt{V_R^2+V_C^2}=\sqrt{90^2+120^2}=150 \mathrm{~V}\)

Now, current through the circuit, I = 3 A

1. Impedance of the circuit, \(Z=\frac{\dot{V}}{I}=\frac{150}{3} \Omega=50 \Omega\)
2. Power factor of the circuit, \(\cos \theta=\frac{V_R}{V}=\frac{90}{150}=0.6\)

Example 9. A 200μF capacitor in series with a 50 Ω resistor is connected to a 220 V, 50 Hz ac source.

1. What is the maximum current in the circuit?
2. What is the difference in time when the current and the voltage attain maximum values?

Solution:

Angular frequency of the source,

ω = 2πf=2 x 3.14 x 50Hz

C = 200μF = 2 x l0^-4 F

Maximum current passing through the circuit

∴ \(I_0=\frac{V_0}{\sqrt{R^2+\frac{1}{C^2 \omega^2}}}\)

= \(\frac{\sqrt{2} \times 220}{\sqrt{(50)^2+\frac{1}{\left(2 \times 10^{-4}\right)^2 \times 4 \times(3.14)^2 \times(50)^2}}}\)

= 5.93 A

Now, if θ is the phase angle, then

= \(\tan \theta=\frac{1}{\omega C R}=\frac{1}{2 \pi f C R}=\frac{1}{2 \times 3.14 \times 50 \times 2 \times 10^{-4} \times 50}\)

= 0. 3185

or, \(\theta=\tan ^{-1}(0.3185)=17.67^{\circ}=\frac{17.67 \times \pi}{180} \mathrm{rad}\)

If the voltage and the current attain maximum value at a time different from t, then

θ = ωt

or, \(t=\frac{\theta}{\omega}=\frac{17.67 \times \pi}{180 \times 2 \pi \times 50}=9.82 \times 10^{-4} \mathrm{~s}\)

Example 10. A resistor, R = 300 Ω, and a capacitor, C = 25 μF are connected in series with an ac source. The peak value of voltage ( V₀) and the frequency (f) of the source is 150 V and \(\frac{50}{\pi}\) Hz respectively. Find the peak value of the current and the power dissipated in the circuit.

Solution:

If ω is the angular frequency of the ac source; then

⇒ \(\frac{1}{\omega C}=\frac{1}{2 \pi \times \frac{50}{\pi} \times 25 \times 10^{-6}}=400 \Omega\)

Thus peak value of the current,

⇒ \(I_0=\frac{V_0}{\sqrt{R^2+\frac{1}{\omega^2 C^2}}}=\frac{150}{\sqrt{300^2+400^2}}=0.3 \mathrm{~A}\)

Hence, the power dissipated in the circuit

⇒ \(\frac{1}{2} I_0^2 R=\frac{1}{2} \times(0.3)^2 \times 300=13.5 \mathrm{~W}\)

Example 11. A series LCR circuit containing a resistance of 120 Ω has an angular resonance frequency of 4 x 10⁵ rad/s. At resonance, the voltages across resistance and inductance are 60 V and 40 V, respectively. Find the values of L and C. At what frequency, does the current in the circuit lag behind the voltage by 45°?

Solution:

At resonance, X_L = X_C

∴ \(I=\frac{V_R}{R}\)

= \(\frac{60}{120}\) [voltage across the resistance, V_R = 60 v]

= 0.5 A

Now, voltage across the inductor,

V_L = IX_L = IωL

∴ or, \(L=\frac{V_L}{I \omega}=\frac{40}{0.5 \times 4 \times 10^5}\) [∵ angular frequency, w = 4 x 10⁵ rad/s] [angular frequency, = 4 x 10⁵ rad/s]

= 2 x 10^-4 H

We know that at resonance,

⇒ \(X_L=X_C \quad \text { or, } \omega L=\frac{1}{\omega C}\)

or, \(C=\frac{1}{\omega^2 L}=\frac{1}{\left(4 \times 10^5\right)^2 \times 0.2 \times 10^{-3}}=3.125 \times 10^{-8} \mathrm{~F}\)

Let the angular frequencyÿ ω₁ when the current lags behind the voltage by 45°.

∴ \(\tan 45^{\circ}=\frac{\omega_1 L-\frac{1}{\omega_1 C}}{R}\)  \(C=3.125 \times 10^{-8} \mathrm{~F}=\frac{1}{32} \times 10^{-6} \mathrm{~F}\)

or, \(1 \times 120=\omega_1 \times 2 \times 10^{-4}-\frac{1}{\omega_1\left(\frac{1}{32}\right) \times 10^{-6}}\)

or, \(\omega_1^2-6 \times 10^5 \omega_1-16 \times 10^{10}=0\)

The physically meaningful solution of the above equation is,

∴ \(\omega_1=\frac{6 \times 10^5+10 \times 10^5}{2}=8 \times 10^5 \mathrm{rad} / \mathrm{s}\)

Example 12. In an LR series circuit, a sinusoidal voltage V = V₀Sin ωt is applied. It is given that L = 35 mH, R = 11Ω, V_rms = 220V, \(\frac{\omega}{2 \pi}=50 \mathrm{~Hz} \text { and } \pi=\frac{22}{7}\). Find the amplitude of the current in the steady state and obtain the phase difference between the current and the voltage. Also, plot the variation of current for one cycle on the given graph.

Solution:

Inductive reactance,

∴ \(X_L=\omega L=(2 \pi)(50)\left(35 \times 10^{-3}\right) \approx 11 \Omega\)

Impedance, Z = \(\sqrt{R^2+X_L^2}=\sqrt{11^2+11^2}=11 \sqrt{2} \Omega\)

Given V_rms = 220V

Hence, amplitude of voltage, V₀ = 2 V_rms = 22072 V

∴ Amplitude of current, \(I_0=\frac{V_0}{Z}=\frac{220 \sqrt{2}}{11 \sqrt{2}}=20 \mathrm{~A}\)

The phase difference between the current and the voltage in the circuit,

∴ \(\theta=\tan ^{-1}\left(\frac{X_L}{R}\right)=\tan ^{-1}\left(\frac{11}{11}\right)=\frac{\pi}{4}\)

In the LR circuit, voltage leads the current by phase angle θ. Thus current in the current circuit,

∴ \(I=I_0 \sin (\omega t-\theta)=20 \sin \left(\omega t-\frac{\pi}{4}\right)\)

Class 12 Physics Electromagnetic Induction And Alternating Current

Alternating Current LG Oscillations

In the circuit shown, S₁S₂S₃ is a two-way switch. To charge the capacitor C first, one has to connect the switch S₁S₂. If S₁S₃ is connected, the charged capacitor and the inductance L constitute an LC circuit.

• As soon as the switch is disconnected from S₂, no current flows through the circuit anymore. In this condition if the amount of charge in the capacitor C is Q, the energy stored in the electric field of the capacitor is \(E_C=\frac{1}{2} \frac{Q^2}{C}\), but the inductor possesses no energy, i.e., E_L = 0.
• Due to the accumulation of charge in the capacitor, it acts as a battery. Hence immediately after connecting the switch S₁S₃, the capacitor starts to send current in the LC circuit thus forming.
• As this discharging current gradually rises, a magnetic field starts developing in the inductor. At this stage, it may be said that the capacitor is being discharged gradually through the inductor L.
• After some time, the capacitor C becomes completely discharged and the current of the LC circuit reaches a peak value I₀ i.e., Ec becomes zero and \(E_L=\frac{1}{2} L I_0^2\).

It means that the electrical energy in the capacitor has been transferred fully into the magnetic energy in the inductor.

Again, due to the flow of current in the LC circuit, the capacitor is being charged in the reverse direction i.e., the capacitor is now charged with a polarity opposite to its initial state.

• Obviously with the increase of charge in the capacitor, the energy stored in its electric field begins to increase and the energy stored in the magnetic field of the inductance gradually decreases.
• Hence the magnitude of the current decreases and finally comes to zero. Thus after some time, E_L becomes zero again, and \(E_C=\frac{1}{2} \frac{Q^2}{C}\). This process of discharging and charging of the capacitor occurs alternately.
• We know that a capacitor can store electric energy whereas an inductor can store magnetic energy in it. Now a charged capacitor with an inductor is connected to an ac circuit, and a periodical energy transformation starts.

The energy of the capacitor is converted to the energy of the inductor and back again. This phenomenon is called LC oscillations.

In the above discussion, one half-cycle of this oscillation has been discussed. At the end of the next half-cycle, the circuit comes back to its initial condition.

• The polarity of the plates becomes equal to its initial state and the current circuit becomes zero again. The direction of the current in the second half cycle is just opposite to that in the first half cycle.
• So the current in an LC circuit fluctuates periodically between the peal values I0 and -I₀.
• The resistance of a pure inductor is zero so there is no energy loss due to the Joule effect. So as time elapses, no loss of total energy takes place, i.e., there is no damping of LC oscillations.
• So the peak value of the alternating current in the LC circuit remains unchanged. But practically, the resistance of any coil cannot be ignored. Thus, there is always some resistance in the circuit due to which some energy is lost in the form of heat.
• So the current remains oscillatory, but is damped, To maintain the alternating current, the circuit must be supplied with the same amount of energy as is being lost during each cycle, from some external source.
• If damping is absent, the frequency of LC oscillations, \(f=\frac{1}{2 \pi \sqrt{L C}}\). When this frequency becomes equal to the frequency of the applied alternating emf then resonance occurs in that circuit.

Oscillators convert direct current (dc) from a power supply to an alternating current signal. LC circuits are used in many cases as an important component of an oscillator.

Class 12 Physics Electromagnetic Induction And Alternating Current Chapter 2 Alternating Current LG Oscillations Numerical Examples

Example 1. A 220 V, 50 Hz ac source is connected to an inductance of 0.2H and a resistance of 20 Ω in series. What is the current in the circuit?

Solution:

rms value of current, \(I_{\mathrm{rms}}=\frac{E_{\mathrm{rms}}}{\sqrt{R^2+(\omega L)^2}}\)

Here, = 220 V,

ω = 2πf = 2 x 3.14 x 50 = 314 Hz

∴ \(\sqrt{R^2+(\omega L)^2}=\sqrt{(20)^2+(314 \times 0.2)^2}\)

∴ \(I_{\mathrm{rms}}=\frac{220}{\sqrt{(20)^2+(314 \times 0.2)^2}}=3.34 \mathrm{~A}\).

Example 2. An ac source of frequency 50 Hz is connected with a resistance (R = 36Ω) and L of 0.12 H in series. What is the phase difference between current and voltage?

Solution:

If θ is the phase difference, then

∴ \(\tan \theta=\frac{\omega L}{R}\)

or, \(\theta=\tan ^{-1} \frac{\omega L}{R}=\tan ^{-1} \frac{314 \times 0.12}{36}=\tan ^{-1}(1.047)=46.3^{\circ}\)

[here ω = 2π x 50 = 2 x 3.14 x 50 = 314 Hz]

So, phase difference = 46.3°.

Example 3. A current of 1 A flows in a coil when connected to a 100 V dc source. If the same coil is connected to a 100 V, 50 Hz ac source, a current of 0.5 A flows in the coil. Calculate the inductance of the coil.

Solution:

If R is the resistance of the coil, in dc circuit \(\frac{E}{I}=R\)

or, \(R=\frac{100}{1}=100 \Omega\)

In ac circuit, \(I_{\mathrm{rms}}=\frac{E_{\mathrm{rms}}}{\sqrt{R^2+(\omega L)^2}}\)

or, \(\sqrt{R^2+(\omega L)^2}=\frac{E_{\mathrm{rms}}}{I_{\mathrm{rms}}}=\frac{100}{0.5}=200 \Omega\)

∴ R² + (ωL)² = (200)²

or, \(\omega L=\sqrt{(200)^2-(100)^2}=100 \sqrt{3} \Omega\)

or, \(L=\frac{100 \sqrt{3}}{\omega}=\frac{100 \times 1.732}{2 \times 3.14 \times 50}=0.55 \mathrm{H}\)

Example 4. A lamp in which 10 A current can flow at 15 V is connected with an alternating source of potential 220 V and frequency 50 Hz. What should be the inductance of the choke coil required to light the bulb?

Solution:

Resistance of the lamp, \(R=\frac{15}{10}=1.5 \Omega\)

To send 10 A current through the lamp, the required impedance of the ac circuit, \(Z=\frac{220}{10}=22 \Omega\)

Now if L is the inductance of the choke coil and its resistance is negligible, then

∴ \(Z=\sqrt{R^2+(\omega L)^2} \text { or, } \omega L=\sqrt{Z^2-R^2}\)

∴ \(L=\frac{1}{\omega} \sqrt{Z^2-R^2} ;[\omega=2 \pi f=2 \times 3.14 \times 50=314 \mathrm{~Hz}]\)

= \(\frac{1}{314} \sqrt{(22)^2-(1.5)^2}=0.07 \mathrm{H}\)

Example 5. What will be the peak value of alternating current when a condenser of 1 μF is connected to an alternating voltage of 200 V, 60 Hz?

Solution:

c = 1μF = 10^-6F; ω = 2πf = 2 x 3.14 x 60 Hz

Peak value of current,

\(I_0=I_{\mathrm{rms}} \times \sqrt{2}=\frac{E_{\mathrm{rms}}}{\frac{1}{\omega C}} \cdot \sqrt{2}=E_{\mathrm{rms}} \cdot \omega C \sqrt{2}\)

= 200 x (2 x 3.14 x 60) x 10^-6 x 1.414

= 0.106 A (approx.)

Class 12 Physics Electromagnetic Induction And Alternating Current Alternating Current Transformer

The electrical appliance used to increase or decrease alternating voltage is called a transformer.

The transformer, which increases the voltage is called a step-up transformer, and the transformer used to decrease the voltage is called a step-down transformer.

Transformer works on the principle of mutual induction between a pair of colls.

Description: The core of a transformer is constructed by several thin laminated sheets of soft iron placed one Over the other. It is known as a laminated core.

A core of a special shape is so chosen that no part of the magnetic flux is wasted and hence the density of lines of induction inside the core becomes maximum. Two insulated wires are wound in many turns on the middle arm of the core very close to each other.

Open coil acts as the primary (P) and the other as the secondary (S).

Working Principle: An alternating voltage (V_p) is applied to the primary coil from an alternating current source.

The alternating current in coil P generates induced emf in the secondary coil S, i.e., an alternating voltage Vs is generated between the ends of S.

If the dissipation of magnetic flux and loss of energy due to heating is neglected in this transformer (called an ideal transformer), it can be proved that,

∴ \(\frac{V_s}{V_p}=\frac{N_s}{N_p}=k\)

where N_p and N_s are the total numbers of turps of the primary and secondary coils, respectively, and k is called the turns ratio or transformer ratio.

• Because of the special nature of the winding, it can be assumed fairly correctly that the magnetic flux Φ_B associated with each turn of primary and secondary is the same. If e is the induced emf in each turn, \(e=-\frac{d \phi_B}{d t}\).
• Hence the emf induced in the primary oil \(V_p=N_p e=-N_p \frac{d \phi_B}{d t}\). Similarly, \(V_s=-N_s \frac{d \phi_B}{d t}\). Now, the emf induced in the primary must necessarily be equal to the voltage applied. Then, by dividing the two \(\frac{V_s}{V_p}=\frac{N_s}{N_{\dot{p}}}=k\).

If N_s > N_p, i.e., k > 1, V_s > V_p we get a step-up transformer.

If N_s < N_p, i.e., k < 1, V_f < V_p we get a step-down transformer.

Uses: Transformers are widely used in our daily lives. With its help, a high voltage can be converted into a low voltage and vice versa, whenever necessary. For example,

1. An electrical power station uses a step-up transformer to produce and transmit large amounts of ac electrical energy over long distances.
2. The energy is transmitted at high voltage (such as 66000V- 132000 V) to reduce the loss of energy due to heating.
3. But supply to domestic area needs low voltages (such as 110 V-440 V). Such conversions of voltage at different levels are facilitated by step-down, transformers.
4. Radio, television, electric bell,s and other electrical appliances require small-sized transformers.

Energy Loss In Transformer: In an ideal transformer, the power dissipated in the primary coil (I_pV_p) = power dissipated in the secondary coil (I_sV_s).

But no transformer, in practice, is ideal. Generally, Some input energy is wasted in any transformer and hence V_sI_s < V_pI_p .The term V_sI_s / V_pI_p is called the efficiency of a transformer.

The main causes of energy loss and their remedies are given below.

1. Copper Loss: Generally copper wire is used to make primary and secondary coils. Due to Joule’s heating, some energy is wasted as heat energy.
   • Remedies: Thick wire should be used to reduce this loss of energy.
2. Iron Loss: Iron core should be used in primary and secondary cords.
   1. Due to the eddy current in this core, energy loss is unavoidable.
      • Remedies: To reduce such loss, a laminated core is used.
   2. The change of magnetization cycle in the core fails to synchronize with ac and some energy is necessarily wasted in the core—known as hysteresis loss.
      • Remedies: Due to high coercivity, the core should not be made of steel. The iron core is more effective, in reducing energy loss.
3. Loss due to magnetic flux leakage: The flux generated by the primary coil may not be wholly linked to the secondary coil due to possible defective design of the core.
   • Remedies: Obviously, special care should be taken in the construction of the core.

Classification Of Transformer: The two most common designs of the transformer are given below:

Core-Type Transformer: In this type, primary and secondary coils are wound around the core ring. Here every limb is occupied with both primary and secondary winding placed successively around them.

Shell-Type Transformer: Here the primary and secondary windings pass inside the steel magnetic circuit (core) which forms a shell around the windings, The main frame is constructed with three limbs. Both the primary and secondary windings are wound around the central limb.

Besides these, transformers can be classified based on their and these are audio frequency transformers, radio frequency transformers, etc.

Class 12 Physics Electromagnetic Induction And Alternating Current

Alternating Current Transformer Numerical Example

Example 1. The number of turns in the primary aid secondary coils of an ideal transformer is 140 and 280, respectively. If the current through the primary coil is 4 A, what will be the current in the secondary coil?

Solution:

In an ideal transformer, the secondary and primary coils are equal,

i.e. V_sI_s ± V_pI_p

∴ \(I_s=I_p \cdot \frac{V_p}{V_s}=I_p \cdot \frac{N_p}{N_s}=4 \times \frac{140}{280}=2 \mathrm{~A}\)

Example 2. The initial voltage and Input power of a transformer of efficiency 80% are 100 V and 4 kW, respectively. If the voltage of the secondary coil Is 200 V, determine the currents flowing through the primary and the secondary coil.

Solution:

Power of the primary coil i.e., input power

P_p = V_pI_p

or, \(I_p=\frac{P_p}{V_p}=\frac{4 \times 1000}{100}=40 \mathrm{~A}\)

Power of the secondary coil, \(P_s=P_p \times \frac{80}{100}\)

Again, P_s = V_sI_s

So, \(I_s=\frac{P_s}{V_s}=\frac{80}{100} \times \frac{P_p}{V_s}=\frac{80}{100} \times \frac{4 \times 1000}{200}=16 \mathrm{~A}\)

 

Class 12 Physics Electromagnetic Induction And Alternating Current  Alternating Current Very Short Questions And Answers

Properties Of Alternating Voltage And Current

Question 1. If the frequency of an alternating emf is 50 Hz, how many times the direction of emf will be reversed per second?

Answer: 100

Question 2. What percentage of its peak value is the rms value of an ac?

Answer: 70.7

Question 3. What is the peak value of the voltage of a 220 V ac line?

Answer: 311 V

Question 4. If an alternating current is represented by I = sin l00 π mA, what is its peak value?

Answer: 1 mA

Question 5. If an alternating current is represented by, I = sin 100 πt mA, then what is the frequency of that current?

Answer: 50 Hz

Question 6. After what time will the direction of current in an electric supply of frequency 50 Hz be reversed?

Answer: 0.01 s

Question 7. An alternating source of emf E = E₀ sinwt and negligible resistance is connected directly to an ac voltmeter. What reading will it show?

Answer: \(\left[\frac{E_0}{\sqrt{2}}\right]\)

Question 8. What changes are observed in the rms value of an ac with changes in the frequency

Answer: No change

Question 9. What is the rms value of an alternating current, I = I₀ sin ωt?

Answer: \(\left[\frac{I_0}{\sqrt{2}}\right]\)

Question 10. What is the ratio between the peak value and the average value of a sinusoidal emf?

Answer: \(\left[\frac{\pi}{2}\right]\)

Question 11. The instantaneous i current in an ac circuit is I = 6 sin 314t A. What is the rms value of current?

Answer: 4.24 A

Question 12. An alternating current is I = cos 100 πt A. Find out its frequency, peak value, and rms value.

Answer: 50 Hz, 1 A, 0.707 A

Question 13. Why a dc voltmeter and dc ammeter cannot read ac?

Answer: The average is zero in a cycle

Question 14. What will be the phase difference between current and emf when 220 V, 50 Hz ac source is connected to a circuit containing pure resistor?

Answer: Zero

Series AC Circuits With R, L, C

Question 15. What is the unit of impedance?

Answer: ohm

Question 16. What is the reactance of pressure resistances in an ac circuit?

Answer: Zero

Question 17. If an LCR circuit is connected to a dc source, what will be the current through the circuit?

Answer: zero

Question 18. What will be the reactance if a current of frequency f flows through an inductor of self-inductance L?

Answer: 2 π fL.

Question 19. What will be the reactance if a current of frequency f flows through a capacitor of capacitance C?

Answer: \(\left[\frac{1}{2 \pi f C}\right]\)

Question 20. If the frequency of an ac circuit is increased, how would the reactance of an inductor change?

Answer: Increase

Question 21. If the frequency of an ac circuit is increased, how would the reactance of a capacitor change?

Answer: Decrease

Question 22. In an LR circuit, the alternating current ________ the alternating emf by a phase current the alter angle.

Answer: Lags behind

Question 23. In a CR circuit, the alternating current ________ the alternating emf by a certain phase angle.

Answer: Lead

Question 24. In an alternating series LCR circuit, what is the phase difference between the voltage drops across L and C?

Answer: 180

Question 25. When does the LCR series circuit have minimum impedance?

Answer: At resonance

Question 26. What is the reactance of a capacitor of capacitance C at f Hz?

Answer: \(\left[-\frac{1}{2 \pi f C}\right]\)

Power In Ac Circuits

Question 27. What is the power factor of a circuit having pure resistance only?

Answer: Zero

Question 28. What is the power dissipated in an ac circuitin which voltage and current are given by \(V=230 \sin \left(\omega t+\frac{\pi}{2}\right)\) and I = I₀ sinωt?

Answer: Zero

LC Oscillations

Question 29. What is the natural frequency of an LC oscillator?

Answer: \(\left[\frac{1}{2 \pi \sqrt{L C}}\right]\)

Ac Generator And Transformer

Question 30. Indicate the change in emf produced by an ac dynamo in the following cases:

1. The magnetic field is doubled,
2. The angular velocity of the coil is decreased.

Answer: Will be doubled, will decrease

Question 31. If the area of the coil of an ac dynamo is halved, how would the emf generated change?

Answer: Halved

Question 32. If the angular velocity of the coil of an ac dynamo is doubled, how would the emfproduced change?

Answer: doubled

Question 33. By what factor would the output voltage of an ac generator change, if the number of turns in its coil is doubled?

Answer: 2

Question 34. The turns ratio of an ideal transformer is 4: 1. What will be the current in the secondary if that in the primary is 1.2A?

Answer: 4.8 A

Class 12 Physics Electromagnetic Induction And Alternating Current Alternating Current Synopsis Conclusion

In a dynamo, mechanical energy is converted into electrical energy.

In an electric motor, electric energy is converted into mechanical energy.

• The current whose direction in an electrical circuit reverses periodically in a definite time interval is called alternating current.
• The emf or potential difference whose direction reverses periodically in a definite time interval is called alternating emf or alternating potential difference.
• The electrical machine used to increase or decrease an alternating voltage is called a transformer. The transformer which increases the voltage is called a step-up transformer and which decreases the voltage is called a step-down transformer.
• The amount of power dissipation in any part of an AC circuit depends not only on ac voltage and current but also on their phase difference.
• The inductor or capacitor in an ac circuit resists the current just like a resistor. This resistance is called reactance. The pure resistance, inductive reactance, and capacitive reactance do not remain in the same phase.

The effective resistance against current in an ac circuit due to a combination of pure resistance, inductive reactance, and capacitive reactance is known as the impedance of the circuit. The magnitude of this impedance depends on the ac voltage and current.

• With the change of ac voltage frequency current also changes simultaneously.
• The frequency for which current becomes maximum is known as resonant frequency.
• In an LC circuit periodic interchange occurs between the stored energy in the electric field of the capacitor and that in the magnetic field of the inductor. This is LC oscillation.
• The power factor ofpure resistor is 1 i.e., it is the resistance that dissipates maximum power.
• The power factor of a pure inductor or capacitor is zero, i.e., they do not dissipate any power. Current through them is called wattless.

The emf induced in a coil rotating with uniform angular velocity ω in a uniform magnetic field intensity B, about an axis perpendicular to the field is,

e = NABω sin(ωt+ α) = e₀sin (ωt+α)

[The coil is of cross-sectional area A having N turns]

If the total resistance of the coil and the external circuit is R, the induced current

∴ \(i=\frac{e}{R}=\frac{\omega B A N}{R} \sin (\omega t+\alpha)=\frac{e_0}{R} \sin (\omega t+\alpha)\)

= i₀sin(ωt+ α)

Equation of an alternating emf,

V = V₀sin(ωt+ α) where V₀ = ωABN

Equation of an alternating current,

∴ \(I=\frac{V}{R}=\frac{V_0}{R} \sin (\omega t+\alpha)=I_0 \sin (\omega t+\alpha)\)

The average values of alternating voltage and current are respectively,

∴ \(\bar{V}=\frac{2 V_0}{\pi} \text { and } \bar{I}=\frac{2 I_0}{\pi}\)

rms values of the alternating voltage and current are respectively,

∴ \(V_{\mathrm{rms}}=\frac{V_0}{\sqrt{2}} \text { and } I_{\mathrm{rms}}=\frac{I_0}{\sqrt{2}}\)

Form factor for a sinusoidal wave, \(f=\frac{V_{\mathrm{rms}}}{\bar{V}}=\frac{\pi}{2 \sqrt{2}}=1.11\)

If the number of turns in the primary coil of a transformer = N_p, the number of turns in its secondary coil = N_s, and the ratio of the number of turns = k, then in the case of an ideal transformer,

∴ \(\frac{I_p}{I_s}=\frac{V_s}{V_p}=\frac{N_s}{N_p}=k\)

In an ideal transformer, input power (VpIp) = Output power ( VsIs ).

The efficiency of a transformer = \(\frac{V_s I_s}{V_p I_p}\); the efficiency of an ideal transformer = 1 or 100%.

Inductive reactance, X_L = ωL.

Capacitive reactance, \(X_C=\frac{1}{\omega C}\)

The impedance of an LCR series circuit,

∴ \(Z=\sqrt{R^2+\left(\omega L-\frac{1}{\omega C}\right)^2}\)

In an LCR series circuit, if ac voltage V = V₀ sin ωt then alternating current, I = I₀ sin(ωt- θ),

Where \(\dot{I}_0=\frac{V_0}{Z} \text { and } \tan \theta=\frac{\omega L-\frac{1}{\omega C}}{R}\)

Condition for resonance in an LCR series circuit,

∴ \(\omega L=\frac{1}{\omega C}\)

Resonant frequency, \(f_0=\frac{1}{2 \pi \sqrt{L C}}\)

Effective voltage magnification for series resonance,

∴ \(Q=\frac{V_L}{V_R}=\frac{V_C}{V_R}=\frac{\omega_0 L}{R}=\frac{1}{\omega_0 C R}=\frac{1}{R} \sqrt{\frac{L}{C}}\)

The natural frequency of an LC oscillator,

∴ \(f=\frac{1}{2 \pi \sqrt{L C}}\)

Power dissipation in an ac circuit

= \(V_{\mathrm{rms}} I_{\mathrm{rms}} \cos \theta\)

where cos θ = power factor.

Class 12 Physics Electromagnetic Induction And Alternating Current Alternating Current Assertion Reason Type Question And Answers

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, and 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: The peak values by alternating voltage and alternating current in a circuit are V0 andd0 respectively.

The phase difference between voltage and current is θ. Then the power consumed is V₀ I₀ cosθ.

Statement 2: The consumed power in an alternating circuit depends on the phase difference between the emf and current.

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

Question 2. Statement I: Q-factor of a series LCR circuit is \(\frac{1}{R} \sqrt{\frac{L}{C}}\).

Statement 2: The resonant frequency of an LCR circuit does not depend on the resistance of the circuit.

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

Question 3. Statement 1: If the total energy in an LC oscillator is equally distributed between the magnetic and electric fields then the charge stored in the capacitor is \(\frac{1}{\sqrt{2}}\) fraction of the maximum charge stored in the capacitor during oscillation.

Statement 2: The charge stored in the capacitor becomes maximum at a time when the total energy of the LC circuit is stored in the electric field.

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

Question 4. Statement 1: Form factor becomes different for different waveforms of alternating voltage and current

Statement 2: The mean value of alternating voltage or current = \(\frac{2}{\pi}\) x peak value and rms value = \(\frac{1}{\sqrt{2}}\) x peak value for any waveform.

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

Question 5. Statement 1: A series LCR circuit when connected to an ac source gives the terminal potential difference 50 V across each of resistor R, inductor L and capacitor C. Then the terminal potential difference across LC is zero.

Statement 2: The terminal alternating voltages across the inductor and capacitor in a series LCR Circuit in an opposite phase.

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

Question 6. Statement 1: If the value of the output voltage of an ideal transformer is half the value of the input voltage, then the output current will become twice.

Statement 2: No energy is dissipated in an ideal transformer

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

Question 7. Statement 1: The alternating current lags behind the voltage by a phase angle \(\frac{\pi}{2}\) when ac flows through an inductor.

Statement 2: The inductive reactance increases as the frequency of ac source decreases.

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

Question 8. Statement 1: An inductor acts as a perfect conductor for dc.

Statement 2: dc remains constant in magnitude and direction.

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

Class 12 Physics Electromagnetic Induction And Alternating Current Alternating Current Match the following

Question 1. Match the columns for a series LCR circuit.

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

Question 2. An LCR circuit (R = 40 Ω, L = 100mH, C = 0.242 μF) is connected with an ac voltage source of peak voltage 200 V and frequency 1000 Hz.

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

Question 3. Column I describes some action and column 2 the required device.

 

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

Question 4. In an LR circuit instantaneous voltage and instantaneous current are V = 100 sin100t and i = I₀ sin(100t – \(\frac{\pi}{4}\) respectively.

 

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

Question 5. Referring to the given circuit, match the following.

image

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

Class 12 Physics Electromagnetic Induction And Alternating Current Alternating Current Comprehension Type Questions And Answers

Read the following passage carefully and answer the questions at the end of It.

Question 1. A series combination of an inductor of self-inductance L, capacitor of capacitance C, and resistor of resistance R is connected to an alternating voltage source of V=V₀ sin ωt. The current through the circuit is I = I₀ sin(t-0), where \(I_0=\frac{V_0}{\sqrt{R^2+\left(\omega L-\frac{1}{\omega C}\right)^2}}\) and \(\theta=\tan ^{-1} \frac{1}{R}\left(\omega L-\frac{1}{\omega C}\right)\).

Note that, the frequency of both voltage and, current is \(f=\frac{\omega}{2 \pi}\). The rms value of these parameters during one complete cycle are \(V_{\mathrm{rms}}=\frac{V_0}{\sqrt{2}} \text { and } I_{\mathrm{rms}}=\frac{I_0}{\sqrt{2}}\) respectively. These values are shown in alternating voltmeter and ammeter.

The power consumed by the circuit P = VI. The mean value i.e., the effective power of the circuit in a complete cycle is \(\bar{P}=V_{\mathrm{rms}} I_{\mathrm{rms}} \cos \theta\). This cos θ is termed the power factor.

1. V = V₀ sin ωt electromotive force is applied to an alternating circuit consisting of resistance R’ and an inductor of self-inductance L. The phase difference between the voltage and current is

1. 90°
2. \(\tan ^{-1} \frac{\omega L}{R^{\prime}}\)
3. \(\tan ^{-1} \frac{R^{\prime}}{\sqrt{\left(R^{\prime}\right)^2+\omega^2 L^2}}\)
4. \(\tan ^{-1} \frac{\sqrt{R^{\prime 2}+\dot{\omega}^2 L^2}}{R^{\prime}}\)

Answer: 2. \(\tan ^{-1} \frac{\omega L}{R^{\prime}}\)

2. The power factor of the circuit in question (1) is

1. Zero
2. \(\frac{\omega L}{R^{\prime}}\)
3. \(\frac{R^{\prime}}{\sqrt{R^{\prime 2}+\omega^2 L^2}}\)
4. \(\frac{\sqrt{R^{\prime 2}+\omega^2 L^2}}{R^{\prime}}\)

Answer: 3. \(\frac{R^{\prime}}{\sqrt{R^{\prime 2}+\omega^2 L^2}}\)

3. In the circuit in question (1) the inductor is replaced by a pure capacitor; the phase difference between the current and terminal voltage of the capacitor is

1. -90°
2. zero
3. Between -90° and zero
4. +90°

Answer: 1. -90°

4. The power factor of the circuit in question (3) is

1. -1
2. Zero
3. Between zero and 1
4. 1

Answer: 3. Between zero and 1

5. The voltage applied in an LCR circuit having R = 10Ω, L = 10 mH and C = 1 μF is V = 20 sin ωt volt. For what frequency of the applied voltage will the current reach Its peak value?

1. 159 Hz
2. 1592 Hz
3. 1.59 x 10⁴ Hz
4. 1.59 x 10⁵ Hz

Answer: 2. 1592 Hz

6. The phase difference between the voltage and peak current in question (v) is

1. Zero
2. -90°
3. +90°
4. 180°

Answer: 1. Zero

7. Which element is responsible for the power consumption in an alternating current circuit?

1. Only resistor
2. Only Inductor
3. Only capacitor
4. Resistor, inductor, and capacitor

Answer: 2. Only Inductor

8. The frequency of the applied alternating voltage In an. ac circuit is 50 Hz. The resistance and self-inductance of the circuit are 37.6 fL and 120 mH. The phase difference between the voltage and current is

1. Zero
2. 45°
3. 60°
4. 90°

Answer: 2. 45°

Question 2. A transformer is a device used to increase or decrease the voltage in the transmission line according to requirements. Generally, the input line voltage is fed into a primary coil and the output line voltage is obtained from the terminals of another coil. In an ideal transformer, the primary and secondary coils are linked in such a way that there is no loss of magnetic flux and electrical energy.

In an ideal transformer, if the number of turns and input voltage across the terminals of the primary coil are N₁ and V₁, then the output voltage at the two terminals of the secondary coil \(V_2=V_1 \cdot \frac{N_2}{N_1}\), where N₂ is. the number of turns in the secondary coil.

1. The ratio of the number of turns of the primary and secondary coils of an ideal transformer is 2: 1. If the input voltage is 440 V, then the output voltage is

1. 220 V
2. 440 V
3. 88 W
4. None of these

Answer: 1. 220 V

2. In question (1) if the input power of the transformer is 44 W, then the output power is

1. 22 W
2. 44 W
3. 88 W
4. None of these

Answer: 2. 44 W

3. In the above-mentioned transformer the input and output currents are respectively,

1. 100 mA, 100 mA
2. 200 mA, 200 mA
3. 100 mA, 200 mA
4. 200 mA, 100 mA

Answer: 3. 100 mA, 100 mA

Class 12 Physics Electromagnetic Induction And Alternating Current

Alternating Current Integer Answer Type Questions And Answers

In this type, the answer to each of the questions is a single-digit integer ranging from 0 to 9.

Question 1. A resistance and a capacitor are connected in series with an alternating voltage of rms value 13 V. The terminal voltage of the resistor is 12 V and that across the capacitor is (n + 0.38) V. What is the value of n?

Answer: 6

Question 2. The current and voltage in an ac circuit are \(I=\sin \left(100 t+\frac{\pi}{3}\right) \mathrm{A}\) and V = 20 sin 100 tV. Calculate the power of the circuit in W.

Answer: 5

Question 3. In a series LCR circuit, the capacitance C is replaced by 2C. To keep the resonance frequency unchanged, the inductance has to be replaced by an inductance of L’. Find the ratio of L and L’.

Answer: 2

Question 4. An alternating voltage of 5 V of frequency 50 Hz is connected to a series LCR circuit. The potential difference across the inductor and resistor is 6 V and 4 V respectively. What is the voltage across the capacitor (in V)?

Answer: 3

Question 5. In a series LCR circuit R = 1 kΩ, C = 2μF, and the potential difference across R is 2 V. At resonance ω = 200 rad.s^-1. What is the potential difference (in V) across L at resonance? across

Answer: 5

Question 6. In a series LCR circuit R = 25Ω, L = 10 mH, and C = 1μF. The circuit is connected to an AC source of variable frequency. What is the Q-factor of the circuit?

Answer: 4

Question 7. A current of 50 mA flows through a 4 μF capacitor connected to a 500 Hz ac source. The terminal potential difference (in V) across the capacitor is (η + 0.98). What is the value of η? (π = 3.14)

Answer: 3

Question 8. In the figure, an LCR series circuit is shown. What would be the ammeter reading ampere?

Answer: 8

Class 12 Physics Electromagnetic Waves Introduction

In 1865, James Clerk Maxwell analyzed the nature of light. According to him, light is a progressive wave of an electric and a magnetic field. In other words, light is an electromagnetic wave. At this time, except for visible light, infrared, and ultraviolet, the existence of other electromagnetic waves was unknown.

• As per Maxwell’s electromagnetic theory, an electric field and a magnetic field exist at each point on the path of propagation of an electromagnetic wave. These two fields undergo periodic vibrations.
• These vibrations propagate from one point to the next, and thus electromagnetic waves spread. Electric and magnetic fields can spread in all directions even without any material medium (solid, liquid, or gaseous). Hence, electromagnetic waves can propagate in a vacuum also.

Class 12 Physics Electromagnetic Waves Displacement Current

As per Ampere’s circuital law (See chapter ‘Electromagnetism; section 1.5),

⇒ \(\oint \vec{B} \cdot d \vec{l}=\mu_0 I\) → (1)

But the law has a limitation. What would be the form of the above equation for a varying current I, i.e., for a varying electric field inside a conductor, is not mentioned.

For example, during the charging of a capacitor, varying current flows through the circuit though no current flows in between the plates of the capacitor. Equation (1) is not applicable in such cases.

Maxwell corrected this error by adding another term to the right side of equation (1). This additional term is,

⇒ \(\mu_0 I_d=\mu_0 \epsilon_0 \frac{d \phi_E}{d t}\) → (2)

Maxwell mentioned the quantity Jd as displacement current. Note that there is no valid reason for using the word ‘displacement’ in this context But, the word is entrenched in the language of physics.

\(\epsilon_0 \text { and } \phi_E\left(=\int \vec{E} \cdot d \vec{A}\right)\) in equation (2) is the permittivity of free space and electrical flux through a Gaussian surface respectively.

• Therefore, is the rate of change of electrical flux and clearly, \( I_=\epsilon_0 \frac{d \phi_E}{d t}\) is 1h expression representing varying electrical field.
• This varying electrical field is equivalent to the die current flow between the plates of the capacitor during charging (known as displacement current). So, now there is no discontinuation in the current flow. Hence, the general form of Ampere’s circuital law is,

⇒ \(\oint \vec{B} \cdot d \vec{l}=\mu_0 I+\mu_0 \epsilon_0 \frac{d \phi_E}{d t}\) → (3)

which is also known as Ampere-Maxwell law.

The meaning of this expression can be understood with help. The electric flux through surface s2 is \(\phi_E=\int \vec{E} \cdot d \vec{A}=E A\), where A is the area of the capacitor plates and E is the magnitude of the uniform electric field between the plates.

If Q is the charge on the plates at any instant, then \(E=\frac{Q}{\epsilon_0 A}\).

Therefore, \(\phi_E=E A=\frac{Q}{\epsilon_0}\)

Hence, the displacement current through S₂ is

∴ \(I_d=\epsilon_0 \frac{d \phi_E}{d t}=\frac{d Q}{d t}\)

The displacement current through S₂ is exactly equal to the conduction current through S₁.

The displacement current can be identified as the source of the magnetic field on the surface S₂. The displacement current has its physical origin in the time-varying electric field. The central point of this formalism is that a magnetic field can be produced in two ways—if

1. Current flows through a conductor or
2. The electric field varies with time in a region.

∴ The general form of Faraday’s law of induction is, \(\oint \vec{E} \cdot d \vec{l}=-\frac{d \phi_B}{d t}\).

• It can be concluded from this law that,’ a changing magnetic field creates an electric field (that is an emf in a conductor). Similarly, from the above law, we can conclude that a changing electric field sets up a magnetic field.
• In other words, the concept of displacement current develops an important connection between varying electric fields and magnetic fields. These varying electric and magnetic fields spread out in all directions and are known as electromagnetic waves.

Definition: When varying electric and magnetic fields exist in a place then these varying fields spread out in all directions like a wave which is called an electromagnetic wave.

Class 12 Physics Electromagnetic Waves The Spectrum Of Electromagnetic Waves

Different types of electromagnetic waves, their wavelengths, sources, and uses are given in the following table.

Today we know about the existence of the electromagnetic spectrum. The whole world is immersed in electromagnetic waves of different wavelengths. The sun is the main source of most of these waves.

• In the increasing order of frequencies, i.e., decreasing order of wavelengths, a few important em waves are radio waves or Hertzian rays, microwave, infrared, visible light, ultraviolet rays, X-rays, gamma rays, cosmic rays, etc.
• which has been discussed in the previous table. There is no distinct line of demarcation of different electromagnetic waves which, almost invariably, overlap to some extent.

All electromagnetic waves are generated due to the motion of accelerated charges. German physicist Heinrich Hertz, in 1888, first generated electromagnetic waves in the range of radio waves on an experimental basis.

And also made arrangements for the identification of the wave, ye took the help of an oscillating electric dipole. An oscillating electric dipole of this type is known as a Hertzian oscillator or Hertzian dipole.

1. Most substances absorb infrared waves very easily. On absorbing infrared radiation, the internal energy of matter increases; hence there is a temperature rise.
2. The ozone layer in the upper atmosphere, mainly in the stratosphere, absorbs the ultraviolet rays and converts them to infrared rays. Some chemicals reduce the density of the ozone layer by reacting with it. These substances are called ozone-depleting substances or ODS in short. Most of the ODS are made artificially and used in aerosol spray cans and refrigerants. Uses of this type of matter are trying to be banned.

Class 12 Physics Electromagnetic Waves Characteristics Of Electromagnetic Wave

1. Electric field \(\vec(E)\) and magnetic field \(\vec(B)\) are perpendicular to the direction of propagation of the wave. Hence electromagnetic waves are transverse.

2. Magnetic field being a varying field, an electric field is induced (following Faraday’s laws of electromagnetic induction) perpendicular to it.

At the same time, varying electric fields induce a magnetic field (following Maxwell’s law of displacement current).

3. Let an electromagnetic wave propagate along the positive x direction. In that case, \(\vec{E}\) field and \(\vec{B}\) field oscillate parallel to the y and z-axes. Expressions for \(\vec{E}\) and \(\vec{B}\) fields at a distance x from the origin O are given in scalar form, by

E = E₀ sin(ωt – kx) → (1)

and B = B₀ sin (ωt – kx) → (2)

where E₀, and B₀, are the amplitudes of the electric field and magnetic field, while ω and k are the angular frequency and the magnitude of the propagation vector, respectively.

It is to be noted that none of the two fields exists independently of the other. A time-varying magnetic field induces an electric field.

Simultaneously, a time-varying electric field induces a magnetic field. Thus, self-sustaining electromagnetic waves travel through space.

This depicts an em wave with its components along the y and z-axes oscillating in the xy-plane and zx-plane and representing equations (1) and (2) respectively. They represent the field vectors in magnitude and direction at different distances from the origin at any time t.

Equations (1) and (2) show that the frequencies of the electric and magnetic fields are equal and these two fields are always in the same phase.

Speed of electromagnetic wave in free space,

⇒ \(c=\frac{1}{\sqrt{\mu_0 \epsilon_0}}\) → (3)

where μ₀ and ε₀ are respectively permeability and permittivity of free space.

Here, = μ₀ x 10^-7 N. s².C^-2

ε₀ = 8.854 X 10^-2 C². N^-2. m^-2

Substituting the values in equation (3) we get,

⇒ \(c=\frac{1}{\sqrt{4 \pi \times 10^{-7} \times 8.854 \times 10^{-12}}}=2.998 \times 10^8\)

≈ 3.0 x 10² m. s^-1

In addition, the velocity of an electromagnetic wave can also be obtained from the amplitudes of electric and magnetic fields as,

⇒ \(\frac{E_0}{B_0}=c\) → (4)

From equations (1) and (2) we get,

∴ \(\frac{E}{B}=\frac{E_0}{B_0}=c\)

If the frequency and the wavelength of an electromagnetic wave in a medium are respectively f and λ, then in that medium the velocity of the wave will be fλ. If the medium changes, the wavelength as well as the velocity will change, but the frequency will remain the same because frequency is a characteristic of the source.

Some Quantities Related To EM Waves

Energy density: We have seen that the electric field, E = E₀sin(ωt – kx), and magnetic field, B = B₀ sin (ωt – kx).

Consider a volume element dV of the medium in which an electromagnetic wave is propagating. At a certain moment, the energy carried by the volume is U. As the energy stored is due to both electric and magnetic fields, we have

U= U_E+U_M

Here, U_E = stored energy due to the electric field in the volume element dV

= \(\frac{1}{2} \epsilon_0 E^2 \cdot d V\)

U_M = stored energy due to magnetic field in the volume element dV

= \(\frac{1}{2} \frac{B^2}{\mu_0} \cdot d V\)

So, at any position x of the electromagnetic wave at time t, the energy density of the electric field,

∴ \(u_E=\frac{U_E}{d V}=\frac{1}{2} \epsilon_0 E^2=\frac{1}{2} \epsilon_0 E_0^2 \sin ^2(\omega t-k x)\)

The energy stored in a unit volume of the space between two plates of a charged air capacitor, \( u_=\frac {1}{2} c_0 E^2[/atex].

We can prove that the total stored energy of the charged capacitor = u_E x volume of the space between two plates of capacitor = [latex]\frac{1}{2} C V_{P d}^2\), where, C = capacitance of the capacitor and VP(t = potential difference between the plates of the capacitor.

and the energy density of the magnetic field,

∴ \(u_M=\frac{U_M}{d V}=\frac{1}{2 \mu_0} B^2=\frac{1}{2 \mu_0} B_0^2 \sin ^2(\omega t-k x)\)

The above equation is used to calculate the energy, stored in an inductor.

Average value of sin2# or cos2# in a complete cycle = \(\frac{1}{2}\)

Here, \(\left\langle\sin ^2(\omega t-k x)\right\rangle=\frac{1}{2}\)

Therefore, the average energy density,

⇒ \(\bar{u}_E=\frac{1}{2} \epsilon_0 E_0^2 \times \frac{1}{2}=\frac{1}{4} \epsilon_0 E_0^2\)

and \(\bar{u}_M=\frac{1}{2 \mu_0} B_0^2 \times \frac{1}{2}=\frac{1}{4 \mu_0} B^2=\frac{1}{4} c^2 \epsilon_0\left(\frac{E_0}{c}\right)^2\)

= \(\frac{1}{4} \epsilon_0 E_0^2\) [∵\(c=\frac{1}{\sqrt{\epsilon_0 \mu_0}} \text { or } \frac{1}{\mu_0}=c^2 \epsilon_0\)]

so \(\bar{u}_E=\bar{u}_M\)

It implies that the energy is equally distributed between electric and magnetic fields during the propagation of electromagnetic waves.

The average energy density of electromagnetic waves,

∴ \(\bar{u}=\bar{u}_E+\bar{u}_M=\frac{1}{4} \epsilon_0 E_0^2+\frac{1}{4} \epsilon_0 E_0^2=\frac{1}{2} \epsilon_0 E^2\)

For a complete cycle, the average value can be written as u instead of \(\bar{u}\).

Hence energy density,

⇒ \(u=\frac{1}{2} \epsilon_0 E^2=\frac{1}{2 \mu_0} B_0^2\) → (1)

In general, for an isotropic medium of permittivity ε, the energy density,\(u=\frac{1}{2} \epsilon E_0^2\), where ε = Kε₀, K is the dielectric constant of the medium.

Intensity: In an electromagnetic field, if we consider a unit area around a point, perpendicular to the direction of radiation, then the electromagnetic energy incident on that area per second is called the intensity of electromagnetic radiation at that point.

Let P is a point in an electromagnetic field. The wave propagates toward the point P with velocity c. A unit area is considered around P, which is perpendicular to the direction of propagation of radiation.

In time dt, the wave travels a distance equal to cdt in the direction of propagation. Imagine a cylinder of length cdt and unit cross-sectional area on the path of the wave such that it crosses the cylinder normally.

The energy of the electromagnetic wave incident at point P is equal to the energy stored in the volume of the imaginary cylinder (cdt X 1 = cdt).

The intensity of radiation at P is equal to the energy incident per second i.e., the energy stored in the volume, c x 1 = c.

So, the energy contained in that cylinder =cu, where u = energy density.

Therefore, the intensity of electromagnetic wave at the point P, by definition

I = cu

By using the equation (l), we get,

⇒ \(I=c u=\frac{1}{2} c \epsilon_0 E_0^2=\frac{1}{2 \mu_0} c B_0^2\) → (2)

Incidentally, unlike in mechanical waves, this energy flow per unit area, per unit time, called energy flux is denoted by S, not I.

Hence, \(S=\frac{d u}{A d t}=c u\)

Average value S is given by \(\bar{S}=\frac{1}{2 \mu_0} E_0 B_0=\frac{1}{2} \epsilon_0 c E_0^2\)

\(\vec{S}\) called Poynting vector, is given by \(\vec{S}=\frac{1}{\mu_0}(\vec{E} \times \vec{B})=\vec{E} \times \vec{H}\)

The direction of \(\vec{S}\) is given by the direction of propagation of the wave, as shown.

If the point P is considered in an isotropic medium of electric permittivity ε and permeability μ instead of free space, then the equation (2) will be

∴ \(I=v u=\frac{1}{2} v \epsilon E_0^2=\frac{1}{2 \mu} v B_0^2\)

Where, \(v=\frac{1}{\sqrt{\epsilon \mu}}\) = velocity of the electromagnetic radiation in tha medium.

Unit of the intensity of electromagnetic radiation

= J.m^-2.s^-1 = W.m^-2

Its dimension = \(\frac{M L^2 T^{-2}}{L^2 T}=M T^{-3}\)

Radiation Pressure: EM wave has linear momentum as well as energy. This means that radiation can exert force and hence pressure on any surface on which it is incident during Its propagation through a medium.

The force exerted on a unit area of the surface, perpendicular to the direction of propagation is called radiation pressure.

From thermodynamical analysis, we can show that (this analysis is beyond our syllabus),

Radiation pressure,\(p=\frac{1}{3} u\). where u = energy density of electromagnetic radiation [see the equation (1)].

The unit of pressure (p) = N.m^-2; the unit of energy density

(u) = J.m^-3.

Generally, both the units are the same as N.m = J.

The dimension of both p and u = ML^-1T^-2.

The force due to the radiation pressure is too small to detect under everyday circumstances. In subjects, like astronomy or astrodynamics-related works, its importance cannot be ignored.

For example, if the radiation pressure of the sun had been ignored, the space crafts, Viking-1 and Viking-2 (sent by NASA to Mars’s orbit to collect information about Mars) would have missed Mars’s orbit by about 15000 km.

Class 12 Physics Electromagnetic Waves Numerical Examples

Example 1. If the velocity of the em wave in a vacuum is 3 x 10⁸ m.s-1 and the -magnetic permeability of the vacuum is 4π x 10^-7 N.C^-2. s², find the electric permittivity of the vacuum.

Solution:

If permittivity and permeability of free space are ε₀ and μ₀, respectively,

∴ \(c=\frac{1}{\sqrt{\epsilon_0 \mu_0}} \quad \text { or, } c^2=\frac{1}{\epsilon_0 \mu_0}\)

∴ \(\epsilon_0=\frac{1}{\mu_0 c^2}=\frac{1}{4 \pi \times 10^{-7} \times 9 \times 10^{16}}\)

[∵ μ₀ = 4π x 10^-7 N.C^-2. s² and c = 3 x 10⁸ m.s^-1]

= 8.842 x 10^-12 C².N^-1 m^-2

Example 2. The amplitude of the electric field of a plane electromagnetic wave is 48 V.m^-1. What is the amplitude of the magnetic field of the wave?

Solution:

In this case, the amplitude of the electric field

E₀ = 48 V.m^-1

As \(\frac{E_0}{B_0}=c\) amplitude of magnetic field, \(B_0=\frac{E_0}{c}\)

∴ \(B_0=\frac{48}{3 \times 10^8}\) [∵ c = 3 x 10⁸ m.s^-1]

= 16 x 10^-8 Wb.m^-2

Example 3. The electric field of an electromagnetic wave is, E = 10^-5 sin(12 x 10¹⁵ t-4 x 10⁷x). Find the frequency, velocity, and wavelength of the wave. Write the equation of the magnetic field corresponding to this wave. Assume all quantities in SI unit.

Solution:

The electric field of the given electromagnetic wave,

E = 10^-5 sin(12 x 10¹⁵ t – 4 x 10⁷x)

Compared with the expression for the electric field,

E = E₀sin(ωt – kx) we get,

ω = 12 x 10¹⁵ rad.s^-1

∴ 2πf = 12 x 10¹⁵

or \(f=\frac{12 \times 10^{15}}{2 \pi}=1.9 \times 10^{15} \mathrm{~Hz}\)

Hence, the frequency of the wave, f = 1.9 x 10¹⁵ Hz.

The velocity of the wave,

⇒ \(v=\frac{\omega}{k}=\frac{12 \times 10^{15}}{4 \times 10^7}=3 \times 10^8 \mathrm{~m} \cdot \mathrm{s}^{-1}\)

Wavelength \(\lambda=\frac{c}{f}=\frac{3 \times 10^8}{1.9 \times 10^{15}}=1.58 \times 10^{-7} \mathrm{~m}\)

Equation of the corresponding magnetic field,

⇒ \(B=B_0 \sin (\omega t-k x)=\frac{E_0}{c} \sin (\omega t-k x)\) [∵\(\frac{E_0}{B_0}=c\)]

⇒ \(\frac{10^{-5}}{3 \times 10^8} \sin \left(12 \times 10^5 t-4 \times 10^7 x\right)\)

∴ 3.33 x 10^-14 sin(12 x 10¹⁵ t-4 x 10⁷x)T

Example 4. In an electromagnetic Held, the amplitude of the electric field at a point is 3 V m^-1. Calculate the energy density and Intensity of the wave at that point. Given μ₀ = 4π x 10^-7H. m^-1

Solution:

The amplitude of the electric field, E₀ = 3 V. m^-1

⇒ \(B_0=\frac{E_0}{c}=\frac{3}{3 \times 10^8}=10^{-8} \mathrm{~Wb} \cdot \mathrm{m}^{-2}\)

So, the energy density,

⇒ \(u=\frac{1}{2 \mu_0} B_0^2=\frac{1}{2 \times 4 \pi \times 10^{-7}} \times\left(10^{-8}\right)^2\)

⇒  3.98 x 10^-11J . m^-3

Intensity, I = cu = (3 x 10⁸) x (3.98 x 10^-11)

= 0.012 W. m^-2

Example 5. A rectangular parallel plate capacitor of dimension 5 cm x 4 cm is charged in such a way that the rate of change of electric field between the two plates is 5.65 x 10¹¹V. m-1 . s^-1. Calculate the displacement current for this capacitor.

Given ε₀ = 8.85 x 10^-12 F. m^-1

Solution:

Area of a rectangular plate,

A = (5 x 4) cm² = 20 cm² = 0.002 m²

If E is the electric field between two plates, then the electric flux,

⇒ \(\phi_E=E A ; \frac{d \phi_E}{d t}=A \frac{d E}{d t}\)

Therefore, the displacement current,

⇒ \(I_d=\epsilon_0 \frac{d \phi_E}{d t}=\epsilon_0 A \frac{d E}{d t}\)

= (8.85 x 10^-12) x 0.002 x (5.65 x 10¹¹)

= 0.01 A = 10 mA

Example 6. Calculate the intensity and the rms value of an electric field of an electromagnetic wave at a distance of 10 m from a 100 W electric bulb.

Given ε₀ = 8.85 x 10^-12 F. m^-1

Solution:

Area of a spherical surface of radius 10 m

A = 4π (10)² = 400 m²

∴ Intensity, I, \(\frac{100 \mathrm{~W}}{400 \pi \mathrm{m}^2}=\frac{1}{4 \pi} \mathrm{W} \cdot \mathrm{m}^{-2}=0.08 \mathrm{~W} \cdot \mathrm{m}^{-2}\)

Here, \(I=\frac{1}{2} c \epsilon_0 E_0^2=c \epsilon_0 E_{\mathrm{rms}}^2\) [latex]E_{\mathrm{rms}}=\frac{E_0}{\sqrt{2}}[/latex]

∴ \(E_{\mathrm{rms}}=\sqrt{\frac{I}{c \epsilon_0}}=\sqrt{\frac{0.08}{\left(3 \times 10^8\right) \times\left(8.85 \times 10^{-12}\right)}}\)

= 5.5 V. m^-1

 

Class 12 Physics Electromagnetic Waves Very Short Answer Type Questions And Answers

Displacement Current

Question 1. The current which comes into play in a region where the electric flux is changing with time is called___________.

Answer: Displacement Current

Question 2. Write down the mathematical form of Ampere-Maxwell’s law.

Answers: \(\oint \overrightarrow{B \cdot} \cdot d \vec{l}=\mu_0 I+\mu_0 \epsilon_0 \frac{d \phi_E}{d t}\)

Electromagnetic Spectrum

Question 3. The wavelength of a radio wave is 10 m. Is this statement true or false?

Answer: True

Question 4. Wavelengths of visible light, X-rays, and infrared rays are λ₁, λ₂, and λ₃, respectively. Arrange them in ascending order of their magnitudes.

Answer: λ₂, λ₁, λ₃

Question 5. Of X-rays, γ-rays, and ultraviolet rays, which one has the maximum frequency?

Answer: γ – ray

Question 6. Which part of the electromagnetic spectrum has the largest penetrating power?

Answer: Gamma rays

Question 7. Name the electromagnetic waves that have frequencies greater than those of ultraviolet light but less than those of gamma rays.

Answer: x-rays

Question 8. Find the wavelength, in angstrom unit, of electromagnetic waves of frequency 5 x 10¹⁹Hz in free space.

Answer: 0.06 A

Production And Propagation Of Electromagnetic Waves

Question 9. An electromagnetic wave is propagating along the positive z-axis. At any moment, if the direction of the magnetic field at a point is along the positive x-axis, what will be the direction of the electric field at that point?

Answer: Along the negative y-axis

Question 10. What is the velocity of a radio wave in a vacuum?

Answer: 3 x 10⁸ m. s^-1

Question 11. Light is a progressive wave of _______ and _______ fields.

Answer: Electric, magnetic

Question 12. An electromagnetic wave is a _______ progressive wave

Answer: Transverse

Question 13. Write down the relation among the velocity of electromagnetic waves, the magnetic permeability of vacuum, and the electric permittivity of vacuum.

Answer: \(c=\frac{1}{\sqrt{\mu_0 \epsilon_0}}\)

Question 14. A Physical quantity is expressed by the ratio of the amplitudes of electric and magnetic fields for an electromagnetic wave.

Answer: velocity of the wave

Question 15. What is the ratio of velocities of light rays of wavelengths 4000A and 8000A in a vacuum?

Answer: 1:1

Question 16. What is the ratio of speeds of infrared rays and ultraviolet rays in a vacuum?

Answer: 1:1

Question 17. Which physical quantity, if any, has the same value waves belonging to the different parts of the electromagnetic spectrum?

Answer: Speed in vacuum

 

Class 12 Physics Electromagnetic Waves Synopsis

Electromagnetic waves exhibit wave wave-like nature as they travel through space.

Electromagnetic wave has both oscillating electric and magnetic field components.

• Electromagnetic waves are transverse waves.
• Electromagnetic waves were first experimentally produced by Hertz and the waves were radio waves.
• When varying electric and magnetic fields exist in a place then these varying fields spread out in all directions like a wave, which is called an electromagnetic wave.
• The Velocity of electromagnetic waves in a vacuum is equal to that of light.
• Some electromagnetic waves according to increasing order of frequency (i.e., decreasing order of wavelength):
• Radio waves (range 10⁴m-0.1m), microwaves (range 0.3 m – 10^-4 m ), infrared waves (range 10^-3 m – 7 x 10^-7 m), visible light (range 7 x 10^-7 m – 4 x 10^-7 m), ultraviolet rays (range 4 x 10^-7 m – 6 x 10^-8 m), X-rays (range 10^-8 m- 10^-11 m), gamma rays (range 10^-11 m – 10^-14 m).

An LC circuit connected with an AC source is known as an LC oscillator. With the help of an LC Oscillator, electromagnetic waves can be created.

The energy is equally distributed between electric and magnetic fields at the time of propagation of electromagnetic waves in a medium.

• When an electromagnetic wave propagates along the positive X-axis, then electric field \(\vec{E}\) and magnetic field \(\vec{B}\) oscillate parallel to the y and z -axis respectively.
• Expression of electric field and magnetic field at a distance x from the origin at the time t are given by
• E = E₀sin(ωt – kx) and B = B0sin(ωt – kx)

The velocity of electromagnetic waves in free space,

⇒ \(c=\frac{1}{\sqrt{\mu_0 \epsilon_0}}\)

where μ₀ and ε₀ are permeability and permittivity of free space respectively.

If μ₀ = 4π x 10^-7N.C^-2. s²;

and ε₀ = 8.854 x 10^-12 C².N^-1.m^-2 then, c = 3.0 x 10⁸m.s^-1

If the amplitudes of electric and magnetic fields are E₀ and B₀ respectively, then

⇒ \(\frac{E}{B}=\frac{E_0}{B_0}=c\)

The energy density of an electromagnetic field,

⇒ \(u=\frac{1}{2} \epsilon_0 E_0^2=\frac{1}{2 \mu_0} B_0^2\)

The intensity of radiation at an OOOpoint in an electromagnetic field,

I = cu

and radiation pressure,

∴ \(p=\frac{1}{3} u\)

Class 12 Physics Electromagnetic Waves Assertion Reason Type Questions And Answers

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, and statement 2 is true; statement 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 electric field 1 and the magnetic field 3 are mutually perpendicular at a point in the electromagnetic wave.

Statement 2: Electromagnetic waves are transverse waves.

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

Question 2. Statement 1: Electromagnetic waves are transverse waves.

Statement 2: Electromagnetic waves have the property of polarisation.

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

Question 3. Statement 1: The ratio of the amplitudes of electric and magnetic fields at a point in the electromagnetic wave is the same as the velocity of the wave.

Statement 2: If a medium, μ, and ε are the magnetic permeability and the electric permittivity respectively, then 1/ √με is the velocity of an electromagnetic wave in that medium.

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

Question 4. Statement 1: During the propagation of visible light and X-rays as electromagnetic waves, X-rays carry more energy than visible light despite both of them having the same amplitudes of electric fields at a point.

Statement 1: The frequency of X-rays is much higher than that of visible light.

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

Question 5. Statement 1: During the propagation of electromagnetic wave along the z-axis, if the electric field \(\vec{E}\) at a point is along the x-axis, then the magnetic field B at that point will be along the y-axis.

Statement 2: In the direction of propagation of the electromagnetic wave, the electric field \(\vec{E}\) and the magnetic field \(\vec{B}\) both form a right-handed cartesian coordinate system.

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

Class 12 Physics Electromagnetic Waves Match The Columns

Question 1. Some types of electromagnetic waves and their corresponding frequencies are given in column 1 and column 2 respectively.

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

Question 2. Some quantities and their corresponding dimensions are given in Column 1 and Column 2 respectively.

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

Class 12 Physics Electromagnetic Waves Comprehension Type Questions And Answers

Question 1. Electromagnetic waves propagate through free space or a medium as transverse waves. The electric and magnetic fields are perpendicular to each other as well as perpendicular to the direction of propagation of waves at each point. In the direction of wave propagation, electric field \(\vec{E}\) and magnetic field \(\vec{B}\) form a right-handed cartesian coordinate system. During the propagation of electromagnetic waves, the total energy of electromagnetic waves is distributed equally between electric and magnetic fields. Since e0 and are permittivity and permeability of free space, the velocity of electromagnetic wave, c = (ε₀μ₀)^-1/2. Energy density i.e., energy volume due to electric field at any point, \(u_E=\frac{1}{2} \epsilon_0 E^2\); Similarly, energy density due to magnetic field, \(u_M=\frac{1}{2 \mu_0} B^2\). If the electromagnetic wave propagates along x -direction, then the equations of electric and magnetic field are respectively,
E = E₀sin(ωt- kx) and B = B₀sin(ωt-kx)

Here, the frequency and the wavelength of oscillating electric and magnetic fields are \(f=\frac{\omega}{2 \pi}\) and \(\lambda=\frac{2 \pi}{k}\) respectively. Thus \(E_{\mathrm{rms}}=\frac{E_0}{\sqrt{2}}\) and \(B_{\mathrm{rms}}=\frac{B_0}{\sqrt{2}}\), where \(\frac{E_0}{B_0}=c\). Therefore, average energy density, \(\bar{u}_E=\frac{1}{2} \epsilon_0 E_{\mathrm{rms}}^2\) and \(\bar{u}_M=\frac{1}{2 \mu_0} B_{\mathrm{rms}}^2\). The intensity of the electromagnetic wave at a point, \(I=c \bar{u}=c\left(\bar{u}_E+\bar{u}_B\right)\).

To answer the following questions, we assume that in the case of propagation of an electromagnetic wave through free space, c = 3 x 10⁸ m. s^-1 and μ₀ = 4 π x 10^-7 H. m^-1

Question 1. If the electromagnetic wave propagates along the x-axis, then the electric field \(\vec{E}\) will be

1. Along y-axis
2. Along z-axis
3. On xy – plane
4. On yz – plane

Answer: 4. Along y-axis

Question 2. If the peak value of the electric field at a point in an electromagnetic wave is 15 V.m^-1, then the average electrical energy density (in J.m^-3)

1. 4.98 x l0^-9
2. 9.95 x 10^-9
3. 4.98 x 10^-10
4. 9.95 x 10^-10

Answer: 3. 4.98 x 10^-10

Question 3. The peak value of the magnetic field (in Wb.m^-2) at that point

1. 5 x 10^-8
2. 45 x 10^-8
3. 5 x 10⁸
4. 45 x 10⁸

Answer: 1. 5 x 10^-8

Question 4. The average energy density (in J.m^-3 )of the electromagnetic wave at that point

1. 4.98 x 10^-9
2. 9.95 x 10^-9
3. 4.98 x 10^-10
4. 9.95 x 10^-10

Answer: 4. 9.95 x 10^-10

Question 5. The intensity (in W.m^-2) of the electromagnetic wave at that point is almost

1. 0.15
2. 0.3
3. 0.45
4. 0.6

Answer: 2. 0.3

Question 6. If the wavelength is 1000A, then the frequency (in Hz)

1. 10¹³
2. 3 x 10¹³
3. 10¹⁵
4. 3 x 10¹⁵

Answer: 4. 3 x 10¹⁵

Question 7. Relation between ω and k

1. \(\omega=c k\)
2. \(\omega=c k\)
3. \(\omega=\frac{c}{k}\)
4. \(\omega=\frac{c}{2 \pi k}\)

Answer: 1. \(\omega=c k\)

Class 12 Physics Electromagnetic Waves Integer Type Questions And Answers

In this type, the answer to each of the equations is a single-digit integer ranging from 0 to 9.

Question 1. If the dielectric constant and relative magnetic permeability of a medium are 4 and 2.25 receptively, then the velocity of an electromagnetic wave in that medium is n x 108m.s^-1. Find the value of n.

Answer: 1.

Question 2. When the frequency of an electromagnetic wave and the amplitude of its associated magnetic field at a point are 10⁸HZ and 10^-10 T respectively, then the amplitude of the electric field is n x 10^-2 V . m^-1. Find the value of n.

Answer: 3.

Question 3. Due to the sunlight, if the amplitude of the electric field at the earth’s surface is 900 V m^-1, then the amplitude of the magnetic field is n x 10^-6 T. Find the value of n.

Answer: 3.

Question 4. What is the wavelength (in meters) of a radio wave of frequency 1.5 x 10⁸ HZ?

Answer: 2.

Question 5. If the intensity of an electromagnetic wave at a point is 0.085 W m^-2, then find the average amplitude (in V.m^-1) of the electric field at that point.

Answer: 8.

 Electromagnetic Induction And Alternating Current

 Electromagnetic Induction Induced Emf Or Induced Induction

When a current-carrying wire is placed in a magnetic field, it is deflected, i.e., a motion is generated. An opposite phenomenon was first observed by Michael Faraday.

He saw that if a closed wire loop is moved in a magnetic field, an electric current is generated in that loop as long as the motion continues.

Definition: If there is a relative motion between a magnetic field and a conductor, an electromotive force is generated in the conductor. It is called induced electromotive force.

It should be noted carefully that, to get electric current from this induced emf, the conductor should be in the form of a closed circuit.

This is because, the resistance of an open circuit is infinite and hence, despite the presence of induced emf, no current passes through the conductor.

Definition: If there is a relative motion between a magnetic field and a closed circuit conductor, the current flowing through that conductor is known as induced current.

Electromagnetic induction: The phenomenon involving the generation of electrical energy due to relative motion between a magnetic field and a conductor is called electromagnetic induction.

Electromagnetic Induction And Alternating Current

Electromagnetic Induction Experimental Demonstration

Magnitude And Direction Of Induced Current

Induced current with the help of a magnet: In Fig. 1.1, C is a circular coil having one or more turns. M is a permanent magnet kept along the axis of C. G is a galvanometer connected to C.

By observing the deflection of the pointer in G towards left or right, the direction of current in the coil C (i.e., anticlockwise or clockwise) can be determined.

Again, noting the extent of deflection of the pointer (i.e.,
low or high), the magnitude of the current, low or high, can be ascertained.

For different motions of the magnet along the axis of the coil, the directions of induced current through the coil are given in the following table.

Note that, no induced current is noticed in the coil when

1. The magnet is at rest, the magnet rotates about its axis or
2. Both the magnet and this coil move in such a way that no relative motion exists between them.
3. Keeping the magnet at rest, if the coil is moved towards or away from the magnet, the same results will be obtained.

Direction Of The Induced Electromotive Force:

This direction depends on

1. The nature of the magnetic pole facing the conductor and
2. The direction of motion of the magnet concerning the conductor.

Magnitude Of The Induced Electromotive Force:

1. Increases with an increase in the relative velocity between the coil and the magnet.
2. Increases with the increase in pole strength of the magnet.
3. Increases with an increase in the number of turns of the coil.

Magnitude And Direction Of The Induced Current:

1. The direction is identical to the direction of induced electromotive force.
2. Magnitude is directly proportional to the magnitude of the induced emf
3. Magnitude is directly proportional to the magnitude of the induced emf.
4. Inversely proportional to the resistance of the coil.

Note that, induced electromotive force does not depend on the resistance of the circuit. In an open circuit, the resistance is infinite, hence no current passes but the induced emf associated with the circuit does exist.

Current Induced With The Help Of Current Carrying Conducting Coil Or Current Carrying Solenoid:

A current-carrying circular coil or a current-carrying solenoid is equivalent to a permanent magnet. The rule for determination of their poles are

1. If the current view from a face is anticlockwise, that face acts as a north pole (N) and
2. If the current view from a face is clockwise, that face acts as a south pole (S).

So, if a circular current-carrying coil or a current-carrying solenoid is used instead of the magnet M shown, the same experimental result will be obtained. Two more important facts should be noted:

1. With the increase in the current through the circular coil or solenoid, the strength of the magnetic field increases.
2. If the direction of current through the circuit is reversed, magnetic polarity is also reversed.

Current Induced With The Help Of Static Electrical Circuits:

In S is a circular coil with one or more turns. A galvanometer G is connected to it. From this galvanometer, the magnitude and direction of the induced current can be determined.

P is another co-axial circular coil connected with an electrical circuit. B is a battery and a key K is used to switch on or off the circuit. Again, with the help of the rheostat, current through the circuit can be increased or decreased.

Now, if the circuit P be switched ‘on’ or if the current in the circuit is increased rapidly, the magnetic field also increases considerably.

It is similar to the situation when a magnet is moved towards the coil S swiftly. So, the results of two phenomena: ‘increase of current in the circuit P’ and ‘relative motion between the coil S and the magnetic field’ are identical.

Naturally, if the circuit P is switched ‘off’ suddenly or if the current in the circuit is decreased rapidly, the effective relative velocity becomes just opposite to the former case.

Observation:

the current through coil P be increased or decreased in different ways using the key K and the rheostat Rh, the induced current found the coil S will be as described below:

Primary And Secondary Coils:

In the experiment shown, the current through coil P is the cause of electromagnetic induction and this coil is called the primary coil.

On the other hand, the current in coil S is the result of electromagnetic induction and this coil is called the secondary coil.

Electromagnetic Induction And Alternating Current

Electromagnetic Induction Magnetic Induction And Magnetic Flux

From the discussion of magnetism we know that, if a magnetic material is placed in an external magnetic field, it gets magnetically induced.

This magnetic induction is a vector quantity, and the most convenient way to specify its magnitude and direction is to draw lines of induction through that substance.

If the magnetic induction is greater at a place, lines of induction get crowded there. Moreover, the direction of lines of induction is to be represented.

According to the direction of magnetic induction at every point. From this we can define magnetic induction as follows:

Definition: The number of lines of induction passing normally through a unit area surrounding a point inside a substance is called magnetic induction at that point.

Magnetic induction is a vector quantity, its symbol is \(\vec{B}\). This vector is identical to the magnetic field \(\vec{B}\) described in the chapter ‘Electromagnetism.

Usually, the lines of induction passing through any medium are called lines of force. These continuous lines of force passing through a medium can be imagined as a kind of stream.

This stream is comparable to the flow of water. We know that, in the case of water flow, the rate of flow of water can be obtained from the mass of water passing through any cross-section. Magnetic flux can be defined from this analogy.

Definition: The number of lines of induction passing normally through any surface placed in a magnetic field is called the magnetic flux linked with that surface.

Magnetic flux is a scalar quantity, and its symbol is Φ.

Let the area of a surface placed in a uniform magnetic field be A. If the magnetic induction \(\vec{B}\) is inclined at an angle θ with the normal to the surface, the component of \(\vec{B}\) in the direction of that normal is Bcosθ (magnitude of \(\vec{B}\) is B).

Hence according to the definition, of magnetic flux,

⇒ \(\phi=B A \cos \theta=\vec{B} \cdot \vec{A}=\vec{B} \cdot \hat{n} A\)

where \(\hat{n}\) is a unit vector normal to the surface.

For a finite surface, we first consider the magnetic flux through an infinitesimal area \(d \vec{A}\). The magnetic flux across this area, \(d \phi=\vec{B} \cdot d \vec{A}\)

A finite surface A can be assumed as the summation of such infinitesimal elements and the magnetic flux through the total surface area,

∴ \(\phi=\int_S \vec{B} \cdot d \vec{A}\)

Special Cases:

If the lines of induction are along the surface, then θ = 90° and hence Φ = 0.

If the lines of induction are perpendicular to the surface, θ = 0° and hence Φ = BA. In this case, \(B=\frac{\phi}{A}\) = magnetic flux linked with unit area.

Thus, magnetic induction \(\vec{B}\) is also called magnetic flux density. According to this flux is positive.

If θ = 180°, Φ = -BA i.e., flux is negative.

Electromagnetic Induction And Alternating Current

Electromagnetic Induction Laws Of Electromagnetic Induction

The phenomenon of electromagnetic induction is completely described by three laws. The exact information obtained from these laws is as follows:

1. Whether or not an emf would be induced in a coil (Faraday’s 1st law).
2. The magnitude of this induced emf (Faraday’s 2nd law).
3. The direction of the induced emf (Lenz’s law)

Faraday’S Laws

First law: Whenever the magnetic flux linked with a coil changes with time, an electromotive force is induced in the coil. Induced EMF lasts as long as the magnetic flux linked with the coil continues to change.

Second law: The magnitude of the induced is directly proportional to the time rate of change of magnetic flux linked with the coil.

If the change in magnetic flux linked with a coil in time dt be dtb, according to Faraday’s second law,

induced emf \(e \propto \frac{d \phi}{d t}\) → (1)

Explanation Of Faraday’s Two Laws: The magnetic lines of force or lines of. induction adjacent to the north pole (IV) of a bar magnet M is shown.

Explanation Of The First Law:

If a closed coil is moved from A to A’ or from A’ to A, the number of lines of force through that coil, i.e., magnetic flux linked with it decreases or increases, respectively.

According to Faraday’s first law, during the motion of the coil in both cases an emf will be induced in the coil: On the other hand, if the coil is at rest, no change in the magnetic flux will occur, and hence induction will not take place.

From the experiment of electromagnetic induction, the same result is obtained. So, from the first law, the cause of the electromagnetic induction and the time of existence of the induced emf can be ascertained.

Explanation Of The Second Law: If the relative velocity between the coil and the magnet is increased, the strength of induced emf also increases. This phenomenon is by Faraday’s second law, because if the coil is moved quickly the rate of change of the number of lines of force, i.e., the rate of change of magnetic flux also increases. So, the second law determines the magnitude of the induced emf.

Lenz’s Law

From Faraday’s laws, the direction of the induced emf cannot be ascertained. The necessary law for the determination of this direction is Lenz’s law. This may be called the third law of electromagnetic induction.

Statement: In the case of electromagnetic induction, the direction of the induced emf is such that, it always opposes the cause of induction in the circuit.

Explanation of Lenz’s law:

Relative motion between a magnet and a closed coil:

Let C be a circular conducting coil. Along its axis, the north pole (AO of a bar magnet M is moved towards the coil.

According to Lenz’s law, the induced current in C will oppose the motion of the magnet M, i.e., it will repel the magnet. For this, a north pole would be generated on the front face of the coil.

So, if viewed from the side of the magnet, the induced current will be anticlockwise. Similarly, if the magnet is moved in the opposite direction the current in the coil C will be clockwise.

If magnet M is moved forward beyond the surface of coil C, the induced current in the coil will attract the south poles of the magnet.

Hence, a north pole should be generated on the back face of the coil and a south pole on the front face. So, if Viewed from the front face, the induced current is clockwise.

Primary And Secondary Coils:

let a current be passed through the primary coil P by switching on the circuit. At that moment.

The instantaneous effect on the secondary coil becomes similar to the effect due to a sudden rapid movement of a current-carrying coil towards the secondary coil.

So, according to Lenz’s law, the induced current in the secondary coil S will repel the primary coil. Since two unlike parallel currents repel each other.

We can conclude that the induced current in the secondary coil is opposite in direction concerning the primary coil.

On the other hand, if the primary circuit is switched off; just at that moment, a like current will be induced in the secondary coil.

Falling Of A Magnet Through A Coil: when a body is allowed to fall freely in the gravitational field of the earth, it falls with „ acceleration due to gravity (g).

Let us consider a closed circular coil C kept horizontally and along its axis, a bar magnet M is released. Due to the downward motion of the magnet, an induced current is set up in coil C.

According to Lenz’s law, the direction of this induced current will be such that, the downward motion of the magnet will be opposed. As a result, the magnet will fall with a smaller acceleration than that due to gravity (g).

If the coil is replaced by a long cylindrical conductor (say, a copper pipe), the slower motion of the magnet becomes more dearly noticeable.

Lenz’s Law From The Law Of Conservation Of Energy:

Let a bar magnet M be moving towards a closed coil C along its axis. Its north pole faces the coil. As a result, a current is induced in the coll, i.e., electrical energy is developed.

According to the law of conservation of energy, this electrical energy can only be obtained at the cost of other forms of energy and hence some external positive work is to be done against an opposing force.

The current induced in the coil is the source of this opposing force. Naturally, the direction of the induced current has to be anticlockwise to develop.

An N pole in its front face will oppose the motion of the bar magnet. So, Lenz’s law is a natural consequence of the law of conservation of energy.

Law Of Conservation Of Energy From Lenz’s Law:

According to Lenz’s law, the direction of the induced emf is such that, this induced emf can oppose the cause of the generation of electric current in the circuit.

It means that if we gradually bring a magnet near the coil, the induced EMF will oppose the forward motion, and when the magnet is taken away from the coil, it opposes that receding motion.

As a result, to generate relative motion between the magnet and the coil, some work must be done against this opposing force and this work will induce electromotive force to obey the principle of conservation of energy.

Hence, the principle of conservation of energy can be obtained from Lenz’s law.

Expression For Induced Electromotive Force

From Faraday’s second law we know that, if the change in magnetic flux linked with a closed conductor be dtp in time dt, induced emf, \(e \propto \frac{d \phi}{d t}\)

So, according to Lenz’s law, we can write,

\(e=-k \frac{d \phi}{d t}\) → (1)

Here, k is a positive constant. Since the induced emf e opposes the change in magnetic flux dΦ, a negative sign is used in the equation.

Note that, equation (1) expresses the three laws of electromagnetic induction simultaneously.

1. If magnetic flux does not change i.e. if dΦ = 0 then e = 0. This supports Faraday’s first law.
2. \(e \propto \frac{d \phi}{d t}\).It is Faraday’s second law. For a coil having n turns, \(e \propto n \frac{d \phi}{d t}\).
3. The ‘negative’ sign on the right-hand side of equation (1) indicates the opposing nature of induced emf e due to the change in magnetic flux d<t>. This is Lenz’s law.

Units Of Magnetic Flux And Magnetic Induction:

The unit of magnetic flux is so defined that the magnitude of the
constant k in equation (1) becomes 1

For \(e=1, \text { if } \frac{d \phi}{d t}=1, \text { then } k=1\)

In that case \(e=-\frac{d \phi}{d t}\) → (2)

So, the unit of magnetic flux is defined as the change in unit time of magnetic flux linked with a conducting coil, for which 1 unit emf is induced in that coil.

Again, since the magnetic flux linked with the unit area is called magnetic induction or magnetic flux density, the unit of magnetic induction or magnetic flux density = \(\frac{\text { unit of magnetic flux }}{\text { unit of area }}\)

1 Wb: The change of magnetic flux linked with a coil in 1 s, for which 1 V emf is induced in the coil, is called 1 Wb.

In other words, 1 Weber is the magnetic flux which, linking a circuit of one turn, would produce in it an electromotive force of 1 volt if it were reduced to zero at a uniform rate in 1 second.

1 Mx: The change of magnetic flux linked with a coil in 1 s, for which 1 abvolt (1 abvolt = 10~8 V) emfis induced in the coil, is called 1 Mx.

In other words, 1 maxwell is the magnetic flux which, linking a circuit of one turn, would produce in it an electromotive force of 1 aibvolt if it were reduced to zero at a uniform rate in 1 second.

Relation Between Different Units:

As l volt = 10⁸ abvolt

so, 1 Wb = 10⁸ Mx .

Again, 1 T = \(\frac{1 \mathrm{~Wb}}{1 \mathrm{~m}^2}=\frac{10^8 \mathrm{Mx}}{10^4 \mathrm{~cm}^2}\)

= 10⁴ Mx.cm^-2 = 10⁴ G

Relation between web and V: According to the relation,

⇒ \(e=\frac{d \phi}{d t}\),

∴ \(\mathrm{V}=\frac{\mathrm{Wb}}{\mathrm{s}}\) or, Wb = V.s

Amount Of Change Flowing Through A Closed Circuit For Induced Electromotive Force:

Let the number of turns of any closed coil be n, the emf induced in the coil be e and the rate of change of magnetic flux linked with the coil be \(\frac{d \phi}{d t}\). Now, if the resistance of the circuit is R, the current induced in it (taking the magnitude only),

∴ \(i=\frac{e}{R}=\frac{n}{R} \cdot \frac{d \phi}{d t}\) [∵ e = \(n \frac{d \phi}{d t}\)]

Or, \(i d t=\frac{n}{R} \cdot d \phi\) → (3)

If the initial magnetic flux linked with the coil be and the final magnetic flux linked be <p2, then integrating equation (3) we get,

⇒ \(\int_0^t i d t=\frac{n}{R} \int_{\phi_1}^{\phi_2} d \phi=\frac{n}{R}\left(\phi_2-\phi_1\right)\)

So, the amount of charge flowing through the circuit,

∴ \(q=\int_0^t i d t=\frac{n}{R}\left(\phi_2-\phi_1\right)\) → (4)

Electromagnetic Induction And Alternating Current

Electromagnetic Induction Numerical Examples

Example 1: A coil of resistance 100 Ω having 100 turns is placed in a magnetic field. A galvanometer of resistance 400 Ω is connected in series with it If the coil is brought from the present magnetic field to another magnetic field in \(\frac{1}{10}\)s, determine the average emf and the current. Given, the initial and final magnetic flux linked with each turn of the coil are 1 mWb and 0.2 mWb respectively.

Solution:

Change in magnetic flux for each turn

= 0.2-1 = -0.8 mWb

So, a change in magnetic flux for 100 turns

= 100 X (+0.8) mWb = -0.08 Wb

Hence, the magnitude of average emf-induced

= the negative of the rate of change of magnetic flux

= \(-\left(-\frac{0.08 \mathrm{~Wb}}{\frac{1}{10} \mathrm{~s}}\right)=0.8 \mathrm{~V}\).

The equivalent resistance of the circuit = 100 + 400 = 500 Ω

Hence, the average induced current = \(\frac{0.8 \mathrm{~V}}{500 \Omega}\) = 0.0016 A = 1.6mA.

Example 2.  A conducting wire is wound around the great circle of a spherical Balloon. This Circular Loop Can Contract with The Balloon. A hemispherical cross-section of the balloon is shown in the figure. The initial radius of the balloon is 0.60m. A uniform magnetic field B = 0.25T exists along the perpendicular to the plane of the circular loop., in +y – direction. After 5.0 x 10^-2, the balloon is deflated to a radius of 0.30m. What will be the average EMF induced in the loop during this time?

Solution: Initially the flux linked with the conducting loop along the y-axis,

⇒ \(\phi_i=\vec{B} \cdot \overrightarrow{A_i}=B A_i \cos 0^{\circ}=0.25 \pi \times 0.60^2\)

= 0.28 Wb

After deflation of the balloon, the flux linked with the conducting loop of radius 0.30m,

⇒ \(\phi_f=\vec{B} \cdot \vec{A}_f=B A_f \cos 0^{\circ}=0.25 \pi \times 0.30^2\)

= 0.070 Wb

∴ change of flux,

∴ \(\Delta \phi=\phi_f \phi_i=(0.070-0.28) \mathrm{Wb}=-0.21 \mathrm{~Wb}\)

∴ Induced emf, e = \(e=-\frac{\Delta \phi}{\Delta t}=\frac{+0.21}{5 \times 10^{-2}}=4.2 \mathrm{~V}\).

Example 3.  A copper wire of diameter 0.04 in. and length 50 cm is bent in the form of a circular loop. The plane of the loop is normal to a uniform magnetic field which is increasing with time at a constant rate of 100 G.s^-1. What is the rate of Joule heating in the loop? [Resistivity of copper = 1.7 X l0^-8 Ω . m, 1 in. = 2.54 cm]

Solution:

The radius of the wire

= 0.02 in. = 0.02 x 2.54 x 10^-2m = 5.08 x 10^-4 m

Area of cross-section of the wire,

⇒ \(A_1=\pi r^2=\pi\left(5.08 \times 10^{-4}\right)^2=81 \times 10^{-8} \mathrm{~m}^2\)

Resistance of the wire,

⇒ \(R_0=\frac{\rho l}{A_1}=\frac{1.7 \times 10^{-8} \times 0.5}{81 \times 10^{-8}} \Omega\) [l =length of the wire = 0.5 m ]

= 1.05 x 10^-2 Ω

The radius of the loop,

⇒ \(R=\frac{l}{2 \pi}=\frac{0.5}{2 \pi}=7.96 \times 10^{-2} \mathrm{~m}\)

Area of the loop,

⇒ \(A=\pi R^2=3.14 \times\left(7.96 \times 10^{-2}\right)^2\)

= 0.02 m²

Here \(\frac{d B}{d t}=100 \mathrm{G} / \mathrm{s}=10^{-2} \mathrm{~Wb} \cdot \mathrm{m}^{-2} \cdot \mathrm{s}^{-1}\)

∴ Induced emf,

⇒ \(e=A \cdot \frac{d B}{d t}=(0.02) \times 10^{-2}=2 \times 10^{-4} \mathrm{~V}\)

∴ From Joule’s law, the rate of Joule heating,

⇒ \(H=I^2 R_0=\frac{e^2}{R_0}=\frac{\left(2 \times 10^{-4}\right)^2}{1.05 \times 10^{-2}}=3.8 \times 10^{-6} \mathrm{~W}\)

Electromagnetic Induction And Alternating Current

Electromagnetic Induction Motional Electromotive Force

Let us assume a uniform magnetic field \(\vec{B}\) directed along the axis which is vertically downwards.

A rod PQ parallel to the z -z-axis, is moving at a constant velocity \(\vec{v}\) along the x -x-axis through this field.

In this condition, a free charge q in the rod experiences a magnetic force, \(\vec{F}_m=q \vec{v} \times \vec{B}\). If q is a positive charge then this force is directed from Q to P (i.e., along +ve z-axis).

On the other hand, if q is negative then this force is directed from P to Q (i.e., along the -ve z-axis). Hence the free charges continue to accumulate at the ends of the rod creating a gradually increasing potential difference between P and Q.

This, in turn, creates an increasing electric field \(\vec{E}\) within the rod, in the direction from P to Q (opposite to the magnetic force). Due to this growing electric field, the electric force exerted on the charge q at any moment, Fe = qE.

As the magnitude of electric force comes to be equal with that of magnetic force, no net force anymore acts on the charge q, i.e., at equilibrium condition,

⇒ \(\left|\vec{F}_e\right|=\left|\vec{F}_m\right|\)

i.e., \(q|\vec{E}|=q|\vec{v} \times \vec{B}| \quad \text { or, }|\vec{E}|=|\vec{v} \times \vec{B}|\)

∴ \(E=v B \quad\)  [∵ \(\vec{v} \perp \vec{B}]\)→ (1)

Let the potential difference between the ends P and Q be V. If the length of the rod is l then,

V = El = vBl [putting the value of E from (1) ] → (2)

Now, if the ends of the rod are connected with a conductive wire, a current is set up in the circuit. Hence the moving rod can be considered as a source of emf (e), i.e., as a cell. Just like in a cell, this emf directs from the negative end Q to the positive end p inside the road.

For an open circuit, equation (2) can be written as,

V = e = vBl → (3)

This induced EMF in the moving rod is called motional electromotive force.

In general, if θ is the angle between \(\vec{v}\) and \(\vec{B}\) then, equation (3) becomes,

e = vBlsinθ → (4)

Discussions:

Dependence Of Induced Motional Emf On Different Factors:

From the above equation (4), the emf e induced across the straight conductor moving in a magnetic field is

1. Directly proportional to the magnetic induction (B),
2. Directly proportional to the length (l) of the conductor,
3. Directly proportional to the velocity (v) of the conductor and
4. Dependent on the angle 6 between the magnetic field and the direction of motion of the conductor. If the direction of motion of the conductor is parallel to the magnetic field then, θ = 0 and e = 0, and in that case, no electromagnetic induction takes place.

For θ = 90°, the emf induced is maximum.

Emf Induced Between The Two Extremities Of The Wings Of An Aeroplane:

At different places on the earth, the geomagnetic field is not horizontal; from the values of the angle of dip, the actual direction of the geomagnetic field is known.

When an aeroplane flies in a horizontal plane above the earth’s surface, the length of its two wings repeatedly intercepts the lines of force of the geomagnetic field.

As a result, electromagnetic induction takes place, i.e., a potential difference is set up between the two extremities of its wings.

The magnitude of this potential difference depends on

1. The distance between the extremities of the wings,
2. The velocity of the aeroplane,
3. The direction of motion of the aeroplane,
4. The horizontal component and
5. The angle of dip of the geomagnetic field.

The value of the angle of dip at the geomagnetic equator is zero. In this position, a horizontally moving aeroplane does not cut the geomagnetic lines of force, and hence no potential difference is set up between the extremities of the wings.

Emf Induced Between The Two Ends Of A Conductor Rotating With Uniform Angular Velocity In A Uniform Magnetic Field:

Let a conductor of length L be rotating about the point O with uniform angular speed ω in the plane of the paper. The magnetic field B is normally upwards relative to the plane of the paper.

Let us consider a small element dx of the conductor at a distance x from the point 0, which is moving with velocity v perpendicular to the direction of the magnetic field. The emf induced in this element of the conductor,

de = B(dx)v [using the relation, e = Blv]

= Bωxdx [∵ v = ωx]

So, the emf induced between the two ends of the entire conductor,

∴ \(e=\int_0^L B \omega x d x=B \omega\left[\frac{x^2}{2}\right]_0^L=\frac{1}{2} B \omega L^2\)

Induced Current In A Moving Straight Conductor:

In the discussion of motional electromotive force, if R be the total resistance of the circuit then, V = IR. So, from (3) we get,

⇒ \(I R=v B l \quad \text { or, } I=\frac{v B l}{R}\) → (5)

The direction of the induced current can be determined with the help of the following simple rule.

Fleming’s right-hand rule:

The thumb, the forefinger and the middle finger of the right hand are stretched perpendicular to each other.

If the magnetic field forefinger points in the direction of the magnetic field and the thumb in the direction of the motion of the conductor, then the middle finger will point in the direction of the induced current.

This rule is also called the dynamo rule.

Electromagnetic Induction And Alternating Current

Electromagnetic Induction Motional Electromotive Force Numerical Examples

Example 1. The distance between the two endpoints of the wings of an aeroplane is 5m and the aeroplane is flying parallel to the earth’s surface with a velocity of 360 km h^-1. If the geomagnetic intensity is 4 x 10^-4 Wb.m^-2 and the angle of dip at that place is 30°, determine the emf induced between the two end-points of the wings.

Solution:

While flying horizontally, the wings of the aeroplane cut the vertical component of the earth’s magnetic field normally and hence, an emf is induced between the two ends of the wings.

This induced emf, e = Blvsinθ.

Here \(B=4 \times 10^{-4} \mathrm{~Wb} \cdot \mathrm{m}^{-2}, l=5 \mathrm{~m}\)

⇒ \(v=360 \mathrm{~km} \cdot \mathrm{h}^{-1}=360 \times \frac{5}{18} \mathrm{~m} \cdot \mathrm{s}^{-1}=100 \mathrm{~m} \cdot \mathrm{s}^{-1}\)

and θ = 30°.

∴ e = 4 x 10^-4 x 5 X 100 x sin30° = 0.1 V

Example 2. A copper disc of diameter 20 cm is rotating uniformly about its horizontal axis passing through the centre with an angular frequency of 600 rpm. A uniform magnetic field of strength 10^-2 T acts perpendicular to the plane of the disc. Calculate the induced emf between its centre and a point on the rim of the disc.

Solution:

The diameter of the disc = 20 cm

∴ Radius, r = 10 cm = 0.1 m

Angular frequency, n = 600 rpm

∴ Angular speed,

⇒ \(\omega=\frac{600 \times 2 \pi \mathrm{rad}}{60 \mathrm{~s}}=20 \pi \mathrm{rad} \cdot \mathrm{s}^{-1}\)

Let us take a small segment dx on the disc at a distance x from its centre. The length of dx is so small that the speed of all the points on this segment is considered to be the same, which is, v’ = ωx.

Therefore, motional emf across dx,

de = v’Bdx [where B is the magnetic field]

∴ Total induced emf between the centre and a point on the rim of the disc,

∴ \(e=\int_0^r \nu_B^{\prime} B d x=\int_0^r \omega x B d x=\frac{1}{2} \omega B r^2\)

⇒ \(\frac{1}{2} \times 20 \pi \times 10^{-2} \times(0.1)^2\)

∴  0.00314 V = 3.14 mV

Example 3. A pair of parallel horizontal conduct invariable resistance shorted at one end Is fixed on a smooth table. The distance between the rails Is L. A massless conducting rod of resistance R can slide on the rails frictionlessly. The rod is tied to a massless string which passes over a pulley fixed to another edge of the table. A mass m, tied to the other end of the string, hangs vertically. A constant magnetic field B exists along the perpendicular to the plane of the table in an upward 1 direction. If the system Is released from rest, calculate

1. The terminal velocity of the rod,
2. The acceleration of the mass at the instant, when the velocity of the rod is half the terminal velocity.

Solution:

1. Let, rod A’B’ of length L and resistance R slide with a velocity v along the y-axis, while a constant magnetic field B exists along the z-direction.

Induced emf in the rod, e = BLv

The induced current flowing through the rod,

⇒ \(I=\frac{e}{R}=\frac{B L v}{R}\)

The direction of flow of the current through the rod A’B’ will be such that, it opposes the cause of generation of induced emf i.e., the motion of the rod along a positive y-axis. Thus the rod will experience force along the negative y-axis. The magnitude of the force is given by,

⇒ \(F=B I L=B \cdot \frac{B L v}{R} \cdot L=\frac{B^2 L^2 v}{R}\)

Let the rod move with acceleration, along the positive y-axis. Then from the equation of motion,

⇒ \(m a=m g-F=m g-\frac{B^2 L^2 v}{R} \quad \text { or, } a=g-\frac{B^2 L^2 v}{m R}\)

When the rod gets terminal velocity v0, then a = 0

∴ \(0=g-\frac{B^2 L^2 v_0}{m R}\)

∴ \(v_0=\frac{m g R}{B^2 L^2}\)

2. When the rod moves with velocity, \(v=\frac{1}{2} v_0\), then

⇒ \(v=\frac{m g R}{2 B^2 L^2}\)

Hence the acceleration of the mass at the instant

∴ \(a=g-\frac{B^2 L^2}{m R} \cdot \frac{m g R}{2 B^2 L^2}=g-0.5 g=0.5 g\)

Electromagnetic Induction And Alternating Current

Electromagnetic Induction Eddy Current

To enhance the effect of electromagnetic induction, in many cases, solid plates of soft iron or rods are used as the core of the armatures rotating in a magnetic field.

For example, in a dynamo, the armature rotating in the magnetic field is wound on a core of soft iron. Now, iron is a good conductor of electricity.

So electromagnetic induction takes place over the entire volume of the iron rotating in the magnetic field.

Many closed circuits or loops are formed locally within iron and induced current continues to flow in each loop.

Such current is known as eddy current Generally eddy current is produced whenever a conducting metal plate is in motion in a magnetic field.

If there is a relative motion between a conducting metal piece and a magnetic field, an induced current is set up throughout the volume of the metal in different closed loops. This current is known as eddy current.

According to Lenz’s law, eddy current opposes the cause that produces it, i.e., it opposes the relative motion between the metallic conductor and the magnetic field.

Due to this opposition, heat is generated in the conductor. This wasteful thermal energy comes at the cost of useful energy.

To minimise the loss of energy in the form of heat due to eddy current the cores are not taken as a single piece but are made of many thin laminas ofconductors which are insulated from each other with suitable insulating materials.

By doing so, the flux linked with local eddy current circuits is reduced to a minimum, hence eddy current becomes quite weak and its heating effect is minimised.

On calculation, it is observed that if the number of lamina is n instead of a single one, the loss of energy due to eddy current will come down to \(\frac{1}{n}\) times.

On the other hand, eddy current in some cases can be applied in our favour. For example, the coil of a moving coil galvanometer when deflected in the magnetic field begins to oscillate around the equilibrium position.

It takes a long time to come to rest To get rid of this problem the coil is wound on a single soft iron core.

The eddy current thus produced in the core opposes the motion of the coil following Lenz’slaw. Thus, the coil will not oscillate for a long time after the current is cut off.

It will return to the equilibrium position quickly. Besides this, eddy current is employed usefully in induction motors, induction cookers, electric brakes, etc.

Electromagnetic Induction And Alternating Current

Electromagnetic Induction Inductance Of A Coil

Self Induction

When current flows through a coil, a magnetic field is generated around the coil. As a result, the coil becomes linked up with its magnetic flux.

This magnetic flux Increases or decreases with the increase or decrease of current. Hence, an electromagnetic induction takes place and an emf is induced in the coil.

This induced emf opposes the change in electric current through the coil. This phenomenon is known as self-induction.

Definition: An electromagnetic induction in a coil due to a change of current through itself is called self-induction.

The electric current generated due to self-induction flows in the direction opposite to the main current.

Self-Inductance: If current flows through a coil, the magnetic flux linked with the coil becomes directly proportional to the current passing through it. So, if the magnetic flux is <p for current I, then

Φ ∝ I or, Φ = LI →(1)

L is a constant which depends on its construction. It is called the self-inductance or coefficient of self-induction of the coil. If I = 1, then L = Φ.

Definition: For unit current flowing through a coil, the magnetic flux linked with it is called its self-inductance.

Again, we know that the emf induced in the coil, \(e=-\frac{d \phi}{d t}\)

So, using equation (1) we can write,

\(e=-L \frac{d I}{d t}\) → (2)

If \(\frac{d I}{d t}=1\), then e = L (considering the magnitude only). From this, the following alternative definition of self-inductance can be drawn:

If the rate of change of electric current in a coil with time is unity, the emf induced in the coil is called its self-inductance.

Unit Of Self-Inductance:

In SI: In this system, the unit of self-inductance is Henry (H). To define it, we can use either of the equations (1) or (2).

1. If I = 1 A and 0 = 1 Wb, then L = 1 H.
2. If \(\) and e = 1 V, then L = 1 H.

So, the self-inductance of a coil is 1 H if

1. For n current of 1 A flowing through that coll, the magnetic flux linked with It is I Wb; or,
2. For a change of 1 A current in 1 second through that coll, the emf induced In It Is 1 V.

Hence, from equation (1) we get

Wb = H.A → (3)

Again from equation (2) we get

⇒ \(V=H \cdot \frac{A}{s} \quad \text { or, } \frac{V}{A} \cdot s=H\)

∴ H = Ω.s → (4)

The unit of resistance R is Ω and the unit of self-inductance L is H. Hence from equation (4) we infer that the unit of \(\frac{L}{R}\) and the unit of time is the same.

In CG System: Putting \(\frac{d I}{d t}=1 \mathrm{emu} \cdot \mathrm{s}^{-1}\) and e = 1 emu of emf, we get, L = 1 emu of self-inductance. So, if the current is changed at the rate of 1 emu.s^-1 and 1 emu electromotive force is induced in the coil, the self-inductance of that coil is 1 emu.

Relation between the CGS and SI units of self-inductance:

⇒ \(1 \mathrm{H}=\frac{1 \mathrm{~V}}{1 \mathrm{~A} \cdot \mathrm{s}^{-1}}=\frac{10^8 \mathrm{emu} \text { of potential difference }}{\frac{1}{10} \text { emu of current per second }}\)

= 10⁹ emu of inductance

Non-inductive Coil: Self-induction creates disturbances in many electrical circuits. A coil of finite resistance but of zero self-inductance is often required. Such a coil is called a non-inductive coil.

A long insulated conducting wire is given a fold and then coiled up. The free ends of the coil are now on the same side. This kind of winding is called non-inductive winding.

Hence antiparallel currents flow through adjacent wires of any part of the coil. So, the resultant magnetic flux becomes zero. That means, in this coil, no electromagnetic induction takes place.

Effect Of Self-Induction On Electrical Circuit:

In the circuit, if the emf of battery B is E, the effective emf of the circuit due to self-induction [according to equation (2)] = \(E-L \frac{d I}{d t}\)

If the current is zero or a steady current flows through the circuit, \(\frac{d I}{d t}=0\); and in this case, self-induction does not affect the circuit.

On the other hand, self-induction affects a circuit when electric current changes with time in that circuit.

Thus, self-induction plays an important role in an AC circuit. In the case of a DC circuit, self-induction occurs only when the circuit is switched ‘on’ or ‘off.

Choke: A coil of high self-inductance and low resistance is commonly called a choke. If a choke is connected in series in an AC circuit, then due to its high self-inductance, the effective resistance of the circuit increases.

On the other hand, due to the low resistance of the coil, less heat is produced in the circuit. So, the function of a choke is to increase the effective resistance of the circuit lowering the dissipation of heat energy as much as possible.

Uses: In a tube light arrangement, a choke is usually connected in series.

Energy stored In the magnetic field of an inductor: We have induced emf in an inductor carrying a time-varying current i, \(e=-L \frac{d i}{d t}\), where L = self-inductance of the inductive coil.

To increase the current in the inductor against this opposing emf, some energy has to be spent on work. This external work will be stored in the magnetic field of the inductor as magnetic energy.

Power, i.e., rate of work done = ei = \(L i \frac{d i}{d t}\) (the negative sign is not taken in calculation of work done)

So, work done in time dt = ei dt = Lidi

Total work done to increase the current through the inductor from 0 to I,

⇒ \(W=L \int_0^I i d i=L \cdot\left[\frac{i^2}{2}\right]_0^I=\frac{1}{2} L I^2\)

This work done is stored as energy (E_L) in the magnetic field of the inductor, i.e.,

∴ \(E_L=\frac{1}{2} L I^2\)

Mutual Induction

Let two coils, P and S, be kept very close to each other. Now if current is passed through them (say, P), a magnetic field is generated around it.

As a result, the second coil (S) gets linked with the magnetic flux generated by the first coil. If the current through P is increased or decreased concerning time.

The magnetic flux linked with an electromotive force induced in the coil S will also increase or decrease accordingly. Hence, an electromotive force is induced in the coil S.

This phenomenon is known as mutual induction. The first coil P is called the primary coil and the second coil S is called the secondary coil.

Definition: If current changes over time in a coil, the electromagnetic induction that occurs in an adjacent coil is called mutual induction.

Mutual Inductance: Mutual inductance is defined in a similar way as self-inductance. If the magnetic flux linked with the secondary coil is Φ_s for the current I_p in the primary coil then,

⇒ \(\phi_S \propto I_P \quad \text { or, } \phi_S=M I_P\) → (1)

Here, M is a constant, which depends on the geometry of the coils and the distance between them and is called mutual inductance or coefficient of mutual induction.

Naturally for I_p = 1, M = Φ_s

Definition: In the case of two adjacent coils, if one carries a unit current, then the magnetic flux linked with the other is called the mutual inductance between the two coils.

Again, induced emf in the secondary cop,

⇒ \(e=-\frac{d \phi_S}{d t}=-M \frac{d I_P}{d t}\) → (2)

If \(\frac{d I_p}{d t}=1\), then e = M [considering the magnitude only]

From this, the following, an alternative definition of mutual inductance is obtained.

In the case of two adjacent coils, if the rate of change of current concerning time in a coil is unity, the emf induced in the other coil is called the mutual inductance between the two coils.

Unit Of Mutual Inductance:

In SI: In this system, the unit of mutual inductance is Henry (H), exactly that of self-inductance L.

‘Mutual inductance between a pair of coils is 1 H’ means that,

1. If the current passing through one coil is 1 A, the magnetic flux linked with the other becomes 1 Wb; or,
2. If the current changes through one coil at the rate of 1 A per second, the emfinduced in the other becomes 1 V.

In the CGS system: If \(\frac{d I}{d t}=1\) emu of current per second and e = 1 emu of potential, then M = 1 emu of mutual inductance. So, in the case of two adjacent coils, if the current changes at the rate of 1 emu.s^-1 in one coil and as a result if 1 emu of emf is induced in the other coil, the mutual inductance between the coils is said to be 1 emu.

Discussions:

Interchange of two colls: The second coil can be used as the primary coil and the first as the secondary. In that case, due to: a change in current in the second coil, electromagnetic induction will take place in the first coil, and the value of mutual, inductance (M) will remain unchanged.

Relation between self-inductance, and mutual inductance: If the self-inductances of the two coils are, L₁ and L₂, the mutual inductance becomes

⇒ \(M=k \sqrt{L_1 L_2}\) (3)

Here, k is a constant whose value is 1 or less than 1. If the magnetic flux of one coil is linked completely with the other; k≈1. Thus,

∴ \(M \approx \sqrt{L_1 L_2}\)

Electromagnetic Induction And Alternating Current

Electromagnetic Induction Inductance Of A Solenoid

Self-inductance of a solenoid: Let the length of the solenoid = l, area of its circular cross-section = A.

So, if several turns are N, then the number of turns per unit length (n) = \(\frac{N}{l}\).

If current I flow through the solenoid, then the magnetic field thus generated inside it is

B = μ₀nI, where μ₀ = magnetic permeability of vacuum or air.

So, the magnetic flux linked with each, turn of the solenoid,

⇒ \(B A=\mu_0 n A I=\frac{\mu_0 N A}{l} \cdot I\)

Hence; the magnetic flux linked with JV turns,

⇒ \(\phi=N B A=\frac{\mu_0 N^2 A}{l} \cdot I\)

If the self-inductance of the solenoid is L then, Φ = LI

∴ \(L=\frac{\phi}{I}=\frac{\mu_0 N^2 A}{l}\)

For any other medium of magnetic permeability p inside the solenoid, \(L=\frac{\mu N^2 A}{l}\).

Mutual Inductance Of Two Inseparable Solenoids:

PP’ and SS’ are primary and secondary solenoids respectively. They arpÿound over each other on a length. The cross-sectional area of each of them = A

Number of turns in length l in PP’ = N₁ and that in SS’ = N₂.

Therefore, the number of turns per unit length of PP’.

⇒ \(n_1=\frac{N_1}{l}\)

If I₁ current flows through the primary coil PP’, the magnetic field thus developed inside it,

⇒ \(B_1=\mu_0 n_1 I_1=\frac{\mu_0 N_1}{l} I_1\)

[μ₀ = magnetic permeability of vacuum or air]

As the solenoids are closely wound on each other, the magnetic flux linked in each turn of the secondary coil = B₁A and hence the magnetic flux linked with N₂ turns,

⇒ \(\phi_2=N_2 B_1 A=\frac{\mu_0 N_1 N_2 A}{l} \cdot I_1\)

If the mutual, inductance of the solenoid is M then,

⇒ \(\phi_2=M I_1\)

So, \(M=\frac{\mu_0 N_1 N_2 A}{l}\) → (2)

For any other medium used as the core, M = \(\frac{\mu N_1 N_2 A}{l}\), where mu = magnetic permeability of that medium.

In the present context, self-inductances of the coils are

⇒ \(L_1=\frac{\mu_0 N_1^2 A}{l} \text { and } L_2=\frac{\mu_0 N_2^2 A}{l}\)

So, \(L_1 L_2=\left(\frac{\mu_0 A}{l}\right)^2 N_1^2 N_2^2=\left(\frac{\mu_0 N_1 N_2 A}{l}\right)^2=M^2\)

i.e., \(M=\sqrt{L_1 L_2}\)

Comparing this equation with equation (3). we get, k = 1. This is the maximum value of k.

Energy Density In The Magnetic Field Of A Solenoid:

Length of a solenoid = l, cross-sectional area = A

So, its volume = l A.

If N be the total turns ui the solenoid, the number of turns per unit length of it, \(n=\frac{N}{l}\)

If I current flows through the solenoid, the magnetic field thus develops along its axis,

⇒ \(B=\mu_0 n I=\frac{\mu_0 N I}{l} \quad \text { or, } I=\frac{B l}{\mu_0 N}\)

But, self-inductance of the solenoid,

⇒ \(L=\frac{\mu_0 N^2 A}{l}\)

So, the stored energy in its magnetic field,

⇒ \(U=\frac{1}{2} L I^2=\frac{1}{2} \frac{\mu_0 N^2 A}{l}\left(\frac{B l}{\mu_0 N}\right)^2=\frac{1}{2 \mu_0} B^2 L A\)

So, the energy stored per unit volume,

⇒ \(u=\frac{U}{L A}\)

i,e. \(u=\frac{1}{2 \mu_0} B^2\) → (3)

This u is the energy density in the magnetic field. For any other medium, instead of air or vacuum,

⇒ \(u=\frac{1}{2 \mu} B^2\)

Though equation (3) has been established for the magnetic field inside a solenoid, it is a general equation for any magnetic field because the relation is independent of the geometrical features of the inductor.

To find the energy density at a point in any magnetic field the above equation may be applied. Unit of energy density = J.m3 and its dimension = \(\frac{M L^2 T^{-2}}{L^3}=M L^{-1} T^{-2}\)

Electromagnetic Induction And Alternating Current Electromagnetic Induction Inductance Of A Solenoid Numerical Examples

Example 1. The mutual inductance between two adjacent coils is 1.5 H. If the current in the primary coil changes from 0 to 20 A in 0.05 s, determine the average emf. induced in the secondary coil. If the number of turns in the secondary coll is 800, what change in flux will be, observed in it?

Solution:

The average value of induced emf,

⇒ \(e=-M \frac{\Delta I}{\Delta t}=-1.5 \times \frac{(20-0)}{0.05} \stackrel{211}{=}-\frac{1.5 \times 20}{0.05}=-600 \mathrm{~V}\)

Hence, the magnitude of the induced emf = 600 V.

Again, from the relation Φ = MI, we get,

change in flux, ΔΦ = M. ΔI = 1.5 x (20 – 0) = 30 Wb

Example 2.  When the current in a coil changes from + 2 A to -2 A in 0.05 s, an emf of 8 V is induced in the coil. Determine the self-inductance of the coil

Solution:

We know that, \(e=-L \frac{\Delta I}{\Delta t}\)

Self-inductance of the coil, L = \(L=-\frac{e \Delta t}{\Delta I}=\frac{8 \times 0.05}{-2-(+2)}=0.1 \mathrm{H}\)

Example 3.  The mutual inductance of two coils is 0.005H, ac in the primary coil, I = I₀ sin ωt, where I₀ = 10A and ω = 100π rad/s. What is the maximum emf in the secondary coil?

Solution:

ac in the primary coil, I = I₀ sin ωt

∴ \(\frac{d I}{d t}=\omega I_0 \cos \omega t\)

So, \(e_2=-M \frac{d I}{d t}\) [here M = mutual inductance]

= -M ωI₀ cos ωt

So the maximum emf in the secondary coil

= MωI₀ = 0.005 x 100π X 10 = 15.7V

Example 4. If a rate of change of current of 2 A.s^-1 induces an emf of 10 mV in a solenoid, what is the self-inductance of the solenoid?

Solution:

∴ \(e=-L \frac{d I}{d \hbar} \quad \text { or, } L=\frac{|e|}{\frac{d I}{d t}}=\frac{10 \times 10^{-3} V}{2 \mathrm{~A} \cdot \mathrm{s}^{-1}}=5 \mathrm{mH}\)

Example 5.  The resistance of a coil is 10Ω and its self-inductance is 5H. Find the energy stored when it is connected to a 100V battery.

Solution:

Curreht in circuit, \(I=\frac{V}{R}=\frac{100}{10}=10 \mathrm{~A}\)

Stored energy in the inductor = \(\frac{1}{2} L I^2=\frac{1}{2} \times 5 \times(10)^2=250 \mathrm{~J}\)

Example 6. The self-inductance of an air core solenoid increases from 0.01 mH to 10 mH when an iron core is introduced into it. What is the relative magnetic permeability of iron?

Solution:

Self-inductance is proportional to the permeability of the core. So, we get

⇒ \(\frac{L}{L_0}=\frac{\mu}{\mu_0}=\mu_r \quad \text { or }, \mu_r=\frac{L}{L_0}=\frac{10}{0.01}\)

∴ \(\mu_r=1000\)

Hence relative magnetic permeability of iron is 1000.

Example 7.  A small square loop of wire of side y is placed Inside a large square loop of side x (x»y). The loops are coplanar and their centres coincide. Find the mutual inductance of the system.

Solution:

Let I be the current flowing through a square loop of side L.

The magnetic field at the centre of the loop, B = 4B₁

[where B₁ is the magnetic field at the centre of the loop due to one of its sides]

Now, for the large square loop, L = x

∴ \(B_1=\frac{\mu_0}{4 \pi} \cdot \frac{I}{\frac{x}{2}}\left(\sin 45^{\circ}+\sin 45^{\circ}\right)\) [∵ V distance of the centre from each side of the large loop = \(\frac{x}{2}\)]

⇒ \(=\frac{\mu_0}{2 \pi} \cdot \frac{I}{x}\left(\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}\right)=\frac{\mu_0 I}{\sqrt{2} \pi x}\)

∴ \(B=\frac{4 \mu_0 I}{\sqrt{2} \pi x}=\frac{2 \sqrt{2} \mu_0 I}{\pi x}\)

Now, magnetic flux linked with the small square loop,

Φ =B x area of the small square loop

\(B \times y^2=\frac{2 \sqrt{2} \mu_0 I y^2}{\pi x}\) → (1)

If M be the mutual inductance between the two loops, then

Φ = MI → (2)

From equations (1) and (2), \(M=\frac{2 \sqrt{2} \mu_0 y^2}{\pi x}\)

Example 8. Cross sectional area, of a solenoid is 10 cm². Haif of its cross section is filled with iron (μ_r = 450) and the remaining half with air ( μ_r = 1 ). Calculate the self-inductance of the solenoid fits length is 2m and the number of turns is 3000.

Solution:

If the α₁ part of the cross-section of the solenoid is filled with a substance of relative permeability μ_r1 and the remaining part of the cross-section α₂ with another substance of relative permeability μ_r2 then self-inductance of the s inside is,

⇒ \(L=\frac{\mu_0 n^2 A}{l}\left(\mu_{r_1} \alpha_1+\mu_{r_2} \alpha_2\right)\)

Here μ₀ = 4π x 10^-7 H/m; number of Pirns, n = 3000; length of solenoid, l = 2m; area of cross section A = 10cm² = 0.001m²; α₁ = 0.5 and α₁ = 0.5.

∴ Self-inductance of the solenoid,

⇒ \(L=\frac{4 \pi \times 10^{-7} \times(3000)^2 \times 0.001}{2}(1 \times 0.5+450 \times 0.5) \mathrm{H}\)

⇒ 2π x 9 x 10^-4 x (0.5 + 225) H

= 1.27H (approx.)

Electromagnetic Induction And Alternating Current

Electromagnetic Induction Very Short Questions And Answers

Question 1. What is the unit of magnetic induction or magnetic flux density in SI?

Answer: Wb.m-2 or T

Question 2. What is the relation of the magnetic field vector 5 with magnetic induction and magnetic flux density?

Answer: They are the spare physical quantity.

Question 3. Induced emf is direct propohiotfÿ’tti the rate of change with time of magnetic _____ linked with a coil.

Answer: Flux

Question 4. In the case of electromagnetic induction, the _______ always opposes the cause of its generation.

Answer: Current

Question 5. With the help of Fleming’s _______ rule, the direction of induced current in a straight conductor in motion can be determined.

Answer: Right-hand

Question 6. What is the relation between the units: tesla and Weber?

Answer: \(\mathrm{IT}=\frac{1 \mathrm{wb}}{1 \mathrm{~m}^2}\)

Question 7. What is the relation between the units: Weber and Volt?

Answer: 1 Wb = 1 V.1 s

Question 8. Which of the conservation laws would not hold if Lenz’s law was incorrect?

Answer: The law of conservation of energy

Question 9. The self-inductance of a coil is 1H. If 1A current passes through it, what will be the magnetic flux linked with the coil?

Answer: 1 Wb

Question 10. What is the relation between the units: weber and ampere?

Answer: 1 Wb = 1A.1H

Question 11. What is the relation between the units of self-inductance and mutual inductance?

Answer: They have the same unit

Question 12. The self-inductance of an air core inductor increases from 0.01 mH to 10 mH on introducing an iron core into it. What is the relative permeability of the core used?

Answer: 1000

Electromagnetic Induction And Alternating Current

Electromagnetic Induction Synopsis

if there is a relative motion between a magnetic field and a conductor, the electromotive force (emf) generated in that conductor is called induced emf.

• If there is a relative motion between a magnetic field and a closed conductor, the current generated through that conductor is called induced current.
• The number of lines of induction passing normally through the unit area surrounding a print inside a substance is called the magnetic induction (\vec{B}) of that point.
• The number of lines of induction that crosses normally any surface placed in a magnetic field is called the magnetic flux (Φ) linked with that surface.

Faraday’s laws of electromagnetic induction:

• First law: Whenever the magnetic flux linked with a coil changes with time, an emf is induced in the coil. The induced EMF lasts as long as the change in magnetic flux linked with the coil continues.
• Second law: The strength of the induced emf is directly proportional to the time rate of change of magnetic flux linked with a coil.
• Lenz’s law: In the case of electromagnetic induction the direction of induced emf is such that, it always opposes the cause of the generation of current in the circuit.
• The change of magnetic flux linked with a conducting coil in unit time, for which unit electromotive force is induced in the coil, is taken as the unit of magnetic flux.

Units of magnetic flux:

Units of magnetic induction:

• An electromagnetic induction that takes place in a coll due to a change of current through itself is called self-induction. The electric current generated due to self-induction flows in the direction opposite to the main current.
• The magnetic flux linked with a coil for unit current flowing through the coil is called its self-inductance.
• A coil having finite resistance but zero self-inductance is called a non-inductive coil.
• A coil having a high self-inductance but low resistance is called a choke.
• The phenomenon of production (induction) of emf in a coil due to a change in current with time in a neighbouring coil is called mutual induction.
• The magnetic flux linked with a coil due to the flow of unit current through a neighbouring coil is called the mutual inductance between the two colls.
• Units of self-inductance and mutual inductance: Henry (H), in SI.
• If there is a relative motion between a conducting metal piece and a magnetic field, an induced current is set up throughout the volume of the metal in different closed loops. This current is known as eddy current.

To minimise the loss of energy in the form of heat due to eddy current, the core is not taken as a single piece of conductor but is made of many thin laminas of conductors which are insulated from each other with suitable insulating materials.

• Magnetic flux linked with surface \(\vec{A}\),

⇒ \(\phi=B A \cos \theta=\vec{B} \cdot \vec{A}\)

[where \(\vec{B}\) = magnetic induction, 6 = angle between \(\vec{B}\) and \(\vec{A}\)]

The magnetic flux through a finite surface S,

∴ \(\phi=\int_S \vec{B} \cdot d \vec{A}\)

• According to Faraday’s second law,

induced emf, \(e \propto \frac{d \phi}{d t}\). In combination with Lenz’s law and the chosen units of 0, it becomes, \(e=-\frac{d \phi}{d t}\).

• 1 Wb = 108 Mx, IT = 10⁴G, Wb = V.s
• Induced emf in a straight conductor moving in a magnetic field, e = Blvsind [where B = magnetic flux density, l = length of the conductor, v = velocity of the conductor and 0 = angle between \(\vec{B}\) and \(\vec{v}\))
• The emf induced between the two ends of a conductor rotating in a uniform magnetic field,

∴ \(e=\frac{1}{2} B \omega L^2\)

[where L = length of the conductor, co = steady angular velocity]

• For current I, if the magnetic flux is <p, then

Φ ∝ I = 1 or, 0 = LI (L = self-inductance of the coil]

The equation \(e=-\frac{d \phi}{d t} \text { gives, } e=-L \frac{d I}{d t}\)

∴ Wb = H.A, H = n .s

or 1 H = 10⁹ emu of Inductance

• If the flux linked with the secondary coll is 0 for current I in the primary coil, then

Φ ∝ I or, Φ = MI {M = mutual inductance]

• If the self-inductances of two coils are L1 and L2, mutual inductance, \(M=k \sqrt{L_1 L_2}\) where k is a constant whose value is 1 or less.
• Energy stored in the magnetic field of a conducting coil of self-inductance L due to current I flowing through it = \(\frac{1}{2} L I^2\).

A conducting wheel is rotating in a uniform magnetic field with angular velocity co. The length of each spoke is l. If there are n number of spokes in the wheel.

Then total electromotive force between the centre and any point situated on the circumference of the wheel, \(e=\frac{1}{2} B \omega l^2\). Here, the total emf does not depend on the number of spokes.

• A thin semicircular conducting ring (PQR) of radius r is falling with its plane vertical in a horizontal magnetic field B, as shown. If the speed of the ring is v, the value of the induced emf between the two ends of the semicircular ring, c = B x (2r) x v.
• Here, the induced emf depends upon the distance between the two ends of the conductor, not on the shape of the conductor.

• If any two vectors among the three vectors \(\vec{B}, \vec{v} \text { and } \vec{l}\) are parallel, then the motion emf induced in the conductor, e = 0.

• Values of self-inductance in some special cases:

• Values Of Mutual Inductance In Some Special Cases:

Electromagnetic Induction And Alternating Current

Electromagnetic Induction Assertion Reason Type Questions And Answers

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: Induced emf in a conductor is proportional to the time rate of change of associated magnetic flux.

Statement 2: In the case of electromagnetic induction transfer of energy takes place in a manner energy is conserved.

Answer: 2. Statement 1 is true, Statement 2 Is true) statement 2 is not a correct explanation for statement 1

Question 2. Statement 1: The north pole of a.bar, magnet is moving towards a closed circular coil along its axis. As a result, the direction of induced currently the front face of the coil will be clockwise.

Statement 2: Any incident connected with electromagnetic induction obeys Lenz’s law.

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

Question 3. Statement 1: The charge passing through a coil in time At is \(\frac{\Delta \phi}{R}\), where R is the resistance of the coil and ΔΦ is the change in flux linked with the coil in time Δt:

Statement 2: The induced emf in a conductor \(e \propto \frac{d \phi}{d t}\), at \(\frac{d \phi}{d t}\) where is the time rate of change of flux linked with the conductor.

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

Question 4. Statement 1: A closed solenoid is placed in an external magnetic field. Both the magnetic field and axis of the solenoid are directed along the z-axis. No electromotive force will be induced if the solenoid is rotated along its axis.

Statement 2: Electromagnetic induction in a conducting coil takes place only when the magnetic flux linked with the coil changes with time.

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

Question 5. Statement 1: Self-inductance of a solenoid having 1000 turns is l00 mH if the associated magnetic flux linked with each turn is 10^-3 Wb for a current of 1 A passing through the solenoid.

Statement 2: The self-inductance L of a coil is defined as Φ = LI when <p is the magnetic flux linked with the coil for a current I through it.

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

Question 6. Statement 1: When two coils are wound on each other, the mutual induction between the coils is maximum.

Statement 2: Mutual induction does not depend on the orientation of the coils.

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

Question 7. Statement 1: When the number of turns of a coil. doubled, the coefficient of self-inductance of the coil becomes 4 times.

Statement 2: Self-inductance oc (number of turns)2

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

Electromagnetic Induction And Alternating Current

Electromagnetic Induction Match The Following

Question 1. Some physical quantities are given in column 1 while their units are given in column 2.

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

Question 2. Column 1 gives the value of self-inductance (L) and the current (I) through some coils. Column 2 gives the energy stored in them.

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

Question 3. Column 1 describes,s some solenoids kept in the air while column 2 gives, their self-inductance. Magnetic permeability of air = 4π x 10^-7 H.m^-1.

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

Electromagnetic Induction And Alternating Current

Electromagnetic Induction Comprehension Types

Read The Following Passage Carefully And Answer The Questions At The End Of It.

If the current through a solenoid changes with time electromagnetic induction takes place in the solenoid. This is known as self-induction, In general, for a current I, the induced emf in the coil is \(e=-L \frac{d I}{d t}\).

L is the self-inductance of the solenoid. On the other hand, such a change in the current in a solenoid can produce electromagnetic induction, in another adjacent solenoid. The induced emf in the other solenoid \(e=-M \frac{d I}{d t}\), M is called the mutual inductance of the solenoids.

If L₁ and L₁ are the self-inductance of the adjacent coils then their mutual inductance \(M=k \sqrt{L_1 L_2}\). If the magnetic flux produced by the current in one coil is linked with the other coil then k = 1.

Question 1. Unit of self-inductance L is Henry (H). Its relation with other known units is

1. V.s
2. A.s^-1
3. Ω .s
4. Ω .s^-1

Answer: 3. Ω .s

Question 2. The self-inductance (in H) of a coil when the Induced emf is 5.0μV for a change of 1 mA.s^-1 in current through it, is

1. 50
2. 5
3. 0.5
4. 0.05

Answer: 4. 0.05

Question 3. If the induced emf in a coil linked with the coil in question (2) is 20μV, the mutual inductance (in H) of die two coils is

1. 0.002
2. 0.02
3. 0.2
4. 2

Answer: 2. 0.02

Question 4. Self-inductance (in H) of the coil in question (3) is

1. 0.1
2. 0.08
3. 0.01
4. 0.008

Answer: 4. 0.008

Question 5. The negative sign in the expression of induced emf is explained by

1. Faraday’s first law
2. Faraday’s second law
3. Law of conservation of energy
4. Law of conservation of charge

Answer: 3. Law of conservation of energy

Electromagnetic Induction And Alternating Current

Electromagnetic Induction Integer Answer Type

In this type, the answer to each of the questions Is a single-digit integer ranging from 0 to 9.

Question 1. The current through a coil of self-inductance 500 mH is 4 A. What amount of magnetic energy (in J) is stored in its magnetic field?

Answer: 4

Question 2. A straight conductor of length 50 cm is moving with a velocity of 5 m.s^-1 in a magnetic field of strength 2T in a direction perpendicular to the field. What is die emf(in V) induced between the two ends of the conductor?

Answer: 5

Question 3. The magnetic flux Φ (in Wb) linked with a 100H coil changes with time t (in s) according to the relation Φ = 8t² – 2t + 1. What is the value of induced current (in A) in the coil at t = 2 s?

Answer: 3

Question 4. A straight conductor of length 10 cm Is rotating In a vortical plane with one of Its ends fixed. The angular velocity Is 10 rad.s^-1. What is the value of one (In μV ) Induced between the two ends of the conductor If the horizontal component of earth’s magnetic field at the place Is 10^-4 T?

Answer: 5

Question 5. A circular wire loop of radius is placed In the xy-plane centred at the origin 0. A square loop of side a (\(a \ll R\)) having two turns is placed with its centre at z = √3R along the axis of the circular wire loop, as shown. The plane of the square loop makes an angle of 45° concerning the z-axis. If the mutual Inductance between the loops is given by \(\frac{\mu_0 a^2}{2^{p / 2} R}\), then what is the value of p?

Answer: 7

Question 6. Find the change in magnetic flux (in Wb )in an inductor of 10H in which the emf induced is 300 V in 10^-2 s.

Answer: 3

Question 7. An average induced emf of 0.20 V appears in a coil when the current is changed from 5A in one direction to 5A in the opposite direction in 0.20 s. Find the self-inductance of the coil in mH.

Answer: 4