Electromagnetic Induction

time

Reason — The charge that flows through a coil of N turns and resistance R when the flux linked with it changes by ΔΦ_B is

$\text q = \frac{\text N\Delta\Phi_\text B}{\text R}$

It depends on the change in flux and on the resistance of the coil, but not on the time in which the flux changes.

There is an electric field in the rod

Reason — The magnetic (Lorentz) force qvB on the free electrons drives them towards one end of the rod, so that end becomes negatively charged and the other end positively charged. This separation of charge sets up an electric field E = Bv within the rod, directed from the positive end to the negative end, and the potential falls uniformly from one end of the rod to the other.

10

Reason — The magnitude of the self-induced emf is $|\text e| = \text L\frac{\Delta \text I}{\Delta \text t}$, so

$$\begin{aligned}\text L = \frac{|\text e|\thinspace\Delta \text t}{\Delta \text I} \\[1em] = \frac{2 \times 3 \times 10^{-2}}{8 - 2} \\[1em] = \frac{6 \times 10^{-2}}{6} \\[1em] = 10^{-2}\ \text H \\[1em] \text L = 10\ \text{mH}\end{aligned}$$

zero

Reason — No magnetic flux is linked with a thin straight wire when a current flows through it, so its self-inductance is zero.

remains same

Reason — Inserting the soft-iron core increases the self-inductance of the coil. But with a DC source the current is steady, so the self-induced emf $\text L\frac{\text{dI}}{\text{dt}}$ is zero and the current is decided by the resistance of the circuit alone. Apart from a momentary dip while the core is being moved in, the current, and hence the light from the bulb, remains the same.

Reason — When the switch is put on, the self-induced emf in the solenoid opposes the growth of current. The current therefore does not reach its steady value at once: it rises from zero and grows gradually to the steady value $\frac{\text E}{\text R}$, the rate of rise becoming smaller as the current approaches this value. This is the variation shown in graph (b).

Note: The textbook answer key prints (c). Graph (c) shows the current decaying, which is the case when the switch is put off; on switching on, the current grows as in graph (b).

NAB

Reason — If θ is the angle between the normal to the coil and $\vec{\text B} \space$at any instant, the flux linked with the coil is

$\Phi = \text{NAB}\cos\theta$

This is maximum when cos θ = 1, that is, when the plane of the coil is perpendicular to the field, giving Φ_max = NAB. The rate of rotation n changes the induced emf, not the maximum flux.

0.2 A

Reason — By Faraday's law,

$$\begin{aligned}\text e = -\frac{\text d\Phi_\text B}{\text{dt}} \\[1em] = -\frac{\text d}{\text{dt}}\left(6\text t^2 - 5\text t + 1\right) \\[1em] = -(12\text t - 5)\end{aligned}$$

At t = 0.25 s,

$$\begin{aligned}\text e = -(12 \times 0.25 - 5) \\[1em] = -(3 - 5) \\[1em] \text e = 2\ \text V\end{aligned}$$

Hence, the induced current is

$$\begin{aligned}\text I = \frac{\text e}{\text R} \\[1em] = \frac{2}{10} \\[1em] \text I = 0.2\ \text A\end{aligned}$$

Lenz's law

Reason — According to Lenz's law, the direction of the induced emf or the induced current in any circuit is such as to oppose the cause that produces it.

energy

Reason — Since the induced current opposes the change that produces it, mechanical work has to be done against it, and this work appears as electrical energy in the circuit. If the induced current did not oppose the change, electrical energy would be obtained continuously without doing any work, which is impossible.

the coil is stationary in external spatially varying magnetic field, which does not change with time

Reason — An emf is induced only when the magnetic flux linked with the coil changes. A stationary coil in a field that does not change with time has a constant flux linked with it, so no emf is induced. In the other cases the flux linked with the coil can change, either because the field varies with time or because the coil moves through the field.

decreases

Reason — Inserting the iron core increases the self-inductance of the solenoid and hence the magnetic flux linked with it. While this flux is increasing, a self-induced (back) emf is set up which, by Lenz's law, opposes the increase. This back emf acts against the battery, so the current in the solenoid decreases while the core is being inserted.

zero

Reason — Speed of the coil = $\frac{1}{0.5}$ = 2 m s^-1 and speed of the magnet = $\frac{2}{1}$ = 2 m s^-1.

Both move in the same direction with the same speed, so there is no relative motion between them. The flux linked with the coil does not change, and the induced emf is zero.

0.16 πμV

Reason — The flux linked with the loop of radius r is Φ = Bπr². As B is constant, the magnitude of the induced emf is

$$\begin{aligned}|\text e| = \left|\frac{\text d}{\text{dt}}\left(\text B\pi\text r^2\right)\right| \\[1em] = \text B \times 2\pi\text r\left|\frac{\text{dr}}{\text{dt}}\right|\end{aligned}$$

Since the radius is shrinking, $\frac{\text{dr}}{\text{dt}}$ is negative and $\left|\frac{\text{dr}}{\text{dt}}\right|$ = 0.2 × 10^-3 m s^-1. Substituting this with B = 40 × 10^-3 T and r = 1.0 × 10^-2 m,

$$\begin{aligned}|\text e| = 40 \times 10^{-3} \times 2\pi \times 1.0 \times 10^{-2} \times 0.2 \times 10^{-3} \\[1em] = 0.16\pi \times 10^{-6}\ \text V \\[1em] |\text e| = 0.16\ \pi\mu\text V\end{aligned}$$

2e

Reason — The induced emf is proportional to the rate of change of flux, which depends on the relative speed of the magnet and the coil. In the first case the relative speed is v. When both recede from each other, each with speed v, the relative speed becomes 2v, so the induced emf becomes 2e.

no current flows through the ammeter A

Reason — The field of the cylindrical magnet is symmetrical about its axis of rotation. Rotating the magnet about this axis therefore produces no change in the magnetic flux linked with the circuit formed by the wire and the ammeter. No emf is induced, so no current flows through the ammeter A.

permeability

Reason — From the self-inductance of a long solenoid, $\text L = \frac{\mu_0\text N^2\text A}{\text l}$, we get $\mu_0 = \frac{\text{Ll}}{\text N^2\text A}$. Since L is in henry, l in metre and A in metre², the unit of μ₀ (permeability) is henry/metre (H m^-1).

4B₀L² Wb

Reason — The square lies in the X-Y plane, so its area vector is along the Z-axis, $\vec{\text A} = \text L^2\hat{\text k}$. Hence,

$$\begin{aligned}\Phi = \vec{\text B} \cdot \vec{\text A} \\[1em] = \text B_0\left(2\hat{\text i} + 3\hat{\text j} + 4\hat{\text k}\right) \cdot \text L^2\hat{\text k} \\[1em] \Phi = 4\text B_0\text L^2\ \text{Wb}\end{aligned}$$

l decreases and A increases

Reason — The self-inductance of a solenoid is $\text L = \frac{\mu_0\text N^2\text A}{\text l}$. With N fixed, L ∝ $\frac{\text A}{\text l}$, so L increases when A increases and l decreases.

self-inductance of coil A is greater

Reason — Since NΦ_B = LI, the slope of the Φ-I graph gives the self-inductance. Line A makes 30° with the Φ-axis, that is, 60° with the I-axis, while line B makes 30° with the I-axis. So

$$\begin{aligned}\frac{\text L_\text A}{\text L_\text B} = \frac{\tan 60^\circ}{\tan 30^\circ} \\[1em] = \sqrt{3} \times \sqrt{3} \\[1em] \frac{\text L_\text A}{\text L_\text B} = 3\end{aligned}$$

Hence the self-inductance of coil A is greater.

Their resistances

Reason — The mutual inductance of two coils depends on their number of turns (and hence their self-inductances), their area of cross-section, their relative separation and orientation, and the permeability of the medium between them. It does not depend on their resistances.

A-II, B-III, C-IV, D-I

Reason — Gauss' law of magnetostatics states that the net magnetic flux through any closed surface is zero (II). Faraday's law states that the line integral of $\vec{\text E} \space$round a closed loop equals the negative rate of change of magnetic flux through it (III). Ampere's law relates the line integral of $\vec{\text B} \space$round a closed loop to the current enclosed (IV). Gauss' law of electrostatics relates the electric flux through a closed surface to the charge enclosed (I).

Note: Entry IV is printed with a negative sign; Ampere's circuital law is $\oint \vec{\text B} \cdot \text{d}\vec{\text l} = \mu_0\text I$. The matching is unaffected.

8rvB

Reason — For a ring moving on the conducting surface, the emf between its lowest point (in contact with S) and its highest point is the same as that across its vertical diameter 2R moving with the velocity of the centre, e = B(2R)v.

For the ring of radius r moving with velocity 2v,

$$\begin{aligned}\text e_1 = \text B \times 2\text r \times 2\text v \\[1em] \text e_1 = 4\text{rvB}\end{aligned}$$

For the ring of radius 2r moving with velocity v,

$$\begin{aligned}\text e_2 = \text B \times 4\text r \times \text v \\[1em] \text e_2 = 4\text{rvB}\end{aligned}$$

The lowest points of both rings touch the conducting surface S, so they are at the same potential. As the rings move in opposite directions, the two emfs have opposite polarities: the highest point of one ring is at +4rvB and that of the other at −4rvB with respect to S. Hence the potential difference between the highest points is 4rvB + 4rvB = 8rvB.

is repelled

Reason — As the current in A increases, the magnetic flux linked with B increases. By Lenz's law, the current induced in B flows opposite to the current in A, so as to oppose this increase. Coils carrying currents in opposite directions repel each other, so B is repelled by A.

[ML²T^-3A^-1].

Explanation:
Numerically, $|\text e| = \frac{\text d\Phi_\text B}{\text{dt}}$, and the dimensions of magnetic flux are [ML²T^-2A^-1]. Therefore,

$$\begin{aligned}[\text e] = \frac{[\text{ML}^2\text T^{-2}\text A^{-1}]}{[\text T]} \\[1em] [\text e] = [\text{ML}^2\text T^{-3}\text A^{-1}]\end{aligned}$$

Resistance of the coil.

Explanation:
By Faraday's law, $\text e = -\text N\frac{\text d\Phi_\text B}{\text{dt}}$, so the induced emf depends on the number of turns and the rate of change of magnetic flux. The resistance of the coil affects only the induced current, $\text I = \frac{\text e}{\text R}$.

Yes.

Explanation:
Lenz's law gives the direction of the induced emf, and an emf is induced even when the circuit is open. Its sense is such that, if the circuit were closed, the current would oppose the change in flux that produces it.

[ML²T^-2A^-2].

Explanation:
Numerically, $\text L = \frac{\text e}{\Delta \text I/\Delta \text t}$, so the unit of L is $\frac{\text{volt} \times \text{second}}{\text{ampere}} = \frac{\text{newton} \times \text{metre} \times \text{second}}{\text{coulomb} \times \text{ampere}}$. Therefore,

$$\begin{aligned}[\text L] = \frac{[\text{MLT}^{-2}][\text L][\text T]}{[\text{AT}][\text A]} \\[1em] [\text L] = [\text{ML}^2\text T^{-2}\text A^{-2}]\end{aligned}$$

Number of turns in the coil and its radius (area of cross-section).

Explanation:
For an air-core coil, $\text L = \frac{\mu_0\text N^2\text A}{\text l}$, with A = πr². So L ∝ N² and L ∝ A.

The self-inductance is directly proportional to the square of the number of turns, L ∝ N².

Explanation:
From $\text L = \frac{\mu_0\text N^2\text A}{\text l}$, for a coil of fixed length and area of cross-section, L varies as N².

Number of turns of the solenoid and the permeability of its core material.

Explanation:
For a solenoid wound on a core of relative permeability μ_r,

$\text L = \frac{\mu_\text r\mu_0\text N^2\text A}{\text l}$

so L ∝ N² and L ∝ μ_r.

It becomes 4 times the initial value.

Explanation:
Since L ∝ N², on doubling the number of turns,

$$\begin{aligned}\text L' = \frac{\mu_0(2\text N)^2\text A}{\text l} \\[1em] \text L' = 4\text L\end{aligned}$$

It increases.

Explanation:
With a core of relative permeability μ_r, $\text L = \frac{\mu_\text r\mu_0\text N^2\text A}{\text l}$, so the self-inductance becomes μ_r times its air-core value. For an iron core μ_r is very large, so L increases greatly.

[ML²T^-2A^-2], the same as that of self-inductance L.

Explanation:
Numerically, $\text M = \frac{\text e_2}{\Delta \text I_1/\Delta \text t}$, which has the same form as $\text L = \frac{\text e}{\Delta \text I/\Delta \text t}$.

M will increase.

Explanation:
When the coils are wound tightly one over the other, almost the whole flux produced by one coil is linked with the other. As the separation between the coils increases, the flux linked with the secondary coil decreases, and so does M.

Zero.

Explanation:
The conductor moves along the direction of the magnetic field, so the angle between $\vec{\text v} \space$and $\vec{\text B} \space$is 0° and it cuts no flux lines,

$$\begin{aligned}\text e = \text{Bvl}\sin 0^\circ \\[1em] \text e = 0\end{aligned}$$

Yes.

Explanation:
The horizontal axle, moving with the train, cuts the flux lines of the vertical component B_V of the earth's magnetic field. A potential difference e = B_Vvl is therefore induced between the ends of the axle, l being the length of the axle.

The axle of a train is horizontal and always moves horizontally, perpendicular to its own length. Whatever the direction of motion, it cuts the flux lines of the vertical component B_V of the earth's magnetic field, and the induced potential difference is B_Vvl.

The horizontal component B_H is never cut, being parallel to the velocity (north-south run) or along the axle (east-west run). Hence, for a given speed, the induced potential difference is the same in every direction.

No.

Explanation:
The wire falls vertically, parallel to the vertical component B_V, and it lies along the horizontal component B_H, which points from south to north. So it cuts the flux lines of neither B_H nor B_V, and no potential difference is induced.

Yes, a potential difference is induced across its ends.

Explanation:
The east-west wire falling vertically is perpendicular to the horizontal component B_H, and it moves perpendicular to B_H. It cuts the flux lines of B_H, so a potential difference B_Hvl is induced across its ends.

No.

Explanation:
The pole and its velocity both lie in the plane of the magnetic meridian, which also contains B_H and B_V. Hence the pole cuts the flux lines of neither B_H nor B_V, and no emf is induced between its ends.

As the magnet oscillates, the magnetic flux linked with the metallic plate changes, and eddy currents are induced in the plate. By Lenz's law these eddy currents oppose the motion of the magnet that produces them, so its oscillations are damped and the magnet stops soon.

The oscillations of the plate are damped and the plate soon comes to rest.

Explanation:
As the plate moves in the strong magnetic field, the flux linked with it changes, and eddy currents are produced in the plate. These currents oppose the motion of the plate (electromagnetic damping), so its oscillations stop soon.

Electromagnetic induction : Whenever the magnetic flux linked with a circuit changes, an emf is induced in the circuit; if the circuit is closed, an induced current also flows through it. This phenomenon is called electromagnetic induction. The induced emf and current last only while the flux is changing.

Faraday's law : When the magnetic flux through a circuit is changing, the magnitude of the induced emf set up in the circuit is equal to the rate of change of magnetic flux,

$|\text e| = \frac{\text d\Phi_\text B}{\text{dt}}$

For a closely wound coil of N turns, $|\text e| = \text N\frac{\text d\Phi_\text B}{\text{dt}}$.

Dimensions of emf : The emf is the work done per unit charge, $\text e = \frac{\text W}{\text q}$. With [W] = [ML²T^-2] and [q] = [AT],

$$\begin{aligned}[\text e] = \frac{[\text{ML}^2\text T^{-2}]}{[\text{AT}]} \\[1em] [\text e] = [\text{ML}^2\text T^{-3}\text A^{-1}] \qquad ...(\text i)\end{aligned}$$

Dimensions of rate of change of magnetic flux : Magnetic flux Φ_B = BA, where A is the area. From F = BIl (force on a current-carrying conductor), $\text B = \frac{\text F}{\text{Il}}$, so Φ_B = $\frac{\text F \times \text{area}}{\text{Il}}$ and

$$\begin{aligned}[\Phi_\text B] = \frac{[\text{MLT}^{-2}][\text L^2]}{[\text A][\text L]} \\[1em] [\Phi_\text B] = [\text{ML}^2\text T^{-2}\text A^{-1}]\end{aligned}$$

Therefore,

$$\begin{aligned}\left[\frac{\text d\Phi_\text B}{\text{dt}}\right] = \frac{[\text{ML}^2\text T^{-2}\text A^{-1}]}{[\text T]} \\[1em] \left[\frac{\text d\Phi_\text B}{\text{dt}}\right] = [\text{ML}^2\text T^{-3}\text A^{-1}] \qquad ...(\text{ii})\end{aligned}$$

From Eqs. (i) and (ii), the dimensions of emf are the same as those of the rate of change of magnetic flux.

Taking the two loops to be of the same size, the change in flux and the time-interval are the same for both. Since $\text e = -\frac{\Delta\Phi_\text B}{\Delta \text t}$, the induced emf will be the same in both loops.

The induced current is $\text I = \frac{\text e}{\text R}$. Copper has a smaller resistivity than aluminium, so the copper loop has the smaller resistance. Hence, the induced current will be greater in the copper loop.

Lenz's law : The direction of the induced emf, or of the induced current, in any circuit is such as to oppose the cause that produces it.

For example, when the north pole of a magnet is moved towards a coil, the induced current makes the near face of the coil a north pole, which opposes the approach of the magnet.

Self-inductance : If a current I flowing through a coil of N turns links a flux Φ_B with each turn, then NΦ_B = LI, where the constant L is the coefficient of self-induction or self-inductance of the coil. Thus, the self-inductance of a coil is equal to the number of flux-linkages with the coil when unit current flows through it.

Equivalently, since $\text e = -\text L\frac{\Delta \text I}{\Delta \text t}$, the self-inductance is numerically equal to the emf induced in the coil when the rate of change of current in it is unity.

SI unit : henry (H), where 1 H = 1 V s A^-1 = 1 Wb A^-1.

Since the self-inductance is numerically $\text L = \frac{\text e}{\Delta \text I/\Delta \text t}$, the statement means that an emf of 1 volt is induced in the coil when the current through it changes at the rate of 1 ampere per second.

Equivalently, from NΦ_B = LI, the number of flux-linkages with the coil is 1 weber when a current of 1 ampere flows through it.

Inductor : Every element of an electric circuit which has the property of self-induction, that is, which opposes a change of current through it by inducing a back emf, is an inductor. A coil or a solenoid is a common example.

Connecting wires : As a matter of principle, the connecting wires are also inductors. But a straight wire links practically no flux with itself, so its self-inductance is very feeble and is neglected. It becomes appreciable only when the wire is wound in the form of a coil.

The coils of a resistance box are made of doubled wire to make them non-inductive.

The doubled wire is wound so that the current at every place in the coil flows in two opposite directions. The magnetic fluxes produced by the two halves cancel, so the net flux linked with the coil is almost zero. Its self-inductance therefore becomes negligible, and when the current in the circuit is increased or decreased, no induced current is produced in the resistance coils.

This precaution avoids the effect of self-inductance of the coils of the bridge.

On pressing the cell-key, momentary induced currents are set up in the coils of the different arms, and the main current takes some time to reach its steady value. If the galvanometer-key were pressed first, the galvanometer would show a momentary deflection even when the bridge is exactly balanced, causing confusion. When the cell-key is pressed first, the induced currents die out before the galvanometer is connected, so its deflection is due to the bridge alone. For the same reason, the galvanometer-key is released first and then the cell-key.

Ideal inductor : An inductor having zero (negligible) resistance but a sufficiently large self-inductance is called an ideal inductor. A solenoid made from a thick wire is close to an ideal inductor.

Use of copper : Inductors are made of copper because, after silver, copper has the lowest resistivity. A copper coil therefore has a very small resistance, which keeps the inductor close to ideal.

Mutual induction : When two coils are placed near each other and the current in one of them (the primary coil) is changed, the magnetic flux linked with the other (the secondary coil) changes and an emf is induced in the secondary. This phenomenon is called mutual induction.

Example : A primary coil P is connected to a battery through a tapping key, and a secondary coil S placed near it is connected to a galvanometer. On pressing the key, the galvanometer shows a momentary deflection: the growth of current in P increases the flux linked with S, and the induced current in S flows opposite to the current in P. On releasing the key, the galvanometer deflects momentarily in the opposite direction, the induced current now being in the same direction as the current in P. The transformer works on this principle.

Mutual inductance : If a current I₁ in the primary coil links a flux Φ₂ with each of the N₂ turns of the secondary coil, then N₂Φ₂ = MI₁. The constant M is the mutual inductance of the two coils. It is equal to the number of flux-linkages with the secondary coil when unit current flows in the primary coil.

SI unit : henry (H).

Factors :

(i) the number of turns of the two coils, M ∝ N₁N₂;

(ii) the common area of cross-section (area of the secondary coil), M ∝ A.

M also depends on the relative separation and orientation of the coils and on the permeability of the core material.

Orientation of stable equilibrium : The torque on a current-loop of area A in an external magnetic field B is

$\tau = \text{IAB}\sin\theta$

where θ is the angle between the normal to the loop and the field. In stable equilibrium τ = 0 with θ = 0, that is, the plane of the loop is perpendicular to the external field, with the magnetic moment of the loop along $\vec{\text B}$.

Flux in this orientation : With θ = 0, the flux of the external field through the loop, BA cos θ, has its maximum value BA. Also, the magnetic field produced by the loop over its own area is along its normal, in the same direction as the external field. So the two fluxes add, and the total magnetic flux through the loop is maximum.

A current-loop is in stable equilibrium in an external magnetic field in the orientation in which the magnetic flux through it is maximum. The forces on the flexible wire therefore make the loop enclose the largest possible area. For a given length of periphery, a circle has the maximum area. Hence the loop takes a circular shape, with its plane perpendicular to the field.

Given,

• Radius of the loop = a
• Resistance of the loop = R
• B = B₀ sin ωt, perpendicular to the plane of the loop

The magnetic flux through the loop at any instant is

$$\begin{aligned}\Phi_\text B = \text{BA} \\[1em] \Phi_\text B = (\text B_0\sin\omega\text t)\pi\text a^2\end{aligned}$$

By Faraday's law, the induced emf is

$$\begin{aligned}\text e = -\frac{\text d\Phi_\text B}{\text{dt}} \\[1em] = -\pi\text a^2\text B_0\frac{\text d}{\text{dt}}(\sin\omega\text t) \\[1em] \text e = -\pi\text a^2\text B_0\omega\cos\omega\text t\end{aligned}$$

The induced current is

$$\begin{aligned}\text I = \frac{\text e}{\text R} \\[1em] \text I = -\frac{\pi\text a^2\text B_0\omega}{\text R}\cos\omega\text t\end{aligned}$$

Hence, the induced current is $\text I = -\frac{\pi\text a^2\text B_0\omega}{\text R}\cos\omega\text t$, an alternating current of peak value $\frac{\pi\text a^2\text B_0\omega}{\text R}$.

The space-traveller should connect the coil to the sensitive galvanometer to form a closed circuit and rotate the coil about different axes.

If a magnetic field is present, rotating the coil changes the magnetic flux linked with it, so an induced current flows and the galvanometer deflects. If there is no deflection for any rotation, there is no magnetic field on the planet.

As soon as the current is switched on, the magnetic field around the wire grows rapidly, and the magnetic flux linked with the bird's body changes. Induced currents therefore flow in the bird's body. Its wings, carrying currents in opposite directions, experience mutual repulsion and spread out, and the bird flies away.

More work will be done in case (a), when the coil is removed rapidly.

When the coil is removed rapidly, the rate of change of magnetic flux is greater, so a larger induced emf and a larger induced current are produced. By Lenz's law this current opposes the removal, so a larger opposing force has to be overcome. Hence more work is done, and it appears as a larger amount of electrical energy (heat) in the coil.

Plate P will be positive and plate Q will be negative.

Both magnets move towards the loop: the N-pole of the left magnet approaches from the left and the S-pole of the right magnet approaches from the right. Both motions increase the magnetic flux through the loop in the same direction, from left to right. By Lenz's law, the induced current opposes this increase: the left face of the loop behaves as a north pole (repelling the approaching N-pole) and the right face as a south pole (repelling the approaching S-pole). The induced current flowing round the loop in this sense carries positive charge to plate P, making P positive and Q negative.

Looking from the magnet's side, the induced current in the ring will be anticlockwise.

As the north pole approaches, the magnetic flux through the ring increases. By Lenz's law, the induced current opposes the approach of the magnet, so the face of the ring towards the magnet behaves as a north pole. For this, the current seen from the magnet's side must be anticlockwise.

(a) : With the current as shown, the right end of the solenoid, facing the loop, behaves as a north pole. As it is taken towards the loop, the flux through the loop increases, and by Lenz's law the near face of the loop becomes a north pole to oppose the approach. The other side of the loop then behaves as a south pole, so the current seen by the observer on that side is clockwise.

(b) : Here the solenoid is wound in the opposite sense, so its right end behaves as a south pole. The near face of the loop becomes a south pole and the other side a north pole, so the current seen by the observer is anticlockwise.

Yes, the direction is correctly shown.

The field lines enter the upper face of the coil, so this face is a magnetic south pole. As the ring falls towards the coil, the downward magnetic flux through the ring increases. By Lenz's law, the induced current opposes the fall: the lower face of the ring behaves as a south pole (repelling the south pole of the coil) and its upper face as a north pole. So, looking from above, the current in the ring is anticlockwise, as shown.

The ring will move away from the magnet.

As the N-pole of the magnet is brought near, the flux through the ring increases. By Lenz's law, the current induced in the ring makes its face towards the magnet behave as a north pole, to oppose the approach. The two north poles repel each other, so the ring moves away from the magnet.

(i) Non-conducting coil : a = g. No current can be induced in a non-conducting coil, so there is no force opposing the fall of the magnet.

(ii) Coil with a small cut : a = g. An emf is induced in the coil, but the circuit is open and no current flows. Hence there is no opposing force on the magnet.

(iii) Conducting coil : a < g. An induced current flows in the coil which, by Lenz's law, opposes the fall of the magnet both while it approaches the coil and while it moves away from it. Part of the kinetic energy of the magnet appears as heat in the coil, so its acceleration is less than g.

The current induced in AB will be from B to A.

The magnetic field due to the current I in C, at the position of AB (to the right of C), is perpendicular to the plane of the paper, directed downwards (into the paper). The conductor AB moves upwards, in the direction of the current I. By Fleming's right-hand rule (first finger into the paper, thumb upwards), the middle finger points from B to A.

The induced current in the loop will be clockwise.

The magnetic field due to the current I (from A to B) at the loop, which lies above the wire, is perpendicular to the plane of the paper, pointing upwards (out of the paper). As I increases, this flux through the loop increases. By Lenz's law, the induced current produces a field into the paper to oppose the increase, so it flows clockwise.

The induced current will again be clockwise.

With the current from B to A, the field at the loop is directed into the paper. As this current decreases, the flux into the paper decreases. By Lenz's law, the induced current tends to maintain this flux by producing a field into the paper, so it flows clockwise.

The induced current in the loop will be clockwise.

The field due to the current in OO' at the loop is directed into the paper. As the loop is pulled away, this flux decreases. The induced current opposes the motion of the loop (Lenz's law), so the current in the arm AB flows in the same direction as in OO' (upwards), and AB and OO' attract each other. A current from A to B in the arm AB means the current in the loop ABCD is clockwise.

Yes, a current is induced in B.

As A moves towards B, the magnetic flux linked with B due to the current in A increases. By Lenz's law, the current induced in B opposes this motion, so it flows opposite to the current in A (anticlockwise as seen from A's side), and B repels A.

No current is induced in B due to the current in C, because B and C are fixed and there is no relative motion between them.

The current in each loop will decrease.

Since the loops carry currents in the same direction, the flux of each loop through the other is in the same direction as its own. As the loops are brought nearer, the magnetic field at their centres increases. The induced emf in each loop acts so as to oppose this increase (Lenz's law), that is, opposite to the existing current. Hence the current in each loop decreases.

(i) Loop completely in the field : No. So long as the loop moves within the uniform field, the magnetic flux passing through it does not change. No emf or current is induced, and no work is required to move the loop.

(ii) Loop partially out of the field : Now the area of the loop within the field, and hence the flux through it (into the paper), decreases. By Lenz's law, a clockwise current is induced in the loop to oppose this decrease.

By Fleming's left-hand rule, forces $\vec{\text F}_1$, $\vec{\text F}_2$ and $\vec{\text F}_3$ act on the arms of the loop lying in the field. $\vec{\text F}_2$ and $\vec{\text F}_3$, on the upper and lower arms, are equal and opposite and cancel. $\vec{\text F}_1$, on the vertical arm inside the field, opposes the motion of the loop. Hence an external force equal to $\vec{\text F}_1$ must be applied to move the loop, and the work done by it appears as electrical energy (heat) in the loop.

(iii) Loop stationary, field changing with time : The flux through the loop changes with time, so an emf is induced and a current flows in the loop.

The rectangular loop will have a constant induced emf.

As the rectangular loop moves out with constant velocity v, its side perpendicular to the motion (of length l) stays at the boundary, so the area inside the field decreases at the constant rate lv, and the induced emf e = Blv is constant. For the circular loop, the length of the chord at the boundary changes continuously during its passage out of the field, so the rate of change of area, and hence the emf, is not constant.

The self-inductance of the coil becomes almost zero.

Such winding is called 'non-inductive' winding. The two sets of wires carry currents in opposite directions, so they produce magnetic fluxes in opposite directions and the net flux linked with the coil is almost zero. Hence the self-inductive effect is minimised.

Pulling the copper sheet out of the field, or pushing it into the field, changes the magnetic flux linked with the sheet. Eddy currents are therefore induced in the sheet. By Lenz's law, these currents oppose the cause that produces them, that is, the motion of the sheet, and so we experience an opposing force.

The stone will reach the earth earlier.

As the metal piece falls through the earth's magnetic field, the flux linked with it changes and eddy currents are produced in it. These currents oppose its motion, so the metal piece falls with an acceleration smaller than the acceleration due to gravity. The stone is non-conducting; no eddy currents are produced in it, and it falls with the acceleration due to gravity.

About the bar : The iron bar must be a magnet. As it falls, the magnetic flux linked with the copper tube (due to the non-uniform field of the magnet) changes, and eddy currents are produced in the body of the tube. By Lenz's law these currents oppose the fall of the magnet, which therefore experiences a retarding force. This force increases with the velocity of the magnet, and when it becomes equal to the weight, the magnet falls with a constant terminal velocity.

Effect of heating : On heating, the resistance of the copper tube increases, so the eddy currents become weaker and the retarding force at a given speed is smaller. Hence the terminal velocity of the magnet increases.

The induced emf will be uniform in the rectangular loop.

The induced emf is $\text e = \text B\frac{\Delta \text A}{\Delta \text t}$. As the rectangular loop enters the field with constant velocity v, its leading side of constant length l crosses into the field, so the area inside the field increases at the constant rate lv and the emf Blv is uniform. For the circular, elliptical and triangular loops, the width of the loop at the boundary changes as they enter, so the rate of change of flux, and hence the emf, is not uniform.

A — Anticlockwise; B — Clockwise; C — Induced current is zero.

The field due to the upward current I, at the loops to the right of the wire, is directed into the paper, and it is stronger nearer the wire.

Loop A : It moves towards the wire, so the flux into the paper increases. The induced current produces a field out of the paper to oppose this, so it is anticlockwise.

Loop B : It moves away from the wire, so the flux into the paper decreases. The induced current produces a field into the paper, so it is clockwise.

Loop C : It moves parallel to the wire, so its distance from the wire, and hence the flux through it, does not change. No current is induced.

A current is induced in the loop only while the magnetic flux through it changes, that is, while the loop is entering or leaving the field.

Case 1 (a > b) : As the front arm crosses the field of width b, the flux increases for a time $\frac{\text b}{\text v}$. Then the whole field region lies within the loop and the flux stays constant, so no current flows. As the rear arm crosses the field, the flux decreases for another $\frac{\text b}{\text v}$. Hence,

$$\begin{aligned}\text t = \frac{\text b}{\text v} + \frac{\text b}{\text v} \\[1em] \text t = \frac{2\text b}{\text v}\end{aligned}$$

Case 2 (a < b) : The loop takes a time $\frac{\text a}{\text v}$ to enter the field completely. While it is wholly inside the field the flux is constant. It again takes $\frac{\text a}{\text v}$ to leave the field. Hence,

$$\begin{aligned}\text t = \frac{\text a}{\text v} + \frac{\text a}{\text v} \\[1em] \text t = \frac{2\text a}{\text v}\end{aligned}$$

Hence, the current is induced for a time $\frac{2\text b}{\text v}$ in case 1 and $\frac{2\text a}{\text v}$ in case 2.

First law (Neumann's law) : When the magnetic flux through a circuit is changing, an induced emf is set up in the circuit whose magnitude is equal to the negative rate of change of magnetic flux. If the flux changes by ΔΦ_B in time Δt,

$\text e = -\frac{\Delta\Phi_\text B}{\Delta \text t}$

In the limit Δt → 0, and for a coil of N turns,

$\text e = -\text N\frac{\text d\Phi_\text B}{\text{dt}} \qquad ...(\text i)$

Explanation : When the north pole of a magnet is brought towards a coil, the number of flux lines passing through the coil increases; when it is taken away, the number decreases. In either case an emf is induced in the coil, but only while the magnet is moving. The faster the motion, the greater the rate of change of flux and the larger the induced emf. If the coil is open, the emf is still induced but no current flows, which shows that the change in flux induces an emf, not a current.

Second law (Lenz's law) : The direction of the induced emf, or the current, in any circuit is such as to oppose the cause that produces it. When the north pole of a magnet is moved towards a coil, the induced current makes the near face of the coil a north pole, which opposes the motion of the magnet.

Significance of the negative sign : The negative sign in Eq. (i) expresses Lenz's law. It shows that the induced emf opposes the change in magnetic flux that produces it, which keeps the law consistent with the conservation of energy.

Lenz's law : The direction of the induced emf, or the induced current, in any circuit is such as to oppose the cause that produces it.

Consistency with conservation of energy : Consider a magnet and a coil connected to a galvanometer.

(i) When the north pole of the magnet is moved towards the coil, the induced current makes the near face of the coil a north pole. The repulsion between the two north poles opposes the motion of the magnet towards the coil.

(ii) When the magnet is moved away, the near face of the coil becomes a south pole. The attraction between the poles opposes the withdrawal of the magnet.

In either case, work has to be done against the opposing force to move the magnet, and it is this mechanical work which appears as electrical energy (heat) in the coil.

If the induced current did not oppose the motion of the magnet, the magnet, once given a push, would be pulled on by the coil, and electrical energy would be obtained continuously without doing any work. This is impossible. Thus, Lenz's law is in accordance with the principle of conservation of energy.

Laws of electromagnetic induction :

First law : When the magnetic flux through a circuit is changing, an induced emf is set up in it whose magnitude is equal to the negative rate of change of magnetic flux. For a coil of N turns,

$\text e = -\text N\frac{\Delta\Phi_\text B}{\Delta \text t} \qquad ...(\text i)$

The emf lasts only while the flux is changing, and it is larger when the change is faster.

Second law (Lenz's law) : The direction of the induced emf, or the current, is such as to oppose the cause that produces it. For example, when the north pole of a magnet approaches a coil, its near face becomes a north pole and opposes the approach.

Induced charge : If the total resistance of the closed coil is R, the induced current is

$$\begin{aligned}\text I = \frac{|\text e|}{\text R} \\[1em] \text I = \frac{\text N}{\text R}\frac{\Delta\Phi_\text B}{\Delta \text t}\end{aligned}$$

The charge that flows through the circuit in the time-interval Δt is

$$\begin{aligned}\text q = \text I \times \Delta \text t \\[1em] = \frac{\text N}{\text R}\frac{\Delta\Phi_\text B}{\Delta \text t} \times \Delta \text t \\[1em] \text q = \frac{\text N\Delta\Phi_\text B}{\text R} \qquad ...(\text{ii})\end{aligned}$$

Eq. (ii) contains no Δt. Hence, the induced charge depends only on the change in magnetic flux and the resistance of the circuit, and not upon the rate of change of flux; whether the flux changes rapidly or slowly, the same charge flows.

Consider a thin conducting rod JK of length l situated in a uniform magnetic field $\vec{\text B} \space$directed into the plane of the paper. The rod is moved in the plane of the paper, perpendicular to its own length, with a constant velocity $\vec{\text v} \space$towards the right.

Magnetic force on the charges : The free electrons in the rod move with it, with velocity $\vec{\text v} \space$perpendicular to $\vec{\text B}$. Each charge q experiences a magnetic (Lorentz) force of magnitude

$\text F_\text m = \text{qvB}$

By Fleming's left-hand rule, this force pushes the free electrons towards the end K. So the end K becomes negatively charged and the end J, having a deficiency of electrons, positively charged.

Electric force : This separation of charge produces a potential difference V between the ends of the rod and hence an electric field within the rod, directed from J to K, of magnitude

$\text E = \frac{\text V}{\text l} \qquad ...(\text i)$

This field exerts on each charge a force F_e = qE, opposite to the magnetic force.

Equilibrium : As more electrons reach K, F_e increases until it becomes equal to F_m. Then the net force on each charge is zero and no more electrons move to K. In this condition,

$$\begin{aligned}\text F_\text e = \text F_\text m \\[1em] \text{qE} = \text{qvB} \\[1em] \text E = \text{Bv} \qquad ...(\text{ii})\end{aligned}$$

Comparing Eqs. (i) and (ii),

$$\begin{aligned}\frac{\text V}{\text l} = \text{Bv} \\[1em] \text V = \text{Blv}\end{aligned}$$

Hence, the motional emf induced between the ends of the rod is e = Blv.

Consider a U-shaped stationary conductor MNOL placed in a uniform magnetic field $\vec{\text B} \space$directed perpendicular to the plane of the paper, downwards. A conducting rod JK of length l slides on its arms with a constant velocity $\vec{\text v} \space$towards the right, so that JNOKJ forms a rectangular coil whose area changes with time.

Change in flux : In a small time-interval Δt, the rod moves through a distance Δx = vΔt to the new position J'K'. The area of the coil increases from JNOKJ to J'NOK'J', the increase being the area JJ'K'K = lΔx. The plane of the coil is perpendicular to $\vec{\text B}$, so the change in flux is

$\Delta\Phi_\text B = \text B \times \text l\Delta \text x$

Rate of change of flux :

$\frac{\Delta\Phi_\text B}{\Delta \text t} = \text{Bl}\frac{\Delta \text x}{\Delta \text t}$

But $\frac{\Delta \text x}{\Delta \text t}$ = v, the velocity of the rod. Thus,

$\frac{\Delta\Phi_\text B}{\Delta \text t} = \text{Blv}$

Induced emf : By Faraday's law,

$$\begin{aligned}\text e = -\frac{\Delta\Phi_\text B}{\Delta \text t} \\[1em] \text e = -\text{Blv}\end{aligned}$$

Hence, the induced emf produced by changing the area of the rectangular coil is − Blv. The negative sign signifies Lenz's law: the induced current flows in such a direction that the force on the rod due to it opposes the motion of the rod.

Eddy currents : When a piece of metal is placed in a changing magnetic field, or moved in a magnetic field so that the magnetic flux linked with it changes, induced currents are set up throughout the volume of the metal. These currents flow in closed paths, like eddies in water, and are called eddy currents (or Foucault currents). They oppose the motion of the metal, or the change in flux.

Production : Consider a copper sheet placed perpendicular to a magnetic field $\vec{\text B}$. When the sheet is withdrawn from the field, its area A within the field decreases, so the flux Φ_B (= BA) linked with it also decreases. Because of this flux-change, current-loops are induced in the part of the sheet coming out of the field. They flow in such a direction that their own magnetic field is in the same direction as the original field, so they oppose the decrease of flux, that is, the drawing out of the sheet.

Applications :

(i) Damping : In a dead-beat moving-coil galvanometer, the coil is wound on a light metal frame. Eddy currents set up in the frame oppose its motion, so the coil returns to zero quickly.

(ii) Induction furnace : The metal to be heated is placed in a rapidly changing magnetic field produced by a high-frequency alternating current. The eddy currents in the metal produce enough heat to melt it.

(iii) Electric brakes : A strong magnetic field applied to a rotating drum sets up eddy currents which exert a torque that stops the drum. This is used in stopping electric trains.

Principle of mutual induction : When two coils are placed near each other and the current in one of them is changed, the magnetic flux linked with the other changes, and an emf is induced in the second coil. This phenomenon is called mutual induction. The first coil is called the primary coil and the second the secondary coil.

A primary coil P is connected to a battery, a tapping key and a rheostat, and a secondary coil S placed near it is connected to a galvanometer.

(i) When the key is pressed, a current starts in P and the number of flux lines passing through S increases from zero. An emf is induced in S and the galvanometer shows a momentary deflection. By Lenz's law, the induced current in S is opposite to the current in P, opposing the growth of flux.

(ii) When the key is released, the flux lines through S disappear, and the galvanometer deflects momentarily in the opposite direction. The induced current in S is now in the same direction as in P, tending to maintain the flux.

(iii) Whenever the current in P is varied with the rheostat, a current is induced in S. The effect is much greater when the coils are wound on a common iron core.

Device : The transformer works on the principle of mutual induction (the induction coil also).

Coefficient of mutual induction : Let a current I₁ in the primary coil link a flux Φ₂ with each turn of the secondary coil of N₂ turns. The number of flux-linkages with the secondary is proportional to I₁,

$$\begin{aligned}\text N_2\Phi_2 \propto \text I_1 \\[1em] \text N_2\Phi_2 = \text M\text I_1 \qquad ...(\text i)\end{aligned}$$

where the constant M is the coefficient of mutual induction (mutual inductance) of the two coils. If I₁ = 1, M = N₂Φ₂. Hence, the coefficient of mutual induction of two coils is equal to the number of magnetic flux-linkages in the secondary coil when a unit current flows in the primary coil.

If the current in the primary changes, then by Faraday's law and Eq. (i),

$$\begin{aligned}\text e_2 = -\frac{\Delta(\text N_2\Phi_2)}{\Delta \text t} \\[1em] \text e_2 = -\text M\frac{\Delta \text I_1}{\Delta \text t}\end{aligned}$$

so M is also numerically equal to the emf induced in the secondary coil when the rate of change of current in the primary coil is unity.

SI unit : henry (H). The mutual inductance of two coils is 1 henry when an emf of 1 volt is induced in one of them due to a current changing at the rate of 1 ampere per second in the other, so 1 H = 1 V s A^-1 = 1 Wb A^-1.

Coefficient of self-induction : Let a current I flowing through a coil of N turns link a magnetic flux Φ_B with each turn. In the absence of magnetic materials, the number of flux-linkages NΦ_B is proportional to I,

$\text N\Phi_\text B = \text{LI} \qquad ...(\text i)$

The constant L is the coefficient of self-induction (self-inductance) of the coil. If I = 1, L = NΦ_B; that is, the coefficient of self-induction of a coil is equal to the number of flux-linkages with the coil when a unit current flows through it. Equivalently, it is numerically equal to the emf induced in the coil when the rate of change of current in it is unity.

SI unit : henry (H), where 1 H = 1 V s A^-1 = 1 Wb A^-1.

Self-inductance of a long solenoid : Consider a long air-cored solenoid of length l and area of cross-section A, having N closely wound turns. The number of turns per unit length is $\text n = \frac{\text N}{\text l}$. When a current I flows through it, the magnetic field inside the solenoid is

$\text B = \mu_0\frac{\text N}{\text l}\text I$

where μ₀ is the permeability of free space. The magnetic flux through each turn is

$$\begin{aligned}\Phi_\text B = \text{BA} \\[1em] \Phi_\text B = \mu_0\frac{\text N}{\text l}\text{IA}\end{aligned}$$

So the total flux linkage with the solenoid is

$$\begin{aligned}\text N\Phi_\text B = \text N\mu_0\frac{\text N}{\text l}\text{IA} \\[1em] \text N\Phi_\text B = \frac{\mu_0\text N^2\text{IA}}{\text l}\end{aligned}$$

From Eq. (i), the self-inductance of the solenoid is

$$\begin{aligned}\text L = \frac{\text N\Phi_\text B}{\text I} \\[1em] \text L = \frac{\mu_0\text N^2\text A}{\text l}\end{aligned}$$

Hence proved.

Motional emf : Consider a conducting rod JK of length l moved without friction on the arms of a U-shaped stationary conductor MNOL with a velocity $\vec{\text v} \space$towards the right. The arrangement lies in a uniform magnetic field $\vec{\text B} \space$directed perpendicular to the plane of the paper, downwards. The free electrons in the rod are acted upon by the magnetic (Lorentz) force qvB, which sets up a current in the closed circuit. The emf induced in the moving rod is called motional emf.

In a small time-interval Δt, the rod moves through a distance Δx = vΔt to the position J'K'. The area of the circuit increases by lΔx, so the change in magnetic flux through the circuit is

$\Delta\Phi_\text B = \text B \times \text l\Delta \text x$

Hence, the rate of change of magnetic flux is

$$\begin{aligned}\frac{\Delta\Phi_\text B}{\Delta \text t} = \text{Bl}\frac{\Delta \text x}{\Delta \text t} \\[1em] \frac{\Delta\Phi_\text B}{\Delta \text t} = \text{Blv}\end{aligned}$$

By Faraday's law, the induced emf is

$$\begin{aligned}\text e = -\frac{\Delta\Phi_\text B}{\Delta \text t} \\[1em] \text e = -\text{Blv} \qquad ...(\text i)\end{aligned}$$

So the magnitude of the motional emf is e = Blv.

Mechanical power : Let R be the total resistance of the circuit. The current in the circuit is

$$\begin{aligned}\text I = \frac{\text e}{\text R} \\[1em] \text I = \frac{\text{Blv}}{\text R} \qquad ...(\text{ii})\end{aligned}$$

The rod JK, carrying current I in the magnetic field, experiences a force (opposite to $\vec{\text v}$) of magnitude

$$\begin{aligned}\text F = \text{IlB} \\[1em] \text F = \frac{\text{Blv}}{\text R} \times \text{lB} \\[1em] \text F = \frac{\text B^2\text l^2\text v}{\text R}\end{aligned}$$

To keep the rod moving with constant velocity v, an equal and opposite external force must be applied. The mechanical power required is

$$\begin{aligned}\text P = \text{Fv} \\[1em] \text P = \frac{\text B^2\text l^2\text v}{\text R} \times \text v \\[1em] \text P = \frac{\text B^2\text l^2\text v^2}{\text R}\end{aligned}$$

Hence proved. This power is dissipated as Joule heat in the circuit, since $\text I^2\text R = \left(\frac{\text{Blv}}{\text R}\right)^2 \times \text R = \frac{\text B^2\text l^2\text v^2}{\text R}$.

Induced current : When the magnetic flux linked with a closed circuit changes, the current that flows in it due to the induced emf is called the induced current. If e is the induced emf and R the resistance of the circuit, I = $\frac{\text e}{\text R}$.

Induced charge : The total charge that flows through the circuit due to the induced current, during the time the magnetic flux is changing, is called the induced charge.

Derivation : Let the flux linked with each turn of a coil of N turns change by ΔΦ_B in a time-interval Δt. By Faraday's law, the magnitude of the induced emf is

$\text e = \text N\frac{\Delta\Phi_\text B}{\Delta \text t}$

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

$$\begin{aligned}\text I = \frac{\text e}{\text R} \\[1em] \text I = \frac{\text N}{\text R}\frac{\Delta\Phi_\text B}{\Delta \text t}\end{aligned}$$

The charge that flows through the circuit in the time-interval Δt is

$$\begin{aligned}\text Q_{\text{induced}} = \text I \times \Delta \text t \\[1em] = \frac{\text N}{\text R}\frac{\Delta\Phi_\text B}{\Delta \text t} \times \Delta \text t \\[1em] \text Q_{\text{induced}} = \frac{\text N}{\text R}\Delta\Phi_\text B\end{aligned}$$

Hence, Q_induced = $\frac{\text N}{\text R}$ (change in magnetic flux). With ΔΦ_B in weber and R in ohm, Q_induced is in coulomb. Since Δt does not appear, the induced charge does not depend on the rate of change of flux: whether the change is rapid or slow, the same charge flows.

Faraday's first law : When the magnetic flux through a circuit is changing, an induced emf is set up in the circuit whose magnitude is equal to the negative rate of change of magnetic flux,

$\text e = -\text N\frac{\text d\Phi_\text B}{\text{dt}}$

where N is the number of turns. The emf lasts only while the flux is changing.

Faraday's second law (Lenz's law) : The direction of the induced emf, or the current, in any circuit is such as to oppose the cause that produces it.

Lenz's law and conservation of energy :

Consider a magnet and a coil connected to a galvanometer.

(i) When the north pole of the magnet is moved towards the coil, the induced current flows in such a direction that the near face of the coil acts as a magnetic north pole. The repulsion between the two north poles opposes the motion of the magnet towards the coil.

(ii) When the north pole is moved away from the coil, the induced current makes the near face of the coil a south pole. The attraction between the two poles opposes the motion of the magnet away from the coil.

In either case, mechanical work has to be done in moving the magnet against the opposing force, and it is this work which appears as electrical energy in the coil.

If the induced current did not oppose the motion of the magnet, that is, if the near face became a south pole when the north pole approached, the magnet would be attracted towards the coil and would move on its own once pushed. Electrical energy would then be obtained continuously without doing any work, which is impossible. Hence, Lenz's law is in accordance with the principle of conservation of energy.

(i) Response of the student is correct.

While the key is pressed, a steady current flows through the solenoid and a magnetic flux is linked with it. On releasing the tapping key, the circuit is broken suddenly and the current falls to zero, so the flux linked with the solenoid vanishes rapidly. Because the solenoid has a large number of turns and a soft iron core, its self-inductance is large, and a large self-induced emf, $\text e = -\text L\frac{\Delta \text I}{\Delta \text t}$, is produced. This emf sends a current through the bulb connected in parallel, so the bulb glows brightly for a moment. As the induced emf dies away, the bulb goes out.

(ii) Response of the student is correct.

The self-inductance of a solenoid wound on a core of relative permeability μ_r is

$\text L = \frac{\mu_0\mu_\text r\text N^2\text A}{\text l}$

For soft iron, μ_r > 1 (very large). When the soft iron core is replaced by an air core (μ_r ≈ 1),

$\text L' = \frac{\mu_0\text N^2\text A}{\text l}$

Here L' < L, so the self-inductance decreases.

(iii) Response of the student is incorrect.

For a long solenoid of the same geometrical shape and size, L ∝ N². On increasing the number of turns from N to 4N,

$$\begin{aligned}\frac{\text L'}{\text L} = \left(\frac{4\text N}{\text N}\right)^2 \\[1em] \frac{\text L'}{\text L} = 16\end{aligned}$$

so the self-inductance becomes 16 times, not 4 times. To make it 4 times, the number of turns should be doubled, from N to 2N, since $\left(\frac{2\text N}{\text N}\right)^2 = 4$.

(i) Emf induced in the loop :

Only the arm PQ, of length b, cuts the magnetic flux lines: the horizontal arms move along their own lengths, and the extended part of the circuit lies outside the field. Hence the motional emf induced in the loop is

$\text e_{\text{in}} = \text{vBb}$

(ii) Reading of the milli-ammeter :

Zero. The circuit is not closed because of the gap RS, so no current flows, although the emf vBb is induced.

(iii) Direction of induced current :

Clockwise. As the loop moves to the right, the part of the circuit lying in the field decreases, so the flux into the paper decreases. By Lenz's law, the induced current produces a field into the paper to oppose this decrease, so it flows clockwise. Equivalently, by Fleming's right-hand rule, the current in PQ flows from Q to P.

(iv) Change in the induced charge :

The induced charge is

$\text q_{\text{in}} = \frac{\Delta\Phi}{\text R}$

where R is the total resistance of the circuit. For the same movement of the frame, the change in flux ΔΦ is the same. Since the induced charge does not depend upon the rate of change of magnetic flux, there will be no change in its value when the velocity is reduced to half.

(v) Force to maintain velocity v :

With r' connected between R and S, the total resistance of the circuit is (r + r'), and the induced current is

$$\begin{aligned}\text I = \frac{\text e_{\text{in}}}{\text r + \text r'} \\[1em] \text I = \frac{\text{vBb}}{\text r + \text r'}\end{aligned}$$

The arm PQ, carrying this current in the field, experiences a force IBb opposing the motion. To maintain the velocity v, an equal force must be applied on the frame in the direction of motion,

$$\begin{aligned}\text F = \text{IBb} \\[1em] \text F = \frac{\text{vBb}}{\text r + \text r'} \times \text{Bb} \\[1em] \text F = \frac{\text{vB}^2\text b^2}{\text r + \text r'}\end{aligned}$$

(i) Increasing the stored energy four times :

The energy stored in an inductor is $\text U = \frac{1}{2}\text{LI}^2$. For a given inductor, U ∝ I², so the energy stored becomes four times by doubling the current.

(ii) Linear dimensions tripled, turns per unit length constant :

The self-inductance of a solenoid is

$$\begin{aligned}\text L = \frac{\mu_0\text N^2\text A}{\text l} \\[1em] \text L = \mu_0\text n^2\text{Al}\end{aligned}$$

where n is the number of turns per unit length. When the linear dimensions are tripled, the area A becomes 9 times and the length l becomes 3 times, while n is unchanged. So

$$\begin{aligned}\text L' = \mu_0\text n^2(9\text A)(3\text l) \\[1em] \text L' = 27\text L\end{aligned}$$

The self-inductance becomes 27 times.

Note: The textbook prints "It becomes 3 times" for (ii). With n constant, L = μ₀n²Al varies as the cube of the linear dimensions, so it becomes 27 times.

(iii) Conservation principle :

Lenz's law is based on the principle of conservation of energy.

(iv) Acceleration of the magnet :

No, the acceleration of the magnet will be less than 'g'. As the magnet falls through the ring, an induced current flows in the ring which, by Lenz's law, opposes the motion of the magnet both while it approaches the ring and while it moves away from it.

(v) Phenomenon behind the transformer :

The working of a transformer is based on the phenomenon of mutual induction.

Given,

• Area of the coil, A = 100 cm² = 10^-2 m²
• Magnetic field, B = 10^-1 T
• Angle between the plane of the coil and the field = 60°
• Time taken by the field to become zero, Δt = 10^-3 s

Magnetic flux :

The angle θ between the field and the normal to the plane of the coil is 90° − 60° = 30°. Hence,

$$\begin{aligned}\phi = \text{BA}\cos\theta \\[1em] \phi = 10^{-1} \times 10^{-2} \times \cos 30^\circ \\[1em] \phi = 10^{-3} \times \frac{\sqrt{3}}{2} \\[1em] \phi = \frac{\sqrt{3}}{2} \times 10^{-3}\ \text{Wb}\end{aligned}$$

Induced emf :

When the field is reduced to zero, the flux changes from ϕ to zero. By Faraday's law, the magnitude of the induced emf is

$$\begin{aligned}\text e = \frac{\Delta\phi}{\Delta\text t} \\[1em] \text e = \frac{0.866 \times 10^{-3}}{10^{-3}} \\[1em] \text e = 0.866\ \text V\end{aligned}$$

Hence, the flux through the coil is $\frac{\sqrt{3}}{2} \times 10^{-3}$ Wb (= 8.66 × 10^-4 Wb) and the induced emf is 0.866 V.

Given,

• Number of turns, N = 50
• Change in flux through each turn, Δϕ = 0.3 − 0 = 0.3 Wb
• Time, Δt = 1 s

By Faraday's law, the magnitude of the induced emf is

$$\begin{aligned}\text e = \text N\frac{\Delta\phi}{\Delta\text t} \\[1em] \text e = 50 \times \frac{0.3}{1} \\[1em] \text e = 15\ \text V\end{aligned}$$

Hence, the induced emf across the ends of the coil is 15 V.

Given,

• Area of the coil, A = 0.5 m²
• Number of turns, N = 100
• Magnetic field, B = 0.04 Wb m^-2
• Time, Δt = 0.01 s

As the coil is perpendicular to the field, the initial flux through each turn is BA. On removal from the field the flux becomes zero, so Δϕ = BA.

$$\begin{aligned}\text e = \text N\frac{\Delta\phi}{\Delta\text t} \\[1em] \text e = \frac{\text{NBA}}{\Delta\text t} \\[1em] \text e = \frac{100 \times 0.04 \times 0.5}{0.01} \\[1em] \text e = \frac{2}{0.01} \\[1em] \text e = 200\ \text V\end{aligned}$$

Hence, the average emf induced in the coil is 200 V.

Given,

• Number of turns, N = 28
• Radius of the coil, r = $\frac{0.02}{2}$ = 0.01 m
• Initial field, B₁ = 8000 T; final field, B₂ = 3000 T
• Time, Δt = 4 s

Area of the coil,

$$\begin{aligned}\text A = \pi\text r^2 \\[1em] \text A = \frac{22}{7} \times (0.01)^2 \\[1em] \text A = \frac{22}{7} \times 10^{-4}\ \text m^2\end{aligned}$$

As the coil is perpendicular to the field, the change in flux through each turn is Δϕ = A(B₁ − B₂). By Faraday's law,

$$\begin{aligned}\text e = \text N\frac{\text A(\text B_1 - \text B_2)}{\Delta\text t} \\[1em] \text e = 28 \times \frac{22}{7} \times 10^{-4} \times \frac{8000 - 3000}{4} \\[1em] \text e = 88 \times 10^{-4} \times 1250 \\[1em] \text e = 11\ \text V\end{aligned}$$

Hence, the magnitude of the induced emf is 11 V.

Given,

• Area of the coil, A = 1.0 m²
• Magnetic field, B = 0.10 T
• Time, Δt = 0.01 s
• Resistance of the coil, R = 2.0 Ω

Initially the coil is perpendicular to the field, so ϕ₁ = BA. After rotation through 180°, its normal is opposite to the field, so ϕ₂ = −BA.

$$\begin{aligned}|\Delta\phi| = \text{BA} - (-\text{BA}) \\[1em] |\Delta\phi| = 2\text{BA} \\[1em] |\Delta\phi| = 2 \times 0.10 \times 1.0 \\[1em] |\Delta\phi| = 0.2\ \text{Wb}\end{aligned}$$

Induced emf :

$$\begin{aligned}\text e = \frac{|\Delta\phi|}{\Delta\text t} \\[1em] \text e = \frac{0.2}{0.01} \\[1em] \text e = 20\ \text V\end{aligned}$$

Induced current :

$$\begin{aligned}\text I = \frac{\text e}{\text R} \\[1em] \text I = \frac{20}{2.0} \\[1em] \text I = 10\ \text A\end{aligned}$$

If the coil is open : The induced emf depends only on the rate of change of flux, not on the resistance of the circuit, so it remains 20 V. The circuit is not closed (infinite resistance), so the induced current is zero.

Hence, the induced emf is 20 V and the induced current is 10 A. With the coil open, the emf is still 20 V but the current is zero.

Given,

• Area of the coil, A = 500 cm² = 5 × 10^-2 m²
• Number of turns, N = 2000
• Magnetic field, B = 4 × 10^-5 Wb m^-2
• Time, Δt = 0.1 s
• Total resistance, R = 20 Ω

On rotation through 180°, the flux through each turn changes from BA to −BA, so the total change in flux linkage is

$$\begin{aligned}\text N\Delta\phi = \text N \times 2\text{BA} \\[1em] \text N\Delta\phi = 2000 \times 2 \times 4 \times 10^{-5} \times 5 \times 10^{-2} \\[1em] \text N\Delta\phi = 8 \times 10^{-3}\ \text{Wb}\end{aligned}$$

(i) Average induced emf :

$$\begin{aligned}\text e = \frac{\text N\Delta\phi}{\Delta\text t} \\[1em] \text e = \frac{8 \times 10^{-3}}{0.1} \\[1em] \text e = 0.08\ \text V\end{aligned}$$

(ii) Induced charge :

The induced charge depends only on the change in flux linkage and the resistance,

$$\begin{aligned}\text q = \frac{\text N\Delta\phi}{\text R} \\[1em] \text q = \frac{8 \times 10^{-3}}{20} \\[1em] \text q = 4.0 \times 10^{-4}\ \text C\end{aligned}$$

Hence, the average induced emf is 0.08 V and the induced charge is 4.0 × 10^-4 C.

Given,

• Sides of the loop = 10 cm × 3 cm = 0.10 m × 0.03 m
• Magnetic field, B = 0.5 T
• Velocity, v = 1 cm s^-1 = 0.01 m s^-1

While the loop moves out, only the side that is still inside the field and perpendicular to the velocity cuts the field lines. The motional emf across the cut is e = Blv, where l is that side. It lasts until the loop has completely left the field, that is, for the time taken to move a distance equal to the side parallel to the velocity.

(i) Velocity normal to the longer side :

Here l = 0.10 m, and the loop must move 3 cm to leave the field.

$$\begin{aligned}\text e = \text{Blv} \\[1em] \text e = 0.5 \times 0.10 \times 0.01 \\[1em] \text e = 5 \times 10^{-4}\ \text V \\[1em] \text e = 0.5\ \text{mV}\end{aligned}$$

$$\begin{aligned}\text t = \frac{3\ \text{cm}}{1\ \text{cm s}^{-1}} \\[1em] \text t = 3\ \text s\end{aligned}$$

(ii) Velocity normal to the shorter side :

Here l = 0.03 m, and the loop must move 10 cm to leave the field.

$$\begin{aligned}\text e = \text{Blv} \\[1em] \text e = 0.5 \times 0.03 \times 0.01 \\[1em] \text e = 1.5 \times 10^{-4}\ \text V \\[1em] \text e = 0.15\ \text{mV}\end{aligned}$$

$$\begin{aligned}\text t = \frac{10\ \text{cm}}{1\ \text{cm s}^{-1}} \\[1em] \text t = 10\ \text s\end{aligned}$$

Hence, the induced voltage is 0.5 mV for 3 s in case (i) and 0.15 mV for 10 s in case (ii).

Given,

• Radius of the solenoid, r = 0.01 m
• Turns per unit length of the solenoid, n = 600 m^-1
• Turns in the smaller coil, N = 100
• Change in current, ΔI = 2 − 0 = 2 A
• Time, Δt = 1 ms = 10^-3 s

The field inside the long solenoid is B = μ₀nI, and it is confined to the area of cross-section A = πr² of the solenoid. The flux linked with the smaller coil is therefore

$\text N\phi = \text N\mu_0\text{nIA}$

so the emf induced in it is

$$\begin{aligned}\text e = \text N\mu_0\text{nA}\frac{\Delta\text I}{\Delta\text t} \\[1em] \text e = 100 \times 4\pi \times 10^{-7} \times 600 \times \pi(0.01)^2 \times \frac{2}{10^{-3}} \\[1em] \text e = 4\pi^2 \times 10^{-7} \times 100 \times 600 \times 10^{-4} \times 2000 \\[1em] \text e = 4.8\pi^2 \times 10^{-3} \\[1em] \text e = 4.8 \times 9.86 \times 10^{-3} \\[1em] \text e = 47.3 \times 10^{-3}\ \text V\end{aligned}$$

Hence, the emf induced in the smaller coil is 47.3 mV.

Given,

• Resistance of the loop, R = 10 Ω
• Area of the loop, A = 10 cm² = 10^-3 m²
• From the graph, the current falls uniformly from 0.4 A at t = 0 to zero at t = 1.0 s

(i) Total charge :

Since I = dq/dt, the charge passed is the area under the I-t graph, which is a triangle.

$$\begin{aligned}\text q = \frac{1}{2} \times 1.0 \times 0.4 \\[1em] \text q = 0.2\ \text C\end{aligned}$$

(ii) Change in magnetic flux :

The induced charge is related to the change in flux by

$\text q = \frac{\Delta\phi}{\text R}$

Therefore,

$$\begin{aligned}\Delta\phi = \text{qR} \\[1em] \Delta\phi = 0.2 \times 10 \\[1em] \Delta\phi = 2\ \text{Wb}\end{aligned}$$

(iii) Magnetic field :

The field acts normally and the loop is removed completely, so the flux changes from BA to zero and Δϕ = BA.

$$\begin{aligned}\text B = \frac{\Delta\phi}{\text A} \\[1em] \text B = \frac{2}{10^{-3}} \\[1em] \text B = 2000\ \text T\end{aligned}$$

Hence, the charge passed is 0.2 C, the change in flux is 2 Wb and the magnetic field is 2000 T.

Given,

• Length of the wire, l = 50 cm = 0.5 m
• Magnetic field, B = 0.8 N A^-1 m^-1 = 0.8 T
• Velocity, v = 2 ms^-1

When the wire moves perpendicular to both its length and the field, the motional emf is

$$\begin{aligned}\text e = \text{Blv} \\[1em] \text e = 0.8 \times 0.5 \times 2 \\[1em] \text e = 0.8\ \text V\end{aligned}$$

When the wire moves parallel to the field, it does not cut any field lines (e = Blv sin 0° = 0), so no emf is induced.

Hence, the induced emf is 0.8 V, and it is zero when the wire moves parallel to the field.

Given,

• Length of the wire, l = 50 cm = 0.5 m
• Speed, v = 6 ms^-1
• Magnetic field, B = 1.1 Wb m^-2, directed from east to west

Induced emf :

$$\begin{aligned}\text e = \text{Blv} \\[1em] \text e = 1.1 \times 0.5 \times 6 \\[1em] \text e = 3.3\ \text V\end{aligned}$$

Positive end :

The wire is horizontal and perpendicular to the east-west field, so it lies along the north-south direction. The wire moves downwards and the field points west. The Lorentz force $\text q(\vec{\text v} \times \vec{\text B})$ then drives the free electrons towards the south end, so the north end becomes positive.

Hence, the emf induced is 3.3 V and the north end of the wire is positive.

Given,

• Length of the axle, l = 2.4 m
• Speed of the car, v = 25 ms^-1
• Vertical component of earth's field, B_V = 3 × 10^-5 T

The axle is horizontal and moves horizontally, so it cuts only the vertical component B_V of the earth's field.

$$\begin{aligned}\text e = \text B_\text V\text{lv} \\[1em] \text e = 3 \times 10^{-5} \times 2.4 \times 25 \\[1em] \text e = 1.8 \times 10^{-3}\ \text V\end{aligned}$$

Hence, the potential difference induced between the ends of the axle is 1.8 × 10^-3 V.

Given,

• Length of the conductor, l = 0.4 m
• Magnetic field, B = 0.9 Wb m^-2
• Velocity, v = 7 ms^-1

The induced emf e = Blv sin θ is maximum when the conductor moves perpendicular to the field (θ = 90°).

$$\begin{aligned}\text e_\text{max} = \text{Blv} \\[1em] \text e_\text{max} = 0.9 \times 0.4 \times 7 \\[1em] \text e_\text{max} = 2.52\ \text V\end{aligned}$$

Hence, the maximum emf induced in the conductor is 2.52 V.

Given,

• Speed of the train, v = 180 km h^-1 = 180 × $\frac{5}{18}$ ms^-1 = 50 ms^-1
• Vertical component of earth's field, B_V = 0.2 × 10^-4 Wb m^-2
• Separation of the rails, l = 1 m

The axles of the train, of length equal to the separation of the rails, move horizontally and cut the vertical component B_V. The emf so induced appears across the rails.

$$\begin{aligned}\text e = \text B_\text V\text{lv} \\[1em] \text e = 0.2 \times 10^{-4} \times 1 \times 50 \\[1em] \text e = 10^{-3}\ \text V\end{aligned}$$

Hence, the milli-voltmeter reads 1 mV.

Given,

• Distance between the edges of the wings, l = 30 m
• Velocity of landing, v = 300 km h^-1 = 300 × $\frac{5}{18}$ ms^-1 = $\frac{250}{3}$ ms^-1
• B_H = 0.4 × 10^-4 T

As shown in the figure, the aeroplane is coming down vertically. Its velocity is parallel to the vertical component B_V, so B_V is never cut.

Wings east-west :

The wings (east-west), the velocity (vertical) and B_H (north) are mutually perpendicular, so the wings cut B_H.

$$\begin{aligned}\text e = \text B_\text H\text{lv} \\[1em] \text e = 0.4 \times 10^{-4} \times 30 \times \frac{250}{3} \\[1em] \text e = 0.4 \times 10^{-4} \times 2500 \\[1em] \text e = 0.1\ \text V\end{aligned}$$

Wings north-south :

The wings are now parallel to B_H, and the velocity is parallel to B_V. The wings cut neither component, so no potential difference is induced.

Hence, the potential difference is 0.1 V when the wings are east-west and zero when they are north-south.

Given,

• Length of the rod, l = 20 cm = 0.2 m
• Resistance of the rod, R = 0.1 Ω
• Magnetic field, B = 0.4 T
• Velocity of the rod, v = 10 × 10^-2 cm s^-1 = 0.1 cm s^-1 = 10^-3 m s^-1

(i) Induced emf :

$$\begin{aligned}\text e = \text{Blv} \\[1em] \text e = 0.4 \times 0.2 \times 10^{-3} \\[1em] \text e = 8 \times 10^{-5}\ \text V\end{aligned}$$

(ii) Induced current and its direction :

The ends A and C are short-circuited and the rails have negligible resistance, so the whole resistance of the circuit is that of the rod.

$$\begin{aligned}\text I = \frac{\text e}{\text R} \\[1em] \text I = \frac{8 \times 10^{-5}}{0.1} \\[1em] \text I = 8 \times 10^{-4}\ \text A\end{aligned}$$

The field is into the plane of the paper and the rod moves to the right. By Fleming's right-hand rule, the induced current in the rod flows from Q to P, that is, anticlockwise in the loop PACQ.

(iii) External force :

The current-carrying rod experiences a magnetic force F = BIl opposing its motion. To keep the velocity uniform, an equal external force must be applied.

$$\begin{aligned}\text F = \text{BIl} \\[1em] \text F = 0.4 \times 8 \times 10^{-4} \times 0.2 \\[1em] \text F = 6.4 \times 10^{-5}\ \text N\end{aligned}$$

(iv) Power required :

$$\begin{aligned}\text P = \text{Fv} \\[1em] \text P = 6.4 \times 10^{-5} \times 10^{-3} \\[1em] \text P = 6.4 \times 10^{-8}\ \text W\end{aligned}$$

Hence, (i) e = 8 × 10^-5 V, (ii) I = 8 × 10^-4 A from Q to P (anticlockwise in loop PACQ), (iii) F = 6.4 × 10^-5 N and (iv) P = 6.4 × 10^-8 W.

Note: The printed key gives 8 × 10^-3 V, 8 × 10^-2 A, 6.4 × 10^-3 N and 6.4 × 10^-4 W. These correspond to a speed of 10 × 10^-2 m/s, whereas the question prints cm/s.

Given,

• Side of the loop, a = 20 cm = 0.2 m
• Initial distance of the loop from the field = 30 cm
• Width of the field region = 80 cm
• Magnetic field, B = 0.1 T
• Velocity, v = 10 cm s^-1 = 0.1 m s^-1
• Resistance of the loop, R = 0.1 Ω

Stages of motion :

• t = 0 to 3 s : the loop covers the 30 cm gap; no part of it is in the field.
• t = 3 s to 5 s : the loop enters the field (20 cm); the area inside the field increases uniformly.
• t = 5 s to 11 s : the loop is wholly inside; its front edge reaches the far boundary after moving 80 cm from entry, at t = 3 + 8 = 11 s.
• t = 11 s to 13 s : the loop leaves the field; the area inside decreases uniformly.
• After t = 13 s : the loop is wholly outside.

Maximum flux, when the loop is wholly inside,

$$\begin{aligned}\phi_\text{max} = \text{Ba}^2 \\[1em] \phi_\text{max} = 0.1 \times (0.2)^2 \\[1em] \phi_\text{max} = 4 \times 10^{-3}\ \text{Wb}\end{aligned}$$

While entering or leaving, only one side of length a cuts the field, so the magnitude of the induced emf is

$$\begin{aligned}|\varepsilon| = \text{Bav} \\[1em] |\varepsilon| = 0.1 \times 0.2 \times 0.1 \\[1em] |\varepsilon| = 2 \times 10^{-3}\ \text V\end{aligned}$$

and the corresponding current is

$$\begin{aligned}|\text I| = \frac{|\varepsilon|}{\text R} \\[1em] |\text I| = \frac{2 \times 10^{-3}}{0.1} \\[1em] |\text I| = 2 \times 10^{-2}\ \text A\end{aligned}$$

By Faraday's law, ε = −dϕ/dt. So ε is negative while the flux increases (entering) and positive while it decreases (leaving).

(i) Magnetic flux versus time :

(ii) Induced emf versus time :

(iii) Induced current versus time :

Hence, the flux graph is a trapezium of height 4 × 10^-3 Wb, while the emf and current are rectangular pulses of 2 × 10^-3 V and 2 × 10^-2 A, negative during entry (3 s to 5 s) and positive during exit (11 s to 13 s).

Note: The printed graphs begin changing at their time origin; the question specifies an initial 30 cm gap, so the flux and induced pulses begin after 3 s.

Given,

• Height of the player, l = 2.0 m
• Speed, v = 20 km h^-1 = 20 × $\frac{5}{18}$ ms^-1 = $\frac{50}{9}$ ms^-1
• B_H = 4.0 × 10^-5 Wb m^-2

Running towards east :

The player's body is vertical, his velocity is towards east and B_H is towards north. These three are mutually perpendicular, so the body cuts B_H. The vertical component B_V is parallel to his body and is not cut.

$$\begin{aligned}\text e = \text B_\text H\text{lv} \\[1em] \text e = 4.0 \times 10^{-5} \times 2.0 \times \frac{50}{9} \\[1em] \text e = 4.44 \times 10^{-4}\ \text V\end{aligned}$$

Running towards north :

His velocity is now parallel to B_H, and his body is parallel to B_V. He cuts neither component, so no potential difference is induced.

Hence, the potential difference between his head and feet is 4.4 × 10^-4 V when running east and zero when running north.

Given,

• Magnetic flux, ϕ = 5 μWb = 5 × 10^-6 Wb
• Current, I = 1 mA = 10^-3 A

The flux linked with a coil is proportional to the current through it, ϕ = LI. Therefore,

$$\begin{aligned}\text L = \frac{\phi}{\text I} \\[1em] \text L = \frac{5 \times 10^{-6}}{10^{-3}} \\[1em] \text L = 5 \times 10^{-3}\ \text H\end{aligned}$$

Hence, the self-inductance of the coil is 5 mH.

From the figure, line A makes an angle of 60° and line B an angle of 30° with the current (I) axis.

Since ϕ = LI, the self-inductance L = ϕ/I is the slope of the ϕ-I graph, that is, the tangent of the angle the line makes with the I-axis.

$$\begin{aligned}\frac{\text L_\text A}{\text L_\text B} = \frac{\tan 60^\circ}{\tan 30^\circ} \\[1em] \frac{\text L_\text A}{\text L_\text B} = \frac{\sqrt{3}}{\frac{1}{\sqrt{3}}} \\[1em] \frac{\text L_\text A}{\text L_\text B} = 3\end{aligned}$$

Hence, the ratio of the coefficients of self-inductance of A and B is 3 : 1.

Given,

• Rate of change of current, $\frac{\text{dI}}{\text{dt}}$ = 2 A s^-1
• Induced emf, e = 10 mV = 10 × 10^-3 V

The magnitude of the self-induced emf is e = L $\frac{\text{dI}}{\text{dt}}$. Therefore,

$$\begin{aligned}\text L = \frac{\text e}{\frac{\text{dI}}{\text{dt}}} \\[1em] \text L = \frac{10 \times 10^{-3}}{2} \\[1em] \text L = 5 \times 10^{-3}\ \text H\end{aligned}$$

Hence, the self-inductance of the solenoid is 5 mH.

Given,

• Change in current, ΔI = 2 − 0 = 2 A
• Time, Δt = 0.05 s
• Induced emf, e = 80 V

Rate of change of current,

$$\begin{aligned}\frac{\Delta\text I}{\Delta\text t} = \frac{2}{0.05} \\[1em] \frac{\Delta\text I}{\Delta\text t} = 40\ \text{A s}^{-1}\end{aligned}$$

Using e = L $\frac{\Delta\text I}{\Delta\text t}$,

$$\begin{aligned}\text L = \frac{\text e}{\Delta\text I / \Delta\text t} \\[1em] \text L = \frac{80}{40} \\[1em] \text L = 2\ \text H\end{aligned}$$

Hence, the coefficient of self-induction of the coil is 2 H.

Given,

• Inductance, L = 4.0 H
• Change in current, ΔI = 2.7 A (from 2.7 A to zero)
• Time, Δt = 1.35 s

$$\begin{aligned}\text e = \text L\frac{\Delta\text I}{\Delta\text t} \\[1em] \text e = 4.0 \times \frac{2.7}{1.35} \\[1em] \text e = 4.0 \times 2 \\[1em] \text e = 8.0\ \text V\end{aligned}$$

Hence, the emf induced in the coil is 8.0 V.

Given,

• Turns in the solenoid, N₁ = 2000
• Length of the solenoid, l = 0.3 m
• Area of cross-section, A = 1.2 × 10^-3 m²
• Turns in the coil, N₂ = 300
• Current reversed from +2 A to −2 A, so ΔI = 4 A, in Δt = 0.25 s

Mutual inductance :

The field inside the solenoid is B = μ₀$\frac{\text N_1}{\text l}$I, and the flux linked with the coil is N₂BA. Therefore,

$$\begin{aligned}\text M = \frac{\mu_0\text N_1\text N_2\text A}{\text l} \\[1em] \text M = \frac{4\pi \times 10^{-7} \times 2000 \times 300 \times 1.2 \times 10^{-3}}{0.3} \\[1em] \text M = 4 \times 3.14 \times 10^{-7} \times 2400 \\[1em] \text M = 3.01 \times 10^{-3}\ \text H\end{aligned}$$

Induced emf :

$$\begin{aligned}\text e = \text M\frac{\Delta\text I}{\Delta\text t} \\[1em] \text e = 3.01 \times 10^{-3} \times \frac{4}{0.25} \\[1em] \text e = 3.01 \times 10^{-3} \times 16 \\[1em] \text e = 48.2 \times 10^{-3}\ \text V\end{aligned}$$

Hence, the emf induced in the coil is 48.2 mV ≈ 48 mV.

Given,

• L₁ = 16 mH, L₂ = 12 mH
• $\frac{\text{dI}}{\text{dt}}$ is the same for both coils
• Power supplied, P₁ = P₂

(i) Ratio of induced voltages :

Using V = L $\frac{\text{dI}}{\text{dt}}$, with $\frac{\text{dI}}{\text{dt}}$ the same for both,

$$\begin{aligned}\frac{\text V_1}{\text V_2} = \frac{\text L_1}{\text L_2} \\[1em] \frac{\text V_1}{\text V_2} = \frac{16}{12} \\[1em] \frac{\text V_1}{\text V_2} = \frac{4}{3}\end{aligned}$$

(ii) Ratio of currents :

Using P = VI with P₁ = P₂,

$$\begin{aligned}\text V_1\text I_1 = \text V_2\text I_2 \\[1em] \frac{\text I_1}{\text I_2} = \frac{\text V_2}{\text V_1} \\[1em] \frac{\text I_1}{\text I_2} = \frac{3}{4}\end{aligned}$$

(iii) Ratio of energies stored :

Using U = $\frac{1}{2}$LI²,

$$\begin{aligned}\frac{\text U_1}{\text U_2} = \frac{\text L_1}{\text L_2} \times \left(\frac{\text I_1}{\text I_2}\right)^2 \\[1em] \frac{\text U_1}{\text U_2} = \frac{4}{3} \times \left(\frac{3}{4}\right)^2 \\[1em] \frac{\text U_1}{\text U_2} = \frac{4}{3} \times \frac{9}{16} \\[1em] \frac{\text U_1}{\text U_2} = \frac{3}{4}\end{aligned}$$

Hence, the ratios are (i) 4 : 3, (ii) 3 : 4 and (iii) 3 : 4.

From the figure, ϕ₂ varies linearly with I₁, and ϕ₂ = 15 Wb when I₁ = 6 A.

(i) Mutual inductance :

Since ϕ₂ = MI₁, M is the slope of the ϕ₂-I₁ graph.

$$\begin{aligned}\text M = \frac{\phi_2}{\text I_1} \\[1em] \text M = \frac{15}{6} \\[1em] \text M = 2.5\ \text H\end{aligned}$$

(ii) Rate of change of current :

Given e₂ = 100 V. Using e₂ = M $\frac{\text{dI}_1}{\text{dt}}$,

$$\begin{aligned}\frac{\text{dI}_1}{\text{dt}} = \frac{\text e_2}{\text M} \\[1em] \frac{\text{dI}_1}{\text{dt}} = \frac{100}{2.5} \\[1em] \frac{\text{dI}_1}{\text{dt}} = 40\ \text{A s}^{-1}\end{aligned}$$

Hence, the mutual inductance is 2.5 H and the required rate of change of current is 40 A s^-1.

Given,

• Change in primary current, ΔI = 5 − 3 = 2 A
• Time, Δt = 2 s
• Mutual inductance, M = 0.16 H

$$\begin{aligned}\text e = \text M\frac{\Delta\text I}{\Delta\text t} \\[1em] \text e = 0.16 \times \frac{2}{2} \\[1em] \text e = 0.16\ \text V\end{aligned}$$

Hence, the emf induced in the secondary is 0.16 V.

Given,

• Mutual inductance, M = 0.15 H
• Change in primary current, ΔI = 20 A (from 20 A to zero)
• Time, Δt = 0.05 s
• Turns in the secondary, N = 60

Induced emf :

$$\begin{aligned}\text e = \text M\frac{\Delta\text I}{\Delta\text t} \\[1em] \text e = 0.15 \times \frac{20}{0.05} \\[1em] \text e = 60\ \text V\end{aligned}$$

Flux change :

The change in flux linkage of the secondary is NΔϕ = MΔI. Therefore, the change in flux through each turn is

$$\begin{aligned}\Delta\phi = \frac{\text M\Delta\text I}{\text N} \\[1em] \Delta\phi = \frac{0.15 \times 20}{60} \\[1em] \Delta\phi = 0.05\ \text{Wb}\end{aligned}$$

Hence, the emf induced in the secondary is 60 V and the flux change in it is 0.05 Wb.

Given,

• Radius of the small loop, r₁ = 0.3 cm = 3 × 10^-3 m
• Radius of the big loop, R₂ = 20 cm = 0.2 m
• Distance between the centres, x = 15 cm = 0.15 m
• Current in the small loop, I₁ = 2.0 A

The mutual inductance of a pair of coils is the same whichever coil carries the current. So suppose a current I₂ flows in the big loop. Its field at a point on its axis at distance x is

$\text B_2 = \frac{\mu_0\text I_2\text R_2^2}{2(\text R_2^2 + \text x^2)^{3/2}}$

The small loop is so small that this field is uniform over its area A₁ = πr₁². The flux through it is B₂A₁, so

$$\begin{aligned}\text M = \frac{\text B_2\text A_1}{\text I_2} \\[1em] \text M = \frac{\mu_0\pi\text r_1^2\text R_2^2}{2(\text R_2^2 + \text x^2)^{3/2}}\end{aligned}$$

Here R₂² + x² = 0.04 + 0.0225 = 0.0625 m², so (R₂² + x²)^3/2 = (0.25)³ = 0.015625 m³.

$$\begin{aligned}\text M = \frac{4\pi \times 10^{-7} \times \pi \times (3 \times 10^{-3})^2 \times 0.04}{2 \times 0.015625} \\[1em] \text M = \frac{4\pi^2 \times 10^{-7} \times 3.6 \times 10^{-7}}{0.03125} \\[1em] \text M = 4 \times 9.86 \times 1.152 \times 10^{-12} \\[1em] \text M = 4.54 \times 10^{-11}\ \text H\end{aligned}$$

The flux linked with the big loop when I₁ flows in the small loop is

$$\begin{aligned}\phi_2 = \text{MI}_1 \\[1em] \phi_2 = 4.54 \times 10^{-11} \times 2.0 \\[1em] \phi_2 = 9.1 \times 10^{-11}\ \text{Wb}\end{aligned}$$

Hence, the flux linked with the bigger loop is 9.1 × 10^-11 Wb.

Self-inductance : When a current I flows through a coil of N turns and Φ_B is the magnetic flux linked with each turn, the number of flux-linkages NΦ_B is proportional to I,

$\text N\Phi_\text B = \text{LI}$

The constant L is called the coefficient of self-induction or self-inductance of the coil. Thus, the self-inductance of a coil is equal to the number of flux-linkages with the coil when unit current flows through it. Equivalently, it is numerically equal to the emf induced in the coil when the current through it changes at the rate of 1 ampere per second. Its SI unit is henry (H).

Self-inductance of a long solenoid :

Consider a long air-cored solenoid of length l and area of cross-section A, having N closely-wound turns, so that the number of turns per unit length is n = N/l. Let a current I flow through it.

The magnetic field inside the solenoid is uniform and its magnitude is

$\text B = \mu_0\frac{\text N}{\text l}\text I \qquad ...(\text i)$

The magnetic flux through each turn is

$$\begin{aligned}\Phi_\text B = \text{BA} \\[1em] \Phi_\text B = \mu_0\frac{\text N}{\text l}\text{IA} \qquad ...(\text{ii})\end{aligned}$$

The total magnetic flux (flux-linkages) through the N turns of the solenoid is

$$\begin{aligned}\text N\Phi_\text B = \text N \times \mu_0\frac{\text N}{\text l}\text{IA} \\[1em] \text N\Phi_\text B = \frac{\mu_0\text N^2\text{IA}}{\text l} \qquad ...(\text{iii})\end{aligned}$$

Since NΦ_B = LI, the self-inductance of the solenoid is

$$\begin{aligned}\text L = \frac{\text N\Phi_\text B}{\text I} \\[1em] \text L = \frac{\mu_0\text N^2\text A}{\text l} \qquad ...(\text{iv})\end{aligned}$$

Putting N = nl, this may also be written as L = μ₀n²Al.

Hence, the self-inductance of a long solenoid is $\text L = \frac{\mu_0\text N^2\text A}{\text l}$, which depends only on its geometry.

The bar magnet moves towards the coil with its north pole N leading. Hence the magnetic flux linked with the coil increases and a current is induced in it. By Lenz's law, this induced current opposes the cause producing it, that is, the approach of the N pole.

Magnetic poles : The end of the coil facing the magnet (left end) becomes a north pole, so that it repels the approaching N pole. The far end (right end) of the coil becomes a south pole.

Direction of current : For the left face to be a north pole, the induced current must flow anticlockwise in the coil when viewed from the magnet. With the clockwise winding shown, this current leaves the coil at its left end and flows through the galvanometer G from left to right, re-entering the coil at its right end.

Note: The figure does not show unambiguously from which side the winding is clockwise; the answer takes it as clockwise when viewed from the magnet. With the opposite reading, the current through G would be from right to left, while the poles stay the same.

Given,

• Area of coil, A = 50 cm² = 50 × 10^-4 m²
• Number of turns, N = 100
• Magnetic field, B = 10^-2 Wb/m²
• Time, Δt = 40 ms = 40 × 10^-3 s

As the coil is perpendicular to the field, the flux linked with each turn is initially BA. When the coil is withdrawn from the field, this flux falls to zero. By Faraday's law, the magnitude of the induced emf is

$$\begin{aligned}|\text e| = \text N\frac{\Delta\Phi_\text B}{\Delta\text t} \\[1em] |\text e| = \frac{\text{NBA}}{\Delta\text t}\end{aligned}$$

Substituting the values,

$$\begin{aligned}|\text e| = \frac{100 \times 10^{-2} \times 50 \times 10^{-4}}{40 \times 10^{-3}} \\[1em] = \frac{5 \times 10^{-3}}{40 \times 10^{-3}} \\[1em] |\text e| = 0.125\ \text V\end{aligned}$$

Hence, the emf induced in the coil is 0.125 V.

Given,

• Distance between PQ and MN = l
• Resistance connected = R
• Magnetic field = B, perpendicular to the plane of the conductors
• Velocity of wire XY = v
• Resistances of PQ, MN and XY are negligible

As XY slides with velocity v, the area of the circuit formed by XY, PQ, MN and R changes at the rate lv. Hence a motional emf is induced in XY,

$\text e = \text{Blv}$

The current in the circuit is

$$\begin{aligned}\text I = \frac{\text e}{\text R} \\[1em] \text I = \frac{\text{Blv}}{\text R}\end{aligned}$$

The current-carrying wire XY in the magnetic field experiences a force F = IlB, which opposes its motion (Lenz's law),

$$\begin{aligned}\text F = \text{IlB} \\[1em] \text F = \frac{\text{Blv}}{\text R} \times \text{lB} \\[1em] \text F = \frac{\text B^2\text l^2\text v}{\text R}\end{aligned}$$

To slide XY with constant velocity v, an equal and opposite external force must be applied. The work done per second by this force is

$$\begin{aligned}\frac{\text W}{\text t} = \text{Fv} \\[1em] \frac{\text W}{\text t} = \frac{\text B^2\text l^2\text v^2}{\text R}\end{aligned}$$

This work appears as Joule heat I²R in the resistance R.

Hence, the work done per second to slide the wire XY is $\frac{\text W}{\text t} = \frac{\text v^2\text B^2\text l^2}{\text R}$.

The self-induced emf in a coil is related to the rate of change of current through it by

$\text e = -\text L\frac{\text{dI}}{\text{dt}}$

So, in magnitude,

$\text L = \frac{\text e}{\left(\frac{\text{dI}}{\text{dt}}\right)}$

Writing the units of each quantity,

$$\begin{aligned}1\ \text{henry} = \frac{1\ \text{volt}}{1\ \text{ampere/second}} \\[1em] = \frac{1\ \text{volt}}{1\ \text{ampere}} \times 1\ \text{second}\end{aligned}$$

By Ohm's law, 1 volt/ampere = 1 ohm. Therefore,

$1\ \text{henry} = 1\ \text{ohm} \times 1\ \text{second}$

The same result follows from the defining equation of mutual inductance, $\text M = \frac{\text e_2}{\left(\frac{\text{dI}_1}{\text{dt}}\right)}$, which has the same form.

Hence, 1 henry = 1 ohm second.

Consider a U-shaped conductor MNOL placed in a uniform magnetic field $\vec{\text B}$, directed perpendicular to the plane of the paper. A straight conducting rod JK of length l rests on the parallel arms of MNOL and is moved towards the right with a constant velocity $\vec{\text v}$, perpendicular to its own length and to $\vec{\text B}$. The rod and the U-shaped conductor together form a closed circuit JNOKJ.

Using Faraday's law :

Let x be the distance of the rod JK from the arm NO at any instant. The area of the circuit is lx, so the magnetic flux linked with it is

$\Phi_\text B = \text{Blx} \qquad ...(\text i)$

As the rod moves, x increases and the flux changes. By Faraday's law, the induced emf is

$$\begin{aligned}\varepsilon = -\frac{\text d\Phi_\text B}{\text{dt}} \\[1em] \varepsilon = -\frac{\text d}{\text{dt}}(\text{Blx}) \\[1em] \varepsilon = -\text{Bl}\frac{\text{dx}}{\text{dt}}\end{aligned}$$

Since B and l are constant and $\frac{\text{dx}}{\text{dt}}$ = v, the velocity of the rod,

$\varepsilon = -\text{Blv} \qquad ...(\text{ii})$

The negative sign shows that the induced emf opposes the change of flux (Lenz's law). In magnitude, ε = Blv.

Using Lorentz force :

Each free charge q in the rod moves with velocity v in the field B and experiences a magnetic (Lorentz) force F_m = qvB along the rod. The charges collect at the two ends until the electric field E set up inside the rod balances this force,

$$\begin{aligned}\text{qE} = \text{qvB} \\[1em] \text E = \text{vB}\end{aligned}$$

The potential difference developed across the ends of the rod is therefore

$$\begin{aligned}\varepsilon = \text{El} \\[1em] \varepsilon = \text{Blv}\end{aligned}$$

Hence, the motional emf produced across the ends of the rod is ε = Blv.

Zero.

Explanation: The motional emf across a conductor is e = Bvl sin θ, where θ is the angle between $\vec{\text v} \space$and $\vec{\text B}$. Here the conductor moves along the direction of the magnetic field, so θ = 0° and sin θ = 0. The conductor cuts no flux-lines, and hence no emf is induced across its ends.

Given,

• Initial current in P, I₁ = 5 A
• Final current in P, I₂ = 0 A
• Time, Δt = 0.2 s
• emf induced in Q, e = 60 V

The emf induced in the secondary coil Q is related to the rate of change of current in the primary coil P by

$\text e = -\text M\frac{\Delta\text I}{\Delta\text t}$

The rate of change of current in P is

$$\begin{aligned}\frac{\Delta\text I}{\Delta\text t} = \frac{0 - 5}{0.2} \\[1em] \frac{\Delta\text I}{\Delta\text t} = -25\ \text A\ \text s^{-1}\end{aligned}$$

Therefore,

$$\begin{aligned}\text M = \frac{\text e}{\left|\frac{\Delta\text I}{\Delta\text t}\right|} \\[1em] \text M = \frac{60}{25} \\[1em] \text M = 2.4\ \text H\end{aligned}$$

Hence, the coefficient of mutual inductance of the coils P and Q is 2.4 H.

Given,

• Distance between arms PQ and RS, l = 40 cm = 0.40 m
• Magnetic field, B = 0.1 T, perpendicular to the paper, inwards
• Velocity of rod, v = 5 ms^-1, towards right

(i) emf induced across CD :

The rod CD moves perpendicular to its length and to the field, so the motional emf is

$$\begin{aligned}\text e = \text{Blv} \\[1em] = 0.1 \times 0.40 \times 5 \\[1em] \text e = 0.2\ \text V\end{aligned}$$

(ii) Direction of induced current :

As CD moves to the right, the area of the closed circuit QCDR increases, so the inward magnetic flux through it increases. By Lenz's law, the induced current sets up an outward field inside the circuit, so it flows anticlockwise round QCDR. The same result follows from Fleming's right-hand rule.

Hence, (i) the emf induced across the rod CD is 0.2 V, and (ii) the induced current flows from D to C in the rod (and from C through Q and R back to D in the wire).

Given,

• Induced emf, e = 2 V
• Change in current, ΔI = 10 − 0 = 10 A
• Time, Δt = 0.40 s

The self-induced emf is related to the rate of change of current by

$\text e = -\text L\frac{\Delta\text I}{\Delta\text t}$

In magnitude,

$$\begin{aligned}\text L = \frac{\text e}{\left(\frac{\Delta\text I}{\Delta\text t}\right)} \\[1em] \text L = \frac{2}{\left(\frac{10}{0.40}\right)} \\[1em] \text L = \frac{2}{25} \\[1em] \text L = 0.08\ \text H\end{aligned}$$

Hence, the coefficient of self-inductance of the coil is 0.08 H.

Given,

• Self-inductance, L = 0.05 H
• Rate of change of current, $\frac{\text{dI}}{\text{dt}}$ = 100 A s^-1

The magnitude of the self-induced emf is

$$\begin{aligned}|\text e| = \text L\frac{\text{dI}}{\text{dt}} \\[1em] = 0.05 \times 100 \\[1em] |\text e| = 5\ \text V\end{aligned}$$

Hence, the magnitude of the emf induced in the coil is 5 volt.

Given,

• Distance between AB and CD, l = 1 m
• Resistance, R = 6 Ω
• Magnetic field, B = 3 × 10^-2 T, into the paper
• Velocity of wire MN, v = 2 ms^-1

The motional emf induced in the moving wire MN is

$$\begin{aligned}\text e = \text{Blv} \\[1em] = 3 \times 10^{-2} \times 1 \times 2 \\[1em] \text e = 6 \times 10^{-2}\ \text V\end{aligned}$$

The resistances of AB, CD and MN are negligible, so the induced current is

$$\begin{aligned}\text I = \frac{\text e}{\text R} \\[1em] = \frac{6 \times 10^{-2}}{6} \\[1em] \text I = 10^{-2}\ \text A\end{aligned}$$

As MN moves to the right, the inward flux through the circuit AMNC increases. By Lenz's law, the induced current flows anticlockwise round this circuit, that is, from N to M in the wire and from A to C through the resistor.

Hence, the induced current through the resistor is 10^-2 A, flowing from N to M in the wire MN.

Given,

• Number of turns, N = 60
• Area of coil, A = 0.4 m²
• Magnetic flux density, B = 5 × 10^-5 T

The coil is held at right angles to the field, so the field is along the normal to its plane (θ = 0°). The flux through each turn is BA, and the total flux linked with the N turns is

$$\begin{aligned}\text N\Phi_\text B = \text{NBA} \\[1em] = 60 \times 5 \times 10^{-5} \times 0.4 \\[1em] \text N\Phi_\text B = 1.2 \times 10^{-3}\ \text{Wb}\end{aligned}$$

The flux through each single turn is BA = 2 × 10^-5 Wb.

Hence, the magnetic flux linked with the coil is 1.2 × 10^-3 Wb.

Given,

• Length of rod, l = 80 cm = 0.8 m
• Magnetic field, B = 0.5 T
• Velocity, v = 36 km/hr

Converting the velocity into m/s,

$$\begin{aligned}\text v = 36 \times \frac{5}{18} \\[1em] \text v = 10\ \text{m/s}\end{aligned}$$

The rod moves perpendicular to its length and to the field, so the motional emf is

$$\begin{aligned}\text e = \text{Blv} \\[1em] = 0.5 \times 0.8 \times 10 \\[1em] \text e = 4\ \text V\end{aligned}$$

Hence, the emf induced between the two ends of the rod is 4 V.

Given,

• Change in current, ΔI = 15 − 0 = 15 A
• Time, Δt = 0.2 s
• Induced emf in the adjacent coil, e = 750 V

The emf induced in the secondary coil is

$\text e = -\text M\frac{\Delta\text I}{\Delta\text t}$

In magnitude,

$$\begin{aligned}\text M = \frac{\text e}{\left(\frac{\Delta\text I}{\Delta\text t}\right)} \\[1em] \text M = \frac{750}{\left(\frac{15}{0.2}\right)} \\[1em] \text M = \frac{750}{75} \\[1em] \text M = 10\ \text H\end{aligned}$$

Hence, the coefficient of mutual inductance of the two coils is 10 H.

(a) When a straight conductor moves in a magnetic field so that it cuts the magnetic flux-lines, an emf (potential difference) is induced across its ends. This emf, which is induced due to the motion of the conductor, is called motional emf. For a conductor of length l moving with velocity v perpendicular to a magnetic field B, its magnitude is e = Bvl.

(b) The magnitude of motional emf depends on the speed v of the conductor; it is directly proportional to v (it also depends on B and l).

200 V

Explanation: The magnitude of the self-induced emf is

$$\begin{aligned}|\text e| = \text L\frac{\Delta\text I}{\Delta\text t} \\[1em] = 8 \times \frac{5 - 0}{0.2} \\[1em] |\text e| = 200\ \text V\end{aligned}$$

(i) shows a momentary deflection and comes back to zero

(ii) shows a momentary and a greater deflection and comes back to zero

Explanation: (i) A current is induced in the solenoid only while the magnetic flux linked with it is changing. As the magnet moves from NS to N'S', the flux linked with the solenoid changes and G deflects. Once the magnet comes to rest at N'S', the flux becomes constant, the induced emf becomes zero and the pointer returns to zero.

(ii) When the magnet is moved more rapidly, the rate of change of flux $\left(\frac{\text d\Phi_\text B}{\text{dt}}\right)$ is greater, so a larger emf and current are induced, but for a shorter time. The deflection is therefore momentary and greater, and the pointer again returns to zero when the magnet stops.

voltage applied to the coil P or Q

Explanation: The coefficient of mutual inductance of two coils depends only on their numbers of turns, their geometry (area of cross-section and length), their relative position and orientation, and the permeability of the core material. For coils wound on a common core,

$\text M = \frac{\mu_0\mu_\text r\text N_1\text N_2\text A}{\text l}$

It does not depend on the voltage applied to either coil.

Current in the circuit is minimum

Explanation: At resonance of a series LCR circuit, X_L = X_C, so the impedance $\text Z = \sqrt{\text R^2 + (\text X_\text L - \text X_\text C)^2}$ has its minimum value R and the current is in phase with the supply voltage. Since the impedance is minimum, the current $\text I = \frac{\text V}{\text R}$ is maximum, not minimum. So this statement is not true.

Yes, a magnetic field can produce an electric current, but only when the magnetic flux linked with a closed circuit changes. This phenomenon is called electromagnetic induction. A steady magnetic field linked with a circuit produces no current.

Example : Connect a coil to a sensitive galvanometer, with no cell in the circuit.

• When the north pole of a bar magnet is moved towards the coil, the galvanometer shows a deflection, showing that a current flows in the coil.
• When the magnet is moved away from the coil, the galvanometer deflects in the opposite direction.
• When the magnet is held at rest near the coil, there is no deflection.
• When the magnet is moved faster, the deflection is larger.

In each case the current flows only while the number of magnetic flux-lines passing through the coil is changing. The induced emf is given by Faraday's law, $\text e = -\frac{\text d\Phi_\text B}{\text{dt}}$.

(i) When a piece of metal is placed in a changing magnetic field, or is moved in a magnetic field so that the magnetic flux linked with it changes, induced currents are set up in closed paths throughout the volume of the metal. These currents oppose the change in magnetic flux (or the motion of the metal) and are called eddy currents or Foucault currents.

(ii) The phenomenon of electromagnetic induction in which, on changing the current in a coil, an opposing induced emf is set up in that very coil, is called self-induction.

When the alternating current in the transformer windings changes, the magnetic flux in the core also changes. If the core were a single solid piece of iron, large eddy currents would flow through its entire volume. These currents heat the core, causing loss of energy, and their own flux opposes the main flux.

To minimise this, the core is made of many thin laminas of soft iron, insulated from each other by varnish (a laminated core). The eddy currents are then set up separately in each lamina, the length of the current path is greatly increased and so the resistance increases. Hence the eddy currents and their heating effects are minimised. The magnetisation of the core is not affected, as the core is continuous in the direction of the field.

Given,

• Initial current, I₁ = 5 A
• Final current, I₂ = 0
• Time, Δt = 20 ms = 20 × 10^-3 s
• Induced emf, e = 500 V

(i) An emf is induced in the solenoid itself due to the change of current in it. This phenomenon is called self-induction.

(ii) The self-induced emf is given by

$\text e = -\text L\frac{\Delta\text I}{\Delta\text t}$

In magnitude,

$$\begin{aligned}\text L = \frac{\text e}{\left(\frac{\Delta\text I}{\Delta\text t}\right)} \\[1em] \text L = \frac{500}{\left(\frac{5}{20 \times 10^{-3}}\right)} \\[1em] \text L = \frac{500}{250} \\[1em] \text L = 2\ \text H\end{aligned}$$

Hence, (i) the phenomenon is self-induction, and (ii) the coefficient of self-inductance of the solenoid is 2 H.

(i) Given, rms value of current, I_rms = 5 A.

For a sinusoidal alternating current, the peak value I₀ is related to the rms value by

$\text I_\text{rms} = \frac{\text I_0}{\sqrt{2}}$

Therefore,

$$\begin{aligned}\text I_0 = \sqrt{2}\ \text I_\text{rms} \\[1em] = \sqrt{2} \times 5 \\[1em] \text I_0 = 5\sqrt{2}\ \text A \approx 7.07\ \text A\end{aligned}$$

Hence, the peak value of the current is $5\sqrt{2}$ A.

(ii) A direct current (dc) always flows in one direction with constant magnitude, whereas an alternating current (ac) changes its magnitude continuously and reverses its direction periodically.

13.89 H

Reason — The magnetic energy stored in an inductor is $\text U = \frac{1}{2}\text{LI}^2$. With U = 25 mJ = 25 × 10^-3 J and I = 60 mA = 60 × 10^-3 A,

$$\begin{aligned}\text L = \frac{2\text U}{\text I^2} \\[1em] = \frac{2 \times 25 \times 10^{-3}}{\left(60 \times 10^{-3}\right)^2} \\[1em] = \frac{50 \times 10^{-3}}{3.6 \times 10^{-3}} \\[1em] \text L = 13.89\ \text H\end{aligned}$$

0.016

Reason — The total resistance of the circuit is R = 0.1 + 0.9 = 1 Ω, so the time constant is

$$\begin{aligned}\tau = \frac{\text L}{\text R} \\[1em] = \frac{10 \times 10^{-3}}{1} \\[1em] \tau = 0.01\ \text s\end{aligned}$$

During growth, I = I₀(1 − e^−t/τ). For I = 0.8 I₀,

$$\begin{aligned}0.8 = 1 - \text e^{-\frac{\text t}{\tau}} \\[1em] \text e^{-\frac{\text t}{\tau}} = 0.2 \\[1em] \text e^{-\frac{\text t}{\tau}} = \frac{1}{5} \\[1em] \text t = \tau \ln 5 \\[1em] = 0.01 \times 1.6 \\[1em] \text t = 0.016\ \text s\end{aligned}$$

$\frac{\text{EL}}{2.7\text R^2}$

Reason — After the switch is closed, the current grows as

$\text I = \frac{\text E}{\text R}\left(1 - \text e^{-\frac{\text{Rt}}{\text L}}\right)$

The charge passing through the battery from t = 0 to $\text t = \frac{\text L}{\text R}$ is

$$\begin{aligned}\text q = \int_0^{\frac{\text L}{\text R}} \text I\ \text{dt} \\[1em] = \frac{\text E}{\text R}\int_0^{\frac{\text L}{\text R}} \left(1 - \text e^{-\frac{\text{Rt}}{\text L}}\right)\text{dt} \\[1em] = \frac{\text E}{\text R}\left[\text t + \frac{\text L}{\text R}\text e^{-\frac{\text{Rt}}{\text L}}\right]_0^{\frac{\text L}{\text R}} \\[1em] = \frac{\text E}{\text R}\left[\left(\frac{\text L}{\text R} + \frac{\text L}{\text R}\text e^{-1}\right) - \left(0 + \frac{\text L}{\text R}\right)\right] \\[1em] = \frac{\text{EL}}{\text R^2}\text e^{-1} \\[1em] \text q = \frac{\text{EL}}{2.7\text R^2}\end{aligned}$$

taking e ≈ 2.7.

Reason — From t = 0 to t₀, with S₁ closed, the current in the LR circuit grows exponentially, $\text I = \text I_0\left(1 - \text e^{-\frac{\text{Rt}}{\text L}}\right)$; it cannot rise linearly. At t₀, S₁ is opened and S₂ closed, so the battery is cut off and L and R form a closed loop through S₂. The current then decays exponentially, $\text I = \text I(\text t_0)\ \text e^{-\frac{\text R(\text t - \text t_0)}{\text L}}$, and does not drop suddenly. Only graph (2) shows an exponential rise up to t₀ followed by an exponential decay.

$\frac{1}{\text L}$

Reason — The self-inductance of a solenoid of N turns, area of cross-section A and length L is

$\text L_\text s = \frac{\mu_0\text N^2\text A}{\text L}$

With N and A fixed, L_s ∝ $\frac{1}{\text L}$.

increases by a factor of 3

Reason — The length of copper wire, l_w, is fixed and the turns are closely wound, so the number of turns per unit length of the frame (n = 1/d, d = diameter of wire) is unchanged. If each side of the triangular frame is a, the perimeter of one turn is 3a. Then

$\text N = \frac{\text l_\text w}{3\text a}, \quad \text A = \frac{\sqrt{3}}{4}\text a^2, \quad \text{length of winding} = \text{Nd}$

$$\begin{aligned}\text L = \frac{\mu_0\text N^2\text A}{\text{Nd}} \\[1em] = \frac{\mu_0\text{NA}}{\text d} \\[1em] = \frac{\mu_0}{\text d} \times \frac{\text l_\text w}{3\text a} \times \frac{\sqrt{3}}{4}\text a^2 \\[1em] \text L = \frac{\sqrt{3}\mu_0\text l_\text w}{12\text d}\text a\end{aligned}$$

So L ∝ a. When a becomes 3a, the self-inductance increases by a factor of 3.

2 ln 2

Reason — Rate of dissipation in R is I²R and rate of storage of magnetic energy is $\frac{\text d}{\text{dt}}\left(\frac{1}{2}\text{LI}^2\right) = \text{LI}\frac{\text{dI}}{\text{dt}}$. Equating them,

$$\begin{aligned}\text I^2\text R = \text{LI}\frac{\text{dI}}{\text{dt}} \\[1em] \text{IR} = \text L\frac{\text{dI}}{\text{dt}}\end{aligned}$$

With $\text I = \text I_0\left(1 - \text e^{-\frac{\text t}{\tau}}\right)$ and τ = L/R = 20/10 = 2 s,

$$\begin{aligned}\text L\frac{\text{dI}}{\text{dt}} = \text L\frac{\text I_0}{\tau}\text e^{-\frac{\text t}{\tau}} \\[1em] \text L\frac{\text{dI}}{\text{dt}} = \text I_0\text R\ \text e^{-\frac{\text t}{\tau}}\end{aligned}$$

So

$$\begin{aligned}\text I_0\text R\left(1 - \text e^{-\frac{\text t}{\tau}}\right) = \text I_0\text R\ \text e^{-\frac{\text t}{\tau}} \\[1em] \text e^{-\frac{\text t}{\tau}} = \frac{1}{2} \\[1em] \text t = \tau\ln 2 \\[1em] \text t = 2\ln 2\ \text s\end{aligned}$$

6 A

Reason — Long after the switch is closed, the current is steady and the inductor offers no opposition, acting as a plain conducting wire. The two 5 Ω resistors are then in parallel across the 15 V battery,

$$\begin{aligned}\text R_\text p = \frac{5 \times 5}{5 + 5} \\[1em] \text R_\text p = 2.5\ \Omega \\[1em] \text I = \frac{15}{2.5} \\[1em] \text I = 6\ \text A\end{aligned}$$

437.5 J

Reason — The current changes uniformly by 15 A in 1 s, so $\frac{\text{dI}}{\text{dt}}$ = 15 A s^-1. From |e| = L dI/dt,

$$\begin{aligned}\text L = \frac{25}{15} \\[1em] \text L = \frac{5}{3}\ \text H\end{aligned}$$

The change in magnetic energy is

$$\begin{aligned}\Delta\text U = \frac{1}{2}\text L\left(\text I_2^2 - \text I_1^2\right) \\[1em] = \frac{1}{2} \times \frac{5}{3}\left(25^2 - 10^2\right) \\[1em] = \frac{5}{6} \times 525 \\[1em] \Delta\text U = 437.5\ \text J\end{aligned}$$

0.14 mC

Reason — The induced current is $\text I = \frac{1}{\text R}\frac{\text d\Phi}{\text{dt}}$, so the charge that flows is $\text q = \frac{\Delta\Phi}{\text R}$. Here Φ = AB(t) increases steadily from t = 0 to t = 10 ms.

At t = 0, B = 0. At t = 10 ms, B = 0.4 sin(50π × 0.01) = 0.4 sin $\frac{\pi}{2}$ = 0.4 T.

$$\begin{aligned}\text q = \frac{\text A\ \Delta\text B}{\text R} \\[1em] = \frac{3.5 \times 10^{-3} \times 0.4}{10} \\[1em] \text q = 1.4 \times 10^{-4}\ \text C \\[1em] \text q = 0.14\ \text{mC}\end{aligned}$$

$\frac{\text n_2}{\text n_1}$

Reason — A current I₁ in the inner coil produces field μ₀n₁I₁ only inside it (area πr₁²). The flux linked with the n₂l turns of the outer coil gives

$\text M = \mu_0\text n_1\text n_2\pi\text r_1^2\text l$

The self-inductance of the inner coil is

$\text L_1 = \mu_0\text n_1^2\pi\text r_1^2\text l$

Dividing,

$\frac{\text M}{\text L_1} = \frac{\text n_2}{\text n_1}$

43 V

Reason — By Faraday's law,

$$\begin{aligned}|\text e| = \frac{\text d\Phi}{\text{dt}} \\[1em] = \frac{\text d}{\text{dt}}\left(5\text t^2 + 3\text t + 16\right) \\[1em] = 10\text t + 3\end{aligned}$$

At t = 4 s,

$$\begin{aligned}|\text e| = 10 \times 4 + 3 \\[1em] |\text e| = 43\ \text V\end{aligned}$$

Reason — As the magnet enters the coil (a), the flux linked with the coil increases and G deflects to one side. When the magnet is completely inside (b), moving at constant speed, the flux linked with the coil hardly changes, so there is no induced current and the pointer is at zero. As the magnet leaves (c), the flux decreases and G deflects to the opposite side. Only set (4) shows a deflection, then zero, then an opposite deflection.

0.61 A

Reason — Each side of the square loop is $\frac{30}{4}$ = 7.5 cm, so its area is A = (0.075)² = 5.625 × 10^-3 m².

$$\begin{aligned}\text e = \text A\frac{\text{dB}}{\text{dt}} \\[1em] = 5.625 \times 10^{-3} \times 0.032 \\[1em] \text e = 1.8 \times 10^{-4}\ \text V\end{aligned}$$

The cross-sectional area of the wire is π(2 × 10^-3)² = 1.257 × 10^-5 m², so its resistance is

$$\begin{aligned}\text R = \frac{\rho\text l}{\text a} \\[1em] = \frac{1.23 \times 10^{-8} \times 0.30}{1.257 \times 10^{-5}} \\[1em] \text R = 2.94 \times 10^{-4}\ \Omega\end{aligned}$$

$$\begin{aligned}\text I = \frac{\text e}{\text R} \\[1em] = \frac{1.8 \times 10^{-4}}{2.94 \times 10^{-4}} \\[1em] \text I = 0.61\ \text A\end{aligned}$$

$\frac{\text R_2^2}{\text R_1}$

Reason — Let a current I flow in the large loop. The field at its centre is $\text B = \frac{\mu_0\text I}{2\text R_1}$, which is nearly uniform over the small loop since R₁ >> R₂. The flux through the small loop is

$\Phi = \frac{\mu_0\text I}{2\text R_1}\pi\text R_2^2$

$$\begin{aligned}\text M = \frac{\Phi}{\text I} \\[1em] \text M = \frac{\mu_0\pi\text R_2^2}{2\text R_1}\end{aligned}$$

So M ∝ $\frac{\text R_2^2}{\text R_1}$.

1 cm/s

Reason — The network of 4 Ω resistors reduces to two 4 Ω resistors in parallel twice over, so its equivalent resistance is 2 Ω + 2 Ω = 4 Ω. With the loop's own 1 Ω, the total resistance is 5 Ω. Only the right arm (l = 0.2 m) is in the field, so the motional emf is e = Blv₀.

$$\begin{aligned}\text I = \frac{\text{Blv}_0}{\text R} \\[1em] 2 \times 10^{-3} = \frac{5 \times 0.2 \times \text v_0}{5} \\[1em] \text v_0 = 10^{-2}\ \text{m/s} \\[1em] \text v_0 = 1\ \text{cm/s}\end{aligned}$$

Reason — As the N end of the magnet approaches and enters the loop, the flux through the loop increases, so $\text e = -\frac{\text d\Phi}{\text{dt}}$ is negative (taking Φ positive). While the body of the magnet passes through the loop, the flux stays nearly constant, so e ≈ 0 for a time of about $\frac{\text l}{\text v}$. As the S end leaves, the flux decreases and e becomes positive. Thus a negative pulse, a gap of $\frac{\text l}{\text v}$, then a positive pulse, as in curve (3).

2 V

Reason — At t = 1 s, the centre O has moved 1 m and is at x = 0, so exactly half the ring is in the field. The boundary x = 0 then cuts the ring along a diameter of length 2R = 2 m. The area inside the field increases at the rate (2R)v, so

$$\begin{aligned}\text e = \text B(2\text R)\text v \\[1em] = 1 \times 2 \times 1 \\[1em] \text e = 2\ \text V\end{aligned}$$

outward and decreasing with time

Reason — A field parallel to the plane of the coil gives no flux through it, so it cannot induce a current. The induced current in the figure is anticlockwise, and its own field at the coil is outward. By Lenz's law, an induced current supports the original flux only when that flux is decreasing. Hence $\vec{\text B} \space$is outward and decreasing.

I₁ is in clockwise direction and I₂ is in anticlockwise direction

Reason — The bar moves to the left in an inward field. The loop containing R₁ (left) shrinks, so its inward flux decreases; the induced current supports the inward flux and flows clockwise. The loop containing R₂ (right) grows, so its inward flux increases; the induced current opposes it by producing an outward field and flows anticlockwise.

$\frac{\text B_0\text v_0\text d^2}{\text a}$

Reason — Let the left edge of the loop be at x. The flux through the loop is

$$\begin{aligned}\Phi = \int_\text x^{\text x + \text d}\text B_0\frac{\text x'}{\text a}\ \text d\ \text{dx}' \\[1em] = \frac{\text B_0\text d}{2\text a}\left[(\text x + \text d)^2 - \text x^2\right] \\[1em] = \frac{\text B_0\text d}{2\text a}\left(2\text{xd} + \text d^2\right)\end{aligned}$$

Differentiating with respect to time, with $\frac{\text{dx}}{\text{dt}} = \text v_0$,

$$\begin{aligned}|\text e| = \frac{\text d\Phi}{\text{dt}} \\[1em] = \frac{\text B_0\text d}{2\text a} \times 2\text d\frac{\text{dx}}{\text{dt}} \\[1em] |\text e| = \frac{\text B_0\text v_0\text d^2}{\text a}\end{aligned}$$

108.25 mV

Reason — The wings of a plane flying horizontally cut the vertical component of the earth's field,

$$\begin{aligned}\text B_\text V = \text B\sin 60° \\[1em] = 2.5 \times 10^{-4} \times 0.866 \\[1em] \text B_\text V = 2.165 \times 10^{-4}\ \text T\end{aligned}$$

With v = 180 × $\frac{5}{18}$ = 50 m/s and l = 10 m,

$$\begin{aligned}\text e = \text B_\text V\text{lv} \\[1em] = 2.165 \times 10^{-4} \times 10 \times 50 \\[1em] \text e = 108.25 \times 10^{-3}\ \text V \\[1em] \text e = 108.25\ \text{mV}\end{aligned}$$

0.2

Reason — From the stored energy and the power dissipated at I = 8 A,

$$\begin{aligned}\frac{1}{2}\text{LI}^2 = 64 \\[1em] \text L = \frac{2 \times 64}{64} \\[1em] \text L = 2\ \text H\end{aligned}$$

$$\begin{aligned}\text I^2\text R = 640 \\[1em] \text R = \frac{640}{64} \\[1em] \text R = 10\ \Omega\end{aligned}$$

$$\begin{aligned}\tau = \frac{\text L}{\text R} \\[1em] = \frac{2}{10} \\[1em] \tau = 0.2\ \text s\end{aligned}$$

$\frac{\mu_0}{4\pi}8\sqrt{2}\frac{\text a^2}{\text b}$

Reason — Each side of the large loop is at distance $\frac{\text b}{2}$ from the centre and subtends 45° on either side. The field at the centre due to one side is

$$\begin{aligned}\text B' = \frac{\mu_0}{4\pi}\frac{\text I}{(\text b/2)}\left(\sin 45° + \sin 45°\right) \\[1em] \text B' = \frac{\mu_0}{4\pi}\frac{2\sqrt{2}\ \text I}{\text b}\end{aligned}$$

For four sides, $\text B = 4\text B' = \frac{\mu_0}{4\pi}\frac{8\sqrt{2}\ \text I}{\text b}$. This field is nearly uniform over the small loop (b >> a), so

$$\begin{aligned}\text M = \frac{\text{Ba}^2}{\text I} \\[1em] \text M = \frac{\mu_0}{4\pi}8\sqrt{2}\frac{\text a^2}{\text b}\end{aligned}$$

10 A

Reason — In the steady state the inductors carry steady currents and act as plain conducting wires. The three 3 Ω branches are then in parallel,

$$\begin{aligned}\frac{1}{\text R_\text p} = \frac{1}{3} + \frac{1}{3} + \frac{1}{3} \\[1em] \text R_\text p = 1\ \Omega\end{aligned}$$

This is in series with the 2 Ω resistor, so the total resistance is 3 Ω and

$$\begin{aligned}\text I = \frac{30}{3} \\[1em] \text I = 10\ \text A\end{aligned}$$

$\frac{\text L}{\text R}\ln 2$

Reason — Since $\text U = \frac{1}{2}\text{LI}^2$, U = 25% of U_max means I = $\frac{\text I_0}{2}$. During growth,

$$\begin{aligned}\frac{\text I_0}{2} = \text I_0\left(1 - \text e^{-\frac{\text{Rt}}{\text L}}\right) \\[1em] \text e^{-\frac{\text{Rt}}{\text L}} = \frac{1}{2} \\[1em] \text t = \frac{\text L}{\text R}\ln 2\end{aligned}$$

2.25 A

Reason — Just after the switch is closed, each inductor opposes the sudden growth of current and behaves as an open circuit. So the branches containing L carry no current, and the battery sends current only through the vertical resistor R and the resistor R in the lower wire, in series,

$$\begin{aligned}\text I = \frac{\text E}{2\text R} \\[1em] = \frac{9}{2 \times 2} \\[1em] \text I = 2.25\ \text A\end{aligned}$$

$\frac{\text m}{\text r^3}$

Reason — The flux through the loop due to the dipole is

$\Phi_\text{dipole} = \frac{\mu_0\text m}{2\pi\text r^3} \times \pi\text a^2$

The superconducting loop keeps its net flux zero, so the induced current i must produce an equal and opposite flux. The flux of the loop's own current is proportional to i, hence

$\text i \propto \frac{\text m}{\text r^3}$

$\frac{\text m^2}{\text r^6}$

Reason — The induced current gives the loop a magnetic moment m_loop = iπa² ∝ $\frac{\text m}{\text r^3}$, facing the dipole with like poles. The repulsive force is

$$\begin{aligned}\text F = \frac{\text{km m}_\text{loop}}{\text r^4} \\[1em] \text F = \frac{\text{Cm}^2}{\text r^7}\end{aligned}$$

where C is a constant. The work done against this force in bringing the dipole from infinity to r is

$$\begin{aligned}\text W = -\int_\infty^\text r\frac{\text{Cm}^2}{\text r'^7}\text{dr}' \\[1em] = -\text{Cm}^2\left[-\frac{1}{6\text r'^6}\right]_\infty^\text r \\[1em] \text W = \frac{\text{Cm}^2}{6\text r^6}\end{aligned}$$

So W ∝ $\frac{\text m^2}{\text r^6}$.

doubled

Reason — The emf is proportional to the number of turns, so it becomes ε/2. The wire length is proportional to the number of turns (halved), and doubling the wire's radius makes its cross-section 4 times; hence the resistance becomes $\frac{\text R}{2 \times 4} = \frac{\text R}{8}$.

$$\begin{aligned}\text P' = \frac{(\varepsilon/2)^2}{\text R/8} \\[1em] \text P' = 2\frac{\varepsilon^2}{\text R} \\[1em] \text P' = 2\text P\end{aligned}$$

15.6 A

Reason — By Faraday's law,

$$\begin{aligned}|\text e| = \frac{\text d\Phi}{\text{dt}} \\[1em] = \frac{\text d}{\text{dt}}\left(5\text t^3 + 4\text t^2 + 2\text t - 5\right) \\[1em] = 15\text t^2 + 8\text t + 2\end{aligned}$$

At t = 2 s, |e| = 60 + 16 + 2 = 78 V, so

$$\begin{aligned}\text I = \frac{78}{5} \\[1em] \text I = 15.6\ \text A\end{aligned}$$

$\frac{2\sqrt{2}\mu_0\text l^2}{\pi\text L}$

Reason — Each side of the large loop is at distance $\frac{\text L}{2}$ from O and subtends 45° on either side. For the four sides,

$$\begin{aligned}\text B = 4 \times \frac{\mu_0}{4\pi}\frac{\text I}{(\text L/2)}\left(\sin 45° + \sin 45°\right) \\[1em] \text B = \frac{2\sqrt{2}\mu_0\text I}{\pi\text L}\end{aligned}$$

This field is nearly uniform over the small loop (L >> l), so

$$\begin{aligned}\text M = \frac{\text{Bl}^2}{\text I} \\[1em] \text M = \frac{2\sqrt{2}\mu_0\text l^2}{\pi\text L}\end{aligned}$$

L₁ + L₂ − 2M

Reason — In the figure, the same current I flows through the two coils in opposite senses, so the flux due to mutual induction in each coil opposes its own flux. The total flux linkage is

$$\begin{aligned}\Phi = \text L_1\text I - \text{MI} + \text L_2\text I - \text{MI} \\[1em] \Phi = \left(\text L_1 + \text L_2 - 2\text M\right)\text I\end{aligned}$$

Hence L_eq = L₁ + L₂ − 2M.

Note: The answer table prints option (a), which is incorrect; the textbook's own hint and the figure give L₁ + L₂ − 2M (option 4).

t = 7 ms, U = 1 mJ

Reason — Here τ = L/R = 1/100 = 0.01 s and I₀ = 6/100 = 0.06 A.

(A) For I = I₀/2, $\text e^{-\frac{\text t}{\tau}} = \frac{1}{2}$, so

$$\begin{aligned}\text t = \tau\ln 2 \\[1em] = 0.01 \times 0.693 \\[1em] \text t = 6.93 \times 10^{-3}\ \text s \approx 7\ \text{ms}\end{aligned}$$

(B) At t = 15 ms, $\frac{\text t}{\tau} = \frac{3}{2}$, so

$$\begin{aligned}\text I = 0.06\left(1 - \text e^{-\frac{3}{2}}\right) \\[1em] = 0.06 \times 0.75 \\[1em] \text I = 0.045\ \text A\end{aligned}$$

$$\begin{aligned}\text U = \frac{1}{2}\text{LI}^2 \\[1em] = \frac{1}{2} \times 1 \times (0.045)^2 \\[1em] \text U = 1.01 \times 10^{-3}\ \text J \approx 1\ \text{mJ}\end{aligned}$$

I → Q, II → P, III → S, IV → R

Reason — Let B = μ₀mI be the solenoid's field, along $\hat{\text n}$. The loop's area vector is $\text A\hat{\text k}$, so Φ = BA$(\hat{\text n}\cdot\hat{\text k})$ and the induced current is $\text i = -\frac{1}{\text R}\frac{\text d\Phi}{\text{dt}}$. The loop's moment is $\text{iA}\hat{\text k}$, and the torque is $\vec\tau = \text{iAB}\left(\hat{\text k} \times \hat{\text n}\right)$. Write α = $\frac{\text A^2\text B^2\omega}{2\text R}$ and ωt = π/6.

(I) $\Phi = \frac{\text{BA}}{\sqrt{2}}\cos\omega\text t$, $\text i = \frac{\text{BA}\omega}{\sqrt{2}\text R}\sin\omega\text t$, $\hat{\text k} \times \hat{\text n} = -\frac{\sin\omega\text t}{\sqrt{2}}\hat{\text i}$

$$\begin{aligned}\vec\tau = -\frac{\text A^2\text B^2\omega}{2\text R}\sin^2\omega\text t\ \hat{\text i} \\[1em] \vec\tau = -\frac{\alpha}{4}\hat{\text i} \quad \text{(Q)}\end{aligned}$$

(II) $\hat{\text n}$ lies in the xy-plane, so Φ = 0, i = 0 and τ = 0 (P).

(III) $\Phi = \frac{\text{BA}}{\sqrt{2}}\cos\omega\text t$, $\text i = \frac{\text{BA}\omega}{\sqrt{2}\text R}\sin\omega\text t$, $\hat{\text k} \times \hat{\text n} = \frac{\sin\omega\text t}{\sqrt{2}}\hat{\text j}$

$$\begin{aligned}\vec\tau = \frac{\text A^2\text B^2\omega}{2\text R}\sin^2\omega\text t\ \hat{\text j} \\[1em] \vec\tau = \frac{\alpha}{4}\hat{\text j} \quad \text{(S)}\end{aligned}$$

(IV) $\Phi = \frac{\text{BA}}{\sqrt{2}}\sin\omega\text t$, $\text i = -\frac{\text{BA}\omega}{\sqrt{2}\text R}\cos\omega\text t$, $\hat{\text k} \times \hat{\text n} = -\frac{\cos\omega\text t}{\sqrt{2}}\hat{\text i}$

$$\begin{aligned}\vec\tau = \frac{\text A^2\text B^2\omega}{2\text R}\cos^2\omega\text t\ \hat{\text i} \\[1em] \vec\tau = \frac{3\alpha}{4}\hat{\text i} \quad \text{(R)}\end{aligned}$$

Note: The stem names the constant θ, while List-II uses α; both stand for $\frac{\text A^2\mu_0^2\text m^2\text I^2\omega}{2\text R}$.

1 A

Reason — The coil rotates about a horizontal diameter, so it cuts the vertical component of the earth's field. The peak emf is

$$\begin{aligned}\text e_0 = \text{NBA}\omega \\[1em] = 1000 \times 2 \times 10^{-5} \times \pi(10)^2 \times 2 \\[1em] \text e_0 = 12.56\ \text V\end{aligned}$$

$$\begin{aligned}\text I_0 = \frac{12.56}{12.56} \\[1em] \text I_0 = 1\ \text A\end{aligned}$$

(C) and (D)

Reason — The induced emf is $\text e = -\frac{\text d\Phi}{\text{dt}}$ with Φ = BA cos θ. In a uniform field, moving the coil (with uniform or non-uniform speed) does not change B, A or θ, so no emf is induced. Rotating the coil changes θ, and changing its area changes A; both change the flux and induce an emf.

to bring the coil to rest quickly

Reason — When the coil oscillates in the magnetic field, eddy currents are induced in the metallic core. By Lenz's law, they oppose the motion of the coil, so the coil comes to rest quickly (electromagnetic damping), making the galvanometer dead-beat.

Insulating ball will reach the earth's surface earlier than the metal ball

Reason — As the metallic ball falls through the earth's magnetic field, the flux linked with it changes and eddy currents are induced in it. By Lenz's law, these oppose its motion, so its acceleration is less than g. No currents are induced in the insulating ball, which falls freely and reaches the ground first.

8 μJ

Reason — The energy stored in an inductor is

$$\begin{aligned}\text U = \frac{1}{2}\text{LI}^2 \\[1em] = \frac{1}{2} \times 4 \times 10^{-6} \times (2)^2 \\[1em] \text U = 8 \times 10^{-6}\ \text J \\[1em] \text U = 8\ \mu\text J\end{aligned}$$

P → 3, Q → 4, R → 2, S → 5

Reason — Here m = 0.02 kg, l = 0.25 m, R = 10 Ω and B₀ = 4 T, so $\frac{\text B_0^2\text l^2}{\text R} = \frac{16 \times 0.0625}{10} = 0.1$ kg s^-1. The equation of motion of the rod is

$\text m\frac{\text{dv}}{\text{dt}} = \text{mg} - \frac{\text B_0^2\text l^2}{\text R}\text v$

Terminal velocity (S) : dv/dt = 0 gives

$$\begin{aligned}\text v_\text T = \frac{\text{mgR}}{\text B_0^2\text l^2} \\[1em] = \frac{0.02 \times 10}{0.1} \\[1em] \text v_\text T = 2\ \text{ms}^{-1}\end{aligned}$$

Velocity at t = 0.2 s : Integrating with v = 0 at t = 0, and τ = $\frac{\text{mR}}{\text B_0^2\text l^2}$ = 0.2 s,

$$\begin{aligned}\int_0^\text v\frac{\text{dv}}{\text v_\text T - \text v} = \int_0^\text t\frac{\text{dt}}{\tau} \\[1em] \ln\frac{\text v_\text T}{\text v_\text T - \text v} = \frac{\text t}{\tau} \\[1em] \text v = \text v_\text T\left(1 - \text e^{-\frac{\text t}{\tau}}\right)\end{aligned}$$

At t = 0.2 s = τ, v = 2(1 − 0.4) = 1.2 ms^-1.

Induced emf (P) :

$$\begin{aligned}\text e = \text B_0\text{lv} \\[1em] = 4 \times 0.25 \times 1.2 \\[1em] \text e = 1.20\ \text V\end{aligned}$$

Magnetic force (Q) :

$$\begin{aligned}\text F = \frac{\text B_0^2\text l^2\text v}{\text R} \\[1em] = 0.1 \times 1.2 \\[1em] \text F = 0.12\ \text N\end{aligned}$$

Power dissipated (R) :

$$\begin{aligned}\text P = \frac{\text e^2}{\text R} \\[1em] = \frac{(1.2)^2}{10} \\[1em] \text P = 0.144\ \text W \approx 0.14\ \text W\end{aligned}$$

Hence P → 3, Q → 4, R → 2 and S → 5.

AB and DC

Reason — The magnet moves to the right, its N pole receding from solenoid-1 and its S pole approaching solenoid-2. By Lenz's law, the end B of solenoid-1 becomes a south pole (to attract the receding N pole), and the end C of solenoid-2 becomes a south pole (to repel the approaching S pole). For the windings shown, this requires the induced current to flow from A to B in solenoid-1 and from D to C in solenoid-2.

1 : 2

Reason — For a solenoid filled with a material of relative permeability μ_r, L = μ₀μ_rn²Al, so with μ_r and A the same,

$$\begin{aligned}\frac{\text L_\text A}{\text L_\text B} = \frac{\text n_\text A^2\text l_\text A}{\text n_\text B^2\text l_\text B} \\[1em] = \left(\frac{1}{2}\right)^2 \times 2 \\[1em] = \frac{1}{2}\end{aligned}$$

Given,

• Radius of coil, r = 10 cm = 0.1 m
• Magnetic field, B = 3.0 × 10^-5 T
• Half rotation (π rad) in 0.2 s

The angular speed of the coil is

$$\begin{aligned}\omega = \frac{\pi}{0.2} \\[1em] \omega = 5\pi\ \text{rad s}^{-1}\end{aligned}$$

With the plane initially perpendicular to the field, the flux at time t is Φ = BA cos ωt. Differentiating,

$$\begin{aligned}|\text e| = \left|\frac{\text d\Phi}{\text{dt}}\right| \\[1em] = \left|\frac{\text d}{\text{dt}}\left(\text{BA}\cos\omega\text t\right)\right| \\[1em] = \text{BA}\omega\sin\omega\text t\end{aligned}$$

The maximum value is

$$\begin{aligned}\text e_\text{max} = \text B\pi\text r^2\omega \\[1em] = 3.0 \times 10^{-5} \times \pi \times (0.1)^2 \times 5\pi \\[1em] = 1.5\pi^2 \times 10^{-6} \\[1em] \text e_\text{max} = 14.8 \times 10^{-6}\ \text V \approx 15\ \mu\text V\end{aligned}$$

Hence, the maximum emf induced is close to 15 μV.

Given,

• Radius of coil, r = 8.0 cm = 0.08 m
• Number of turns, N = 20
• Angular speed, ω = 50 rad s^-1
• Magnetic field, B = 3.0 × 10^-2 T

For a coil rotating in a uniform field, the maximum emf is

$$\begin{aligned}\text e_0 = \text{NBA}\omega \\[1em] = 20 \times 3.0 \times 10^{-2} \times \pi(0.08)^2 \times 50 \\[1em] = 20 \times 3.0 \times 10^{-2} \times 0.0201 \times 50 \\[1em] \text e_0 = 0.603\ \text V \approx 60 \times 10^{-2}\ \text V\end{aligned}$$

Hence, the maximum emf induced is 60 × 10^-2 volt.

Given,

• Radius of coil, r = 1 m
• Resistance, R = 2 μΩ = 2 × 10^-6 Ω
• $\text B = \frac{4}{\pi} \times 10^{-3}\left(1 - \frac{\text t}{100}\right)$ T

The field becomes zero when t = 100 s. The rate of change of the field is

$$\begin{aligned}\frac{\text{dB}}{\text{dt}} = \frac{4}{\pi} \times 10^{-3} \times \left(-\frac{1}{100}\right) \\[1em] \frac{\text{dB}}{\text{dt}} = -\frac{4}{\pi} \times 10^{-5}\ \text T\ \text s^{-1}\end{aligned}$$

This is constant, so the induced emf is constant,

$$\begin{aligned}|\text e| = \pi\text r^2\left|\frac{\text{dB}}{\text{dt}}\right| \\[1em] = \pi \times 1^2 \times \frac{4}{\pi} \times 10^{-5} \\[1em] |\text e| = 4 \times 10^{-5}\ \text V\end{aligned}$$

The power dissipated is

$$\begin{aligned}\text P = \frac{\text e^2}{\text R} \\[1em] = \frac{\left(4 \times 10^{-5}\right)^2}{2 \times 10^{-6}} \\[1em] \text P = 8 \times 10^{-4}\ \text W\end{aligned}$$

The energy dissipated in 100 s is

$$\begin{aligned}\text E = \text P \times \text t \\[1em] = 8 \times 10^{-4} \times 100 \\[1em] \text E = 8 \times 10^{-2}\ \text J \\[1em] \text E = 80\ \text{mJ}\end{aligned}$$

Hence, the energy dissipated is E = 80 mJ.

Given,

• Φ_B(t) = 10t² + 20t (in mWb)
• R = 2 Ω
• t = 5 s

By Faraday's law, the magnitude of the induced emf is

$$\begin{aligned}|\text e| = \frac{\text d\Phi_\text B}{\text{dt}} \\[1em] = \frac{\text d}{\text{dt}}\left(10\text t^2 + 20\text t\right) \\[1em] = 20\text t + 20\ \text{(in mV)}\end{aligned}$$

At t = 5 s,

$$\begin{aligned}|\text e| = 20 \times 5 + 20 \\[1em] |\text e| = 120\ \text{mV}\end{aligned}$$

The current through R is

$$\begin{aligned}\text I = \frac{|\text e|}{\text R} \\[1em] = \frac{120}{2} \\[1em] \text I = 60\ \text{mA}\end{aligned}$$

Hence, the magnitude of current through R is 60 mA.

Given,

• Φ = 8t² − 9t + 5 (in Wb)
• R = 20 Ω
• t = 0.25 s

By Faraday's law,

$$\begin{aligned}\text e = -\frac{\text d\Phi}{\text{dt}} \\[1em] = -\frac{\text d}{\text{dt}}\left(8\text t^2 - 9\text t + 5\right) \\[1em] = -(16\text t - 9)\end{aligned}$$

At t = 0.25 s,

$$\begin{aligned}\text e = -(16 \times 0.25 - 9) \\[1em] \text e = 5\ \text V\end{aligned}$$

The induced current is

$$\begin{aligned}\text I = \frac{\text e}{\text R} \\[1em] = \frac{5}{20} \\[1em] \text I = 0.25\ \text A \\[1em] \text I = 250\ \text{mA}\end{aligned}$$

Hence, the magnitude of the induced current is 250 mA.

Given,

• R = 8 Ω
• $\Phi = \frac{2}{3}\left(9 - \text t^2\right)$

The flux becomes zero when 9 − t² = 0, that is, at t = 3 s. The induced emf is

$$\begin{aligned}\text e = -\frac{\text d\Phi}{\text{dt}} \\[1em] = -\frac{2}{3}(-2\text t) \\[1em] \text e = \frac{4\text t}{3}\end{aligned}$$

The heat produced from t = 0 to t = 3 s is

$$\begin{aligned}\text H = \int_0^3\frac{\text e^2}{\text R}\text{dt} \\[1em] = \int_0^3\frac{16\text t^2}{9 \times 8}\text{dt} \\[1em] = \frac{2}{9}\left[\frac{\text t^3}{3}\right]_0^3 \\[1em] = \frac{2}{9} \times 9 \\[1em] \text H = 2\ \text J\end{aligned}$$

Hence, the total heat produced in the coil is 2 J.

Given,

• $\vec{\text B} = 3\text t^3\ \hat{\text j} + 3\text t^2\ \hat{\text k}$ (SI units)
• Radius of loop, r = 1 m, in the X-Y plane
• t = 2 s

The loop lies in the X-Y plane, so its area vector is along $\hat{\text k}$ and only the z-component of $\vec{\text B} \space$gives flux,

$$\begin{aligned}\Phi = 3\text t^2 \times \pi\text r^2 \\[1em] \Phi = 3\pi\text t^2\end{aligned}$$

The induced emf is

$$\begin{aligned}|\text e| = \frac{\text d\Phi}{\text{dt}} \\[1em] = \frac{\text d}{\text{dt}}\left(3\pi\text t^2\right) \\[1em] = 6\pi\text t\end{aligned}$$

At t = 2 s,

$$\begin{aligned}|\text e| = 6\pi \times 2 \\[1em] |\text e| = 12\pi\ \text V\end{aligned}$$

Hence, the value of n is 12.

Given,

• Number of turns, N = 1000
• Area of each turn, A = 1 m²
• Magnetic field, B = 0.07 T
• Frequency, f = 1 rev/s

The angular speed is ω = 2πf = 2π rad s^-1. The maximum emf is

$$\begin{aligned}\text e_\text{max} = \text{NBA}\omega \\[1em] = 1000 \times 0.07 \times 1 \times 2\pi \\[1em] = 140 \times \frac{22}{7} \\[1em] \text e_\text{max} = 440\ \text V\end{aligned}$$

Hence, the maximum voltage generated is 440 V.

Given,

• Self-inductance, L = 2.0 H
• Current changes from 0 to 2 A

The energy spent in building up a current I in an inductor is stored as magnetic energy $\frac{1}{2}\text{LI}^2$; it depends only on the final current, not on how the current varies with time.

$$\begin{aligned}\text U = \frac{1}{2}\text{LI}^2 \\[1em] = \frac{1}{2} \times 2.0 \times (2)^2 \\[1em] \text U = 4\ \text J\end{aligned}$$

Hence, the energy spent is 4 J.

Given,

• Resistance, R = 10 Ω
• Inductance, L = 20 mH = 20 × 10^-3 H
• Battery voltage, V = 20 V
• Time for current to become zero, Δt = 100 μs = 100 × 10^-6 s

The constant current through the coil before the switch is opened is

$$\begin{aligned}\text I = \frac{\text V}{\text R} \\[1em] = \frac{20}{10} \\[1em] \text I = 2\ \text A\end{aligned}$$

The average emf induced as the current falls from 2 A to zero is

$$\begin{aligned}\text e = \text L\frac{\Delta\text I}{\Delta\text t} \\[1em] = 20 \times 10^{-3} \times \frac{2}{100 \times 10^{-6}} \\[1em] \text e = 400\ \text V\end{aligned}$$

Hence, the average emf induced in the coil is 400 V.

Given,

• Side of loop, a = 2.0 cm, so A = (0.02)² = 4 × 10^-4 m²
• Turns per unit length, n = 50 per cm = 5000 m^-1
• Current amplitude, i₀ = 2.5 A
• Angular frequency, ω = 700 rad s^-1

The field inside the solenoid is B = μ₀ni, with i = i₀ sin ωt. The flux through the loop is

$\Phi = \mu_0\text{ni}_0\text A\sin\omega\text t$

Differentiating,

$$\begin{aligned}|\text e| = \frac{\text d\Phi}{\text{dt}} \\[1em] = \mu_0\text{ni}_0\text A\frac{\text d}{\text{dt}}\left(\sin\omega\text t\right) \\[1em] = \mu_0\text{ni}_0\text A\omega\cos\omega\text t\end{aligned}$$

The amplitude of the emf is

$$\begin{aligned}\text e_0 = \mu_0\text{ni}_0\text A\omega \\[1em] = 4 \times \frac{22}{7} \times 10^{-7} \times 5000 \times 2.5 \times 4 \times 10^{-4} \times 700 \\[1em] = 8800 \times 10^{-7} \times 5000 \times 2.5 \times 4 \times 10^{-4} \\[1em] \text e_0 = 44 \times 10^{-4}\ \text V\end{aligned}$$

Hence, the value of x is 44.

Given,

• Side of loop = 10 cm, so A = (0.1)² = 10^-2 m²
• Magnetic field, B = 0.20 T, in north-east direction
• Field falls to zero in Δt = 1 s at a steady rate

The loop is vertical in the east-west plane, so its area vector points north (or south). The field is in the north-east direction, so the angle between $\vec{\text B} \space$and the area vector is 45°. The initial flux is

$$\begin{aligned}\Phi = \text{BA}\cos 45° \\[1em] = 0.20 \times 10^{-2} \times \frac{1}{\sqrt{2}} \\[1em] \Phi = \sqrt{2} \times 10^{-3}\ \text{Wb}\end{aligned}$$

The flux falls to zero in 1 s, so

$$\begin{aligned}|\text e| = \frac{\Delta\Phi}{\Delta\text t} \\[1em] = \frac{\sqrt{2} \times 10^{-3}}{1} \\[1em] |\text e| = \sqrt{2} \times 10^{-3}\ \text V\end{aligned}$$

Hence, the value of x is 2.

Given,

• Mutual inductance, M = 0.002 H
• i = i₀ sin ωt, with i₀ = 5 A and ω = 50π rad/s

The emf induced in the second coil is

$$\begin{aligned}\text e = -\text M\frac{\text{di}}{\text{dt}} \\[1em] = -\text M\frac{\text d}{\text{dt}}\left(\text i_0\sin\omega\text t\right) \\[1em] = -\text{Mi}_0\omega\cos\omega\text t\end{aligned}$$

Its maximum value is

$$\begin{aligned}\text e_\text{max} = \text{Mi}_0\omega \\[1em] = 0.002 \times 5 \times 50\pi \\[1em] \text e_\text{max} = \frac{\pi}{2}\ \text V\end{aligned}$$

Hence, the value of α is 2.

If assertion is true but reason is false.

Explanation

Assertion (A) is correct: By Faraday's law, the induced emf is $\text e = -\frac{\text d\Phi_\text B}{\text{dt}}$. An induced current flows in a coil only while the magnetic flux linked with it is changing, so the current is maintained only so long as the flux keeps changing.

Reason (R) is incorrect: A large but constant magnetic flux through a coil induces no emf, however continuous the circuit may be. It is the rate of change of flux, not the amount of flux, that produces the current.

Therefore, assertion is true but reason is false.

If both assertion and reason are true and reason is the correct explanation of assertion.

Explanation

Assertion (A) is correct: When a metallic surface is moved into or out of a magnetic field, the magnetic flux linked with it changes, and so an emf is induced in it.

Reason (R) is also correct: The emf induced throughout the metal by this change of flux sets up closed loops of current within the metal surface, which are the eddy currents. The eddy currents produced in the moving surface are the evidence of the emf induced in it, so the Reason explains the Assertion.

Therefore, both assertion and reason are true and reason is the correct explanation of assertion.

If assertion is true but reason is false.

Explanation

Assertion (A) is correct: In electromagnetic induction, the mechanical work done in moving a magnet or conductor against the opposing force (Lenz's law) is converted into electrical energy. Faraday's laws, together with Lenz's law, are thus in accordance with the law of conservation of energy.

Reason (R) is incorrect: In a purely resistive a.c. circuit, the current is in phase with the emf; it neither leads nor lags.

Therefore, assertion is true but reason is false.

If assertion is true but reason is false.

Explanation

Assertion (A) is correct: As the bar magnet falls towards the coil, the magnetic flux linked with the coil increases and a current is induced in it. By Lenz's law, this current opposes the approach of the magnet, so the coil exerts an upward force on the magnet and its acceleration is less than g.

Reason (R) is incorrect: The N pole of the magnet faces the coil. To oppose its approach, the upper face of the coil must become a north pole, so the induced current is anticlockwise when viewed from above (from the magnet), not clockwise.

Therefore, assertion is true but reason is false.