Class 12 Physics Chapter 6 Important Questions (2026-27)

Class 12 Physics Chapter 6 Electromagnetic Induction: important questions with answers on Faraday's and Lenz's laws for CBSE 2026-27.

Important Questions for Class 12 Physics Chapter 6 – Electromagnetic Induction

Looking for Class 12 Physics Chapter 6 important questions? This page covers Electromagnetic Induction with exam-focused questions and step-by-step answers, arranged by marks: 1 mark, 2–3 marks and 5 marks. Written for the CBSE 2026–27 syllabus, it focuses on Faraday's law, Lenz's law and self and mutual inductance. Attempt them after finishing the NCERT chapter.

Very Short Answer Questions (1 Mark)

Self-Inductance

Question 1

A plot of magnetic flux (\(\phi\)) versus current (I) is shown for two inductors A and B. Which of the two has a larger value of self-inductance?

Answer: Since \(\phi = LI\), the slope of the graph gives L. The slope of A is greater than the slope of B, so inductor A has a larger self-inductance.

Question 2

Define self-inductance of a coil. Write its SI unit.

Answer: Self-induction is the property of a coil by virtue of which it opposes the growth or decay of the current flowing through it. SI unit: henry (H).

Question 3

Define the term 'self-inductance' of a coil. Write its SI unit.

Answer: The self-inductance of a coil equals the induced emf produced in the coil when the rate of change of current in the coil is unity: \(\varepsilon = -L\dfrac{dI}{dt}\). SI unit: henry.

Question 4

How does the mutual inductance of a pair of coils change when (i) distance between the coils is increased and (ii) number of turns in the coils is increased?

Answer: (i) Mutual inductance decreases, because the flux linked with the secondary coil decreases.

(ii) Mutual inductance increases, since \(M = \mu_0 n_1 n_2 Al\).

Question 5

How does the mutual inductance of a pair of coils change when (i) distance between the coils is decreased and (ii) number of turns in the coils is decreased?

Answer: (i) Mutual inductance increases.

(ii) Mutual inductance decreases, since \(M = \mu_0 n_1 n_2 Al\).

Lenz's Law and Faraday's Law

Question 6

State Lenz's law.

Answer: Lenz's law states that the polarity of induced emf is such that it tends to produce a current which opposes the change in magnetic flux that induced it.

Question 7

State Faraday's law of electromagnetic induction.

Answer: Faraday's law states that the magnitude of emf induced in a circuit is directly proportional to the rate of change of magnetic flux linked with the circuit.

Mathematically: \(e = -\dfrac{d\phi}{dt}\), where \(d\phi\) is the small change in magnetic flux in time \(dt\).

Question 8

State the law that gives the polarity of the induced emf.

Answer: Lenz's law gives the polarity of the induced emf: the direction of induced emf is such that it opposes the cause which produces it.

Predicting Induced Currents

Question 9

Two bar magnets are quickly moved towards a metallic loop connected across a capacitor C. Predict the polarity of the capacitor.

Answer: When both magnets move towards the loop, the A-side plate of the capacitor becomes positive and the lower plate B becomes negative, making the induced current flow clockwise.

Question 10

Predict the polarity of the capacitor when the two magnets are quickly moved in the directions marked by arrows.

Answer: The current in the coil will be anticlockwise (seen from the left), so plate A becomes positive and plate B becomes negative.

Question 11

Predict the directions of induced currents in metal rings 1 and 2 lying in the same plane where current I in the wire is increasing steadily.

Answer: In ring 1, the induced current flows clockwise. In ring 2, the induced current flows anticlockwise.

Question 12

Predict the direction of induced current in a metal ring when the ring is moved towards a conductor carrying current I as shown.

Answer: Clockwise direction.

Question 13

Predict the directions of induced current in metal rings 1 and 2 when current I in the wire is steadily decreasing.

Answer: In ring 1, the induced current flows anticlockwise. In ring 2, the induced current flows clockwise.

Question 14

A bar magnet is moved in the direction indicated by the arrow between two coils PQ and CD. Predict the directions of induced current in each coil.

Answer: By Lenz's law, the ends of both coils closer to the magnet behave as south poles. Hence the induced current in both coils flows clockwise when seen from the magnet side.

Question 15

Predict the direction of the induced current in the rectangular loop abcd as it is moved into the region of a uniform magnetic field \(\vec{B}\) directed normal to the plane of the loop.

Answer: The induced current is anticlockwise, i.e., along cbadc.

Question 16

Predict the polarity of the capacitor in the given situation.

Answer: When the magnets move towards the loop, plate A becomes positive and plate B becomes negative, with the induced current flowing clockwise.

Question 17

The electric current flowing in a wire in the direction from B to A is decreasing. Find the direction of the induced current in the metallic loop kept above the wire.

Answer: The induced current in the loop flows clockwise.

Question 18

A conducting loop is held above a current-carrying wire PQ. Depict the direction of the current induced in the loop when the current in the wire PQ is constantly increasing.

Answer: The induced current is clockwise.

Question 19

A conducting loop is held below a current-carrying wire PQ. Predict the direction of the induced current in the loop when the current in the wire is constantly increasing.

Answer: The induced current is anticlockwise.

Question 20

Predict the polarity of plate A of the capacitor when a magnet is moved towards it.

Answer: Plate A has positive polarity.

Question 21

Predict the polarity of the capacitor in the given situation.

Answer: Plate A is positive, plate B is negative.

Question 22

A bar magnet is moved in the direction indicated by the arrow between two coils PQ and CD. Predict the direction of the induced current in each coil.

Answer: Induced current flows from P to Q through the ammeter, and from D to C through the ammeter.

Question 23

What is the direction of induced currents in metal rings 1 and 2 when current I in the wire is increasing steadily?

Answer: Clockwise in ring 1 and anticlockwise in ring 2.

Question 24

In the given figure, mark the polarity of plates A and B of a capacitor when the magnets are quickly moved towards the coil.

Answer: Plate A is positive, plate B is negative.

Eddy Currents and Applications

Question 25

A light metal disc on the top of an electromagnet is thrown up as the current is switched on. Why? Give reason.

Answer: Due to eddy currents. By Lenz's law, the lower face of the disc acquires the same polarity (say N) as the upper face of the core, and the disc is repelled upward.

Question 26

The motion of a copper plate is damped when it is allowed to oscillate between the two poles of a magnet. What is the cause of this damping?

Answer: The cause of this damping is the eddy current induced in the plate.

Question 27

The motion of a copper plate is damped when it is allowed to oscillate between the two poles of a magnet. If slots are cut in the plate, how will the damping be affected?

Answer: Cutting slots reduces the eddy currents, so the damping is reduced.

Question 28

A metallic piece gets hot when surrounded by a coil carrying high-frequency alternating current. Why?

Answer: Due to the heating effect of eddy currents set up in the metallic piece.

Question 29

Name any two applications where eddy currents are used to advantage.

Answer: (i) Electromagnetic damping, (ii) Magnetic braking, (iii) Induction furnace, (iv) Electric power meters (any two).

Question 30

Two spherical bobs, one metallic and the other of glass, of the same size, are allowed to fall freely from the same height above the ground. Which of the two would reach earlier and why?

Answer: The glass bob reaches earlier. On the metallic bob, eddy currents induced by the earth's magnetic field produce an upward force that slows its fall.

Induced emf and Flux

Question 31

A long straight current-carrying wire passes normally through the centre of a circular loop. If the current through the wire increases, will there be an induced emf in the loop? Justify.

Answer: No. The magnetic field due to the current-carrying wire lies in the plane of the circular loop, so the magnetic flux through the loop remains zero.

Short Answer Questions (2–3 Marks)

Self-Inductance of a Solenoid

Question 32

Derive an expression for the self-inductance of a long air-cored solenoid of length l and number of turns N.

Answer: Consider a long solenoid of length \(l\), radius \(r\) (\(r \ll l\)), and \(n\) turns per unit length. The field inside is \(B = \mu_0 nI\).

Flux through each turn: \(\phi = BA = \mu_0 nIA\).

Total flux linkage: \(N\phi = (nl)(\mu_0 nIA) = \mu_0 n^2 A l I\).

Self-inductance: \(L = \dfrac{N\phi}{I} =\) \(\mu_0 n^2 Al\). With \(N = nl\): \(L = \dfrac{\mu_0 N^2 A}{l}\)

Question 33

Derive the expression for the self-inductance of a long solenoid of cross-sectional area A and length l, having n turns per unit length.

Answer: Same as Question 32. \(L = \mu_0 n^2 Al\). With a core of relative permeability \(\mu_r\): \(L = \mu_r \mu_0 n^2 Al\).

Question 34

Write its SI unit for self-inductance of a coil. Derive the expression for the self-inductance of a long solenoid of cross-sectional area A, length l having n turns per unit length.

Answer: SI unit of self-inductance: henry (H).

Self-inductance of a coil is numerically equal to the magnetic flux linked with the coil when unit current flows through it: \(\phi = LI\).

For a long solenoid: field \(B = \mu_0 nI\). Flux through \(N = nl\) turns: \(N\phi = nl \cdot \mu_0 nIA = \mu_0 n^2 A l I\).

So \(L = \dfrac{N\phi}{I} =\) \(\mu_0 n^2 Al\).

Question 35

(a) Define self-inductance. Write its SI units. (b) Derive an expression for self-inductance of a long solenoid of length l, cross-sectional area A having N number of turns.

Answer: (a) Self-inductance is numerically equal to the magnetic flux linked with the coil when unit current passes through it (or the induced emf when the rate of change of current is unity). SI unit: henry.

(b) With \(n = \dfrac{N}{l}\), \(L = \mu_0 n^2 Al =\) \(\dfrac{\mu_0 N^2 A}{l}\).

Question 36

Define the term self-inductance of a solenoid. Obtain the expression for the magnetic energy stored in an inductor of self-inductance L to build up a current I through it.

Answer: Self-inductance: \(\phi = LI\) where the proportionality constant L is the self-inductance. The induced emf \(= -L\dfrac{dI}{dt}\).

For a solenoid: \(L = \mu_0 n^2 Al\) (or \(\mu_r\mu_0 n^2 Al\) with a core).

Energy stored: The rate of work done against the back emf is \(\dfrac{dW}{dt} = \varepsilon I = LI\dfrac{dI}{dt}\).

Total work done to build current I from 0: \(W = \int_0^I LI\,dI =\) \(\dfrac{1}{2}LI^2\) — this is the magnetic potential energy stored in the inductor.

Question 37

Define self-inductance of a coil. Show that the magnetic energy required to build up the current I in a coil of self-inductance L is given by \(\dfrac{1}{2}LI^2\).

Answer: Same as Question 36. Self-inductance of a coil is the induced emf when the rate of change of current is unity. Work done in building current I: \(W = \int_0^I LI\,dI = \dfrac{1}{2}LI^2\) — stored as magnetic field energy.

Question 38

(a) Define self-inductance of a coil and hence write the definition of 'henry'. (b) Write any two factors on which (i) self-inductance of a coil and (ii) mutual inductance of a pair of coils depend.

Answer: (a) The self-inductance L of a coil equals the magnetic flux linked with it when unit current flows through it. One henry is the self-inductance of a coil for which the magnetic flux linked, due to a current of 1 A, equals one weber.

(b) Self-inductance depends on: (i) geometry (area and length), (ii) number of turns, (iii) medium within the coil.

Mutual inductance depends on: (i) geometry of both coils, (ii) their distance of separation, (iii) number of turns in each coil, (iv) nature of the medium between them.

Question 39

Define the term 'self-inductance' and write its SI unit.

Answer: Self-inductance of a coil equals the magnitude of the emf induced in it when the current in the coil is changing at a unit rate. SI unit: henry.

Mutual Inductance

Question 40

Define mutual inductance between two long coaxial solenoids. Find out the expression for the mutual inductance of an inner solenoid of length l, radius r₁ and n₁ turns per unit length, due to a second outer solenoid of the same length and n₂ turns per unit length.

Answer: Mutual induction is the phenomenon of inducing emf in a coil due to the rate of change of current in a nearby coil.

Consider two coaxial solenoids S₁ (inner) and S₂ (outer), each of length l, S₂ completely surrounding S₁. A current I₂ through S₂ produces a field \(B_2 = \mu_0 n_2 I_2\) inside S₂ (and hence inside S₁).

Flux through each turn of S₁: \(\phi_1 = B_2 A_1 = \mu_0 n_2 I_2 \cdot \pi r_1^2\). Total flux linkage: \(N_1 \phi_1 = (n_1 l)(\mu_0 n_2 I_2 \pi r_1^2)\).

So \(M_{12} = \dfrac{N_1\phi_1}{I_2} = \mu_0 n_1 n_2 \pi r_1^2 l\). (Note: \(M_{12} = M_{21} = M\).)

Question 41

Define the term 'mutual inductance' between two coils. Obtain the expression for mutual inductance of a pair of long coaxial solenoids, each of length l, radii r₁ and r₂ (r₂ >> r₁), total turns N₁ and N₂ respectively.

Answer: Mutual inductance of two coils is numerically equal to the induced emf produced in one coil when the rate of change of current is unity in the other coil.

Same derivation as Question 40, but with N₁ and N₂ as total turns: \(n_1 = \dfrac{N_1}{l}\), \(n_2 = \dfrac{N_2}{l}\).

\(M = \dfrac{\mu_0 N_1 N_2 \pi r_1^2}{l}\)

Question 42

Define the term 'mutual inductance' between the two coils. Obtain the expression for mutual inductance of a pair of long coaxial solenoids each of length l and radii r₁ and r₂ (r₂ >> r₁). Total number of turns in the two solenoids are N₁ and N₂ respectively.

Answer: Same as Question 41.

Question 43

(i) Define mutual inductance. (ii) A pair of adjacent coils has a mutual inductance of 1.5 H. If the current in one coil changes from 0 to 20 A in 0.5 s, what is the change of flux linkage with the other coil?

Answer: (i) Mutual inductance is the magnetic flux linked with the secondary coil due to a unit current flowing in the primary coil: \(\phi_2 = MI_1\). It is also the induced emf in the secondary when the primary current changes at unit rate: \(\varepsilon_2 = -M\dfrac{dI_1}{dt}\).

(ii) \(M = 1.5\ \text{H}\), \(\Delta I = 20 - 0 = 20\ \text{A}\).

Change in flux linkage: \(\Delta\phi = M \Delta I = 1.5 \times 20 =\) 30 Wb

Question 44

Define mutual inductance between a pair of coils. Derive an expression for the mutual inductance of two long coaxial solenoids of the same length wound one over the other.

Answer: Mutual inductance of two coils is equal to the magnetic flux linked with one coil when a unit current is passed through the other. SI unit: henry.

Consider S₁ and S₂ of same length l, S₂ surrounding S₁. A current I₂ through S₂ sets up flux through S₁: total flux linkage \(= N_1\phi_1\). As shown earlier, this gives \(M = \dfrac{\mu_0 N_1 N_2 A}{l} = \mu_0 n_1 n_2 Al\).

Question 45

Define mutual inductance and write its SI units. Derive an expression for the mutual inductance of two long co-axial solenoids of same length wound one over the other. In an experiment, two coils C₁ and C₂ are placed close to each other. Find out the expression for the emf induced in coil C₁ due to a change in the current through coil C₂.

Answer: Same as Question 44 for the derivation. SI unit: henry (or Wb A⁻¹, or V s A⁻¹).

Emf in C₁: If current I₂ in C₂ changes at rate \(\dfrac{dI_2}{dt}\), the emf induced in C₁ is \(\varepsilon_1 = -M_{12}\dfrac{dI_2}{dt}\).

Question 46

Define mutual inductance of a pair of coils and write on which factors does it depend. A square loop of side 20 cm is initially kept 30 cm away from a region of uniform magnetic field of 0.1 T. It is then moved towards the right with a velocity of 10 cm/s till it goes out of the field. Plot a graph showing the variation of (i) magnetic flux through the loop with time, (ii) induced emf in the loop with time, (iii) induced current in the loop if it has resistance of 0.1 Ω.

Answer: Mutual inductance: It equals the magnetic flux linked with one coil when unit current is passed in the other, or the induced emf set up in one coil when the rate of change of current in the other is unity. SI unit: henry.

Factors: (i) separation between coils, (ii) relative orientation, (iii) number of turns.

Graphs:
(i) Flux: zero until the loop enters the field, then increases linearly as more area enters the field, stays constant while fully inside, then decreases as it exits.
(ii) Emf: zero when outside/fully inside, non-zero constant value (of one sign) while entering, and non-zero constant value (of opposite sign) while exiting.
(iii) Current: same shape as emf divided by R, i.e., ± \(\dfrac{B l v}{R}\).

Question 47

(a) Explain the meaning of the term mutual inductance. Consider two concentric circular coils, one of radius r₁ and the other of radius r₂ (r₁ < r₂) placed coaxially with centres coinciding. Obtain the expression for the mutual inductance of the arrangement. (b) A rectangular coil of area A, having number of turns N, is rotated at f revolutions per second in a uniform magnetic field B, the field being perpendicular to the coil. Prove that the maximum emf induced in the coil is \(2\pi f N B A\).

Answer: (a) Mutual inductance is the flux linkage in one coil due to unit current in the other. For concentric circular coils (r₁ < r₂), the field produced by the outer coil at its centre is approximately uniform over the inner coil's area:

\(B_2 = \dfrac{\mu_0 N_2 I_2}{2r_2}\) (at the centre). Flux through inner coil: \(\phi_1 = B_2 \cdot \pi r_1^2\).

\(N_1\phi_1 = N_1 \cdot \dfrac{\mu_0 N_2 I_2}{2r_2} \cdot \pi r_1^2\). So \(M = \dfrac{\mu_0 N_1 N_2 \pi r_1^2}{2r_2}\).

(b) Flux through the coil at time t: \(\phi = NBA\cos\omega t\) where \(\omega = 2\pi f\). Emf: \(\varepsilon = -\dfrac{d\phi}{dt} = NBA\omega\sin\omega t\).

Maximum emf: \(\varepsilon_{max} = NBA\omega = NBA(2\pi f) =\) \(2\pi f NBA\) — proved.

Faraday's Laws and Moving Coils

Question 48

(i) When primary coil P is moved towards secondary coil S, the galvanometer shows momentary deflection. What can be done to have larger deflection in the galvanometer with the same battery? (ii) State the related law.

Answer: (i) Move coil P faster towards S, so the rate of change of magnetic flux is greater.

(ii) The related law is Faraday's law of electromagnetic induction: the induced emf is proportional to the rate of change of magnetic flux.

Question 49

A coil Q is connected to a low-voltage bulb B and placed near another coil P. Give reasons to explain the following observations: (a) The bulb B lights. (b) Bulb gets dimmer if the coil Q is moved towards the left.

Answer: (a) The bulb lights because of the emf induced in coil Q due to mutual induction between P and Q.

(b) When Q is moved left, the magnetic flux linked with Q decreases (and may even fall to zero at large distance), so the induced emf and current decrease — the bulb gets dimmer.

Question 50

Two identical loops, one of copper and the other of aluminium, are rotated with the same angular speed in the same magnetic field. Compare (i) the induced emf and (ii) the current produced in the two coils. Justify your answer.

Answer: (i) Induced emf \(\varepsilon = NBA\omega\sin\omega t\). Same N, B, A, \(\omega\) → same emf in both.

(ii) Current \(I = \dfrac{\varepsilon}{R}\). Since copper has lower resistivity than aluminium, the copper loop has lower resistance and hence carries larger current.

Question 51

A rectangular loop and a circular loop are moving out of a uniform magnetic field to a field-free region with a constant velocity v. Explain in which loop you expect the induced emf to be constant during the passage out of the field region.

Answer: In the rectangular loop, the induced emf is constant because the rate of change of area (and hence flux) is uniform. In the circular loop, the rate of change of area is not uniform as it exits, so the emf varies.

Question 52

A current is induced in coil C₁ due to the motion of a current-carrying coil C₂. (a) Write any two ways by which a large deflection can be obtained in the galvanometer G. (b) Suggest an alternative device to demonstrate the induced current in place of a galvanometer.

Answer: (a) (i) Use a soft iron rod inside C₂. (ii) Use a powerful battery. (iii) Move C₁ rapidly towards C₂ (or C₂ towards C₁).

(b) Replace the galvanometer with a small torch bulb.

Question 53

A magnet is quickly moved in the direction indicated by the arrow between two coils C₁ and C₂. What will be the direction of induced current in each coil as seen from the magnet? Justify your answer.

Answer: Both coils have induced current clockwise (as seen from the magnet). By Lenz's law, C₁ (magnet receding) becomes S to attract the magnet, and C₂ (magnet approaching) becomes S to repel it — both require the same sense of current.

Question 54

What are eddy currents? Write any two applications of eddy currents.

Answer: Eddy currents are induced currents set up in the bulk volume of a conductor when the magnetic flux through it changes.

Applications: Electric brakes, induction furnaces, dead-beat galvanometers, magnetic braking (any two).

Question 55

State Lenz's law. A metallic rod held horizontally along the east-west direction is allowed to fall under gravity. Will there be an emf induced at its ends? Justify your answer.

Answer: Lenz's law: the polarity of induced emf is such that it tends to produce a current which opposes the change in magnetic flux that induced it.

Yes, an emf will be induced. The rod, the Earth's horizontal component of magnetic field, and the velocity of the rod are all mutually perpendicular, so a motional emf \(\varepsilon = B_H l v\) is induced. The flux linked with the rod changes as it falls.

Question 56

State Faraday's law of electromagnetic induction. A jet plane is travelling towards the west at a speed of 1800 km/h. What is the voltage difference developed between the ends of the wing having a span of 25 m, if the Earth's magnetic field at the location has a magnitude of \(5 \times 10^{-4}\ \text{T}\) and the dip angle is \(30^\circ\)?

Answer: First law: Whenever the magnetic flux linked with a closed circuit changes, an emf (and hence current) is induced in it, which lasts only as long as the change in flux continues.

Second law: The magnitude of the induced emf equals the rate of change of magnetic flux linked with the closed circuit: \(e = -\dfrac{d\phi}{dt}\).

Calculation: \(v = 1800\ \text{km/h} = 500\ \text{m/s}\). Only the vertical component of B contributes (v is horizontal, so v ⊥ B_V).

\(B_V = B\sin\delta = 5 \times 10^{-4} \times \sin 30^\circ = 2.5 \times 10^{-4}\ \text{T}\).

\(\varepsilon = B_V l v = 2.5 \times 10^{-4} \times 25 \times 500 =\) 3.125 V

Motional emf in Rotating Rods

Question 57

A metallic rod of length L is rotated with angular frequency \(\omega\) with one end hinged at the centre and the other end at the circumference of a circular metallic ring of radius L, about an axis passing through the centre and perpendicular to the plane of the ring. A constant uniform magnetic field B parallel to the axis is present everywhere. Deduce the expression for the emf between the centre and the metallic ring.

Answer: Consider a small element \(dr\) of the rod at distance \(r\) from the centre. Its velocity \(v = \omega r\). Emf across it: \(d\varepsilon = Bv\,dr = B\omega r\,dr\).

Total emf: \(\varepsilon = \int_0^L B\omega r\,dr =\) \(\dfrac{1}{2}B\omega L^2\)

Question 58

A metallic rod of length l is rotated at a constant angular speed \(\omega\), normal to a uniform magnetic field B. Derive an expression for the current induced in the rod, if the resistance of the rod is R.

Answer: From Question 57: \(\varepsilon = \dfrac{1}{2}B\omega l^2\).

Current: \(I = \dfrac{\varepsilon}{R} = \dfrac{B\omega l^2}{2R}\)

Question 59

A metallic rod of length l is rotated with a uniform angular speed \(\omega\) with one end hinged at the centre and the other end at the circumference of a circular metallic ring of radius R = l, about an axis passing through the centre and perpendicular to the plane of the ring. A constant and uniform magnetic field B parallel to the axis is present everywhere. Deduce the expression for the emf induced in the rod. If r is the resistance of the rod and the metallic ring has negligible resistance, obtain the expression for the power generated.

Answer: \(\varepsilon = \dfrac{1}{2}B\omega l^2\) (same as Question 57).

Power: \(P = \dfrac{\varepsilon^2}{r} = \dfrac{B^2 \omega^2 l^4}{4r}\)

Question 60

A metallic rod of length 'l' is rotated with a frequency v with one end hinged at the centre and the other end at the circumference of a circular metallic ring of radius r, about an axis passing through the centre and perpendicular to the plane of the ring. A constant uniform magnetic field B parallel to the axis is present everywhere. Using Lorentz force, explain how emf is induced between the centre and the metallic ring and hence obtain the expression for it.

Answer: Free electrons in the rod at distance \(x\) from the pivot move with velocity \(v = \omega x\). The Lorentz force \(F = -evB\) pushes them along the rod. This creates an electric field along the rod, giving emf.

\(d\varepsilon = B v\,dx = B\omega x\,dx\). Integrating: \(\varepsilon = \dfrac{1}{2}B\omega l^2\). Since \(\omega = 2\pi v\): \(\varepsilon = \pi B v l^2\).

Question 61

A metallic rod of length l is moved horizontally with uniform velocity v in a direction perpendicular to its length through a region where a uniform magnetic field acts vertically downward. Derive the expression for the emf induced across the ends of the rod. How does one understand this motional emf by invoking the Lorentz force acting on the free charge carriers of the conductor? Explain.

Answer: Consider a rectangular loop LMNO of length l, with side ON = x at some instant. Area = lx. Flux \(\phi = Blx\).

\(\varepsilon = -\dfrac{d\phi}{dt} = -Blv\). Magnitude: \(\varepsilon = Blv\).

Lorentz force interpretation: Free charge carriers in the rod move with velocity v in the field B. They experience force \(F = qvB\). This force does work \(W = qvBl\) in moving a charge through the rod. Emf \(= \dfrac{W}{q} = vBl\) — same result.

Question 62

A rectangular loop PQMN with movable arm PQ of length 10 cm and resistance 2 Ω is placed in a uniform magnetic field of 0.1 T acting perpendicular to the plane of the loop. The resistances of arms MN, NP, MQ are negligible. Calculate (i) the emf induced in arm PQ and (ii) the current induced in the loop when arm PQ is moved with velocity 20 m/s.

Answer: (i) \(\varepsilon = Blv = 0.1 \times 0.1 \times 20 =\) 0.2 V

(ii) \(I = \dfrac{\varepsilon}{R} = \dfrac{0.2}{2} =\) 0.1 A

Question 63

A rectangular loop PQMN with movable arm PQ of length 20 cm and resistance 5 Ω is placed in a uniform magnetic field of 0.2 T acting perpendicular to the plane of the loop. The resistances of arms MN, NP, MQ are negligible. Calculate (i) the emf induced in arm PQ and (ii) the current induced in the loop when arm PQ is moved with velocity 15 m/s.

Answer: (i) \(\varepsilon = 0.2 \times 0.2 \times 15 =\) 0.6 V

(ii) \(I = \dfrac{0.6}{5} =\) 0.12 A

Question 64

A wheel with 8 metallic spokes, each 50 cm long, is rotated with a speed of 120 rev/min in a plane normal to the horizontal component of the Earth's magnetic field. The Earth's magnetic field at the place is 0.4 G and the angle of dip is 60°. Calculate the emf induced between the axle and the rim of the wheel. How will the value of emf be affected if the number of spokes were increased?

Answer: \(\omega = \dfrac{120 \times 2\pi}{60} = 4\pi\ \text{rad/s}\).

\(B_H = B\cos\delta = 0.4 \times 10^{-4} \times \cos 60^\circ = 0.2 \times 10^{-4}\ \text{T}\) (wait, B given as 0.4 G = \(0.4 \times 10^{-4}\ \text{T}\)).

\(B_H = 0.4 \times 10^{-4} \times 0.5 = 2 \times 10^{-5}\ \text{T}\).

\(\varepsilon = \dfrac{1}{2}B_H \omega l^2 = \dfrac{1}{2} \times 2 \times 10^{-5} \times 4\pi \times (0.5)^2 = 10^{-5} \times 4\pi \times 0.25 = \pi \times 10^{-5} \approx\) \(3.14 \times 10^{-5}\ \text{V}\)

Increasing the number of spokes does not change the emf, as all spokes are in parallel between the axle and the rim.

Question 65

State Lenz's law. A metallic rod held horizontally along the east-west direction, is allowed to fall under gravity. Will there be an emf induced at its ends? Justify your answer.

Answer: Same as Question 55.

Long Answer Questions (5 Marks)

Eddy Currents and Motional EMF

Question 66

(a) What are eddy currents? Write their two applications. (b) A rectangular conducting loop PQSR has arm RS of length l movable, kept in a uniform magnetic field B directed downward perpendicular to the plane of the loop. The arm RS is moved with uniform speed v. Deduce an expression for (i) the emf induced across the arm RS, (ii) the external force required to move the arm, and (iii) the power dissipated as heat.

Answer: (a) Eddy currents are induced currents set up in the bulk volume of a conductor due to changing magnetic flux. Applications: electric brakes, induction furnaces.

(b) (i) Emf induced: \(dA = lv\,dt\), so \(d\phi = B\,lv\,dt\). Thus \(\varepsilon = \dfrac{d\phi}{dt} =\) \(Blv\).

(ii) Current \(I = \dfrac{Blv}{R}\) (R = resistance of the loop). Magnetic force on RS: \(F_m = IlB = \dfrac{B^2 l^2 v}{R}\). The external force must equal this (to maintain constant v): \(F_{ext} = \dfrac{B^2 l^2 v}{R}\).

(iii) Power dissipated: \(P = F_{ext} v =\) \(\dfrac{B^2 l^2 v^2}{R}\) (same as \(I^2 R\)).

Question 67

State Faraday's law of electromagnetic induction. A rectangular conductor PQRS has conductor PQ free to move in a uniform magnetic field B perpendicular to the plane of the paper, extending from x = 0 to x = b (zero for x > b). Only arm PQ has resistance r. When arm PQ is pulled outward from x = 0 to x = 2b and then moved backward to x = 0 with constant speed v, obtain the expressions for flux and induced emf. Sketch their variations with x for 0 ≤ x ≤ 2b.

Answer: Faraday's law: The magnitude of induced emf in a circuit equals the time rate of change of magnetic flux through the circuit.

Flux: While 0 ≤ x ≤ b: \(\phi = Blx\). For x > b, the loop's portion in the field stops growing, so \(\phi = Blb\) (constant).

Emf: For 0 ≤ x < b (moving outward): \(\varepsilon = -Blv\) (constant magnitude). For b < x ≤ 2b (still outward but no field change): \(\varepsilon = 0\).

On the return path, the emf reverses sign while x goes from 2b to b (still no flux change, ε = 0) and then from b to 0 (ε = +Blv).

Sketch: Flux increases linearly to Blb, stays flat, then decreases back to zero. Emf is a square pulse (positive while entering, zero while fully out, negative while exiting on return).

Lenz's Law and Energy Conservation

Question 68

State Lenz's law. Explain, by giving examples, that Lenz's law is a consequence of conservation of energy.

Answer: Lenz's law states that the polarity of induced emf is such that it tends to produce a current which opposes the change in magnetic flux that produced it.

Example: When the N pole of a bar magnet is pushed towards a coil, the induced current makes the coil's near face a N pole (opposing the approach). Work must be done against this repulsion. This mechanical work is converted into the electrical energy of the induced current — this is why the induced current appears only while the magnet is moving. If Lenz's law were reversed, the magnet would be attracted and would accelerate without any external work — violating conservation of energy.

Question 69

(a) Show that in an a.c. circuit containing a pure inductor, the voltage is ahead of current by \(\pi/2\) in phase. (b) A horizontal straight wire of length L extending from east to west is falling with speed v at right angles to the horizontal component of Earth's magnetic field B. (i) Write the expression for the instantaneous value of the emf induced in the wire. (ii) What is the direction of the emf? (iii) Which end of the wire is at the higher potential?

Answer: (a) When a.c. is applied to an ideal inductor, current lags behind the voltage in phase by \(\pi/2\) radian. So we can write the instantaneous values as \(V = V_0\sin\omega t\) and \(I = I_0\sin(\omega t - \pi/2)\). The average power dissipated per cycle in an inductor is zero. Thus, the voltage is ahead of current by \(\pi/2\).

(b) (i) \(\varepsilon = Blv = B_H L v\).
(ii) According to Fleming's right-hand rule, the emf is directed from west to east.
(iii) The eastern end is at higher potential.

Question 70

(a) Starting from the expression for the Lorentz magnetic force acting on the free charge carriers of a conductor moving in a perpendicular magnetic field, obtain the expression for the motional emf induced. (b) Hence deduce the expressions for the power delivered by the source and the power dissipated as Joule heat.

Answer: (a) Free electrons moving with the conductor (velocity v) in field B experience force \(F = evB\). Work done on a charge e to move through length L: \(W = evBL\). Emf = \(W/e =\) \(vBL\).

(b) Induced current \(I = \dfrac{Blv}{R}\). Power delivered by source = mechanical power = \(F_{ext} v\) with \(F_{ext} = IlB\):

\(P_{source} = IlB v = \dfrac{B^2 l^2 v^2}{R}\).

Power dissipated as Joule heat: \(P_{heat} = I^2 R = \dfrac{B^2 l^2 v^2}{R}\). Thus \(P_{source} = P_{heat}\).

Question 71

(a) Describe a simple experiment to show that the polarity of emf induced in a coil is always such that it tends to produce a current which opposes the change of magnetic flux that produces it. (b) The current flowing through an inductor of self-inductance L is continuously increasing. Plot a graph showing the variation of (i) magnetic flux versus the current, (ii) induced emf versus dI/dt, (iii) magnetic potential energy stored versus the current.

Answer: (a) Activity: When the N pole of a bar magnet is pushed towards a coil, the flux linked increases. The induced current makes the nearer face a N pole, opposing the approach. When the N pole is pulled away, flux decreases and the induced current makes the near face a S pole (attracting the magnet). This is Lenz's law in action.

(b) (i) \(\phi\) vs I: straight line through origin (slope = L).
(ii) \(\varepsilon\) vs dI/dt: straight line through origin with negative slope (magnitude proportional to dI/dt).
(iii) U vs I: parabola (U = ½LI²).

Question 72

Define Lenz's law. Use it to predict the polarity of the capacitor in the given situation. A jet plane is travelling towards west at a speed of 1800 km/h. (i) Estimate the voltage difference developed between the ends of the wing having a span of 25 m if the Earth's magnetic field at the location has a magnitude of \(5 \times 10^{-4}\ \text{T}\) and dip angle is \(30^\circ\). (ii) How will the voltage developed be affected if the jet changes its direction from west to north?

Answer: Lenz's law: The direction of induced emf is such that it tends to produce a current which opposes the change in magnetic flux producing it.

Capacitor polarity: Current induced in the coil opposes the approach of the magnet, so the left face becomes N-pole and the right face S-pole. The current is anticlockwise (seen from left). Plate A is positive, plate B is negative.

(i) Using \(v = 500\ \text{m/s}\) and \(B_V = B\sin 30^\circ = 2.5 \times 10^{-4}\ \text{T}\): \(\varepsilon = B_V L v = 2.5 \times 10^{-4} \times 25 \times 500 =\) 3.125 V

(ii) If the jet changes direction from west to north, the voltage developed is not affected (since only the vertical component contributes and the velocity remains horizontal and perpendicular to B_H).

Self-Inductance and Mutual Inductance Derivations

Question 73

(a) Define self-inductance. Write its SI unit. (b) Obtain the expression for the energy stored in an inductor L connected across a source of emf.

Answer: (a) Self-inductance of a coil equals the magnetic flux linked with it when a unit current flows through it. SI unit: henry.

(b) Instantaneous emf in the inductor: \(\varepsilon = -L\dfrac{dI}{dt}\). Work done against this back emf in time dt: \(dW = -\varepsilon I\,dt = LI\,dI\).

Total work done to raise current from 0 to I: \(W = \int_0^I LI\,dI =\) \(\dfrac{1}{2}LI^2\) — stored as magnetic potential energy.

Question 74

Define mutual inductance and write its SI units. Derive an expression for the mutual inductance of two long co-axial solenoids of same length wound one over the other.

Answer: Same as Question 44. \(M = \mu_0 n_1 n_2 Al\).

Question 75

(a) State Lenz's law. Give one example to illustrate this law. "Lenz's law is a consequence of the principle of conservation of energy." Justify this statement. (b) Deduce an expression for the mutual inductance of two long coaxial solenoids but having different radii and different number of turns.

Answer: (a) Same as Question 68 for Lenz's law and the energy conservation justification.

(b) As in Question 41, \(M = \dfrac{\mu_0 N_1 N_2 A_1}{l}\) where \(A_1 = \pi r_1^2\) (inner solenoid's area, since B from outer solenoid is effectively only within the inner solenoid for the flux linkage).

Question 76

State Faraday's law of electromagnetic induction. Figure shows a rectangular conductor PQRS in which the conductor PQ is free to move in a uniform magnetic field B perpendicular to the plane of the paper. The field extends from x = 0 to x = b and is zero for x > b. Assume that only the arm PQ possesses resistance r. When arm PQ is pulled outward from x = 0 to x = 2b and is then moved backward to x = 0 with constant speed v, obtain the expressions for the flux and induced emf. Sketch the variations of these quantities with distance 0 ≤ x ≤ 2b.

Answer: Same as Question 67.

Mutual Inductance of Coils and Solenoids

Question 77

(a) Define mutual inductance and write its SI units. (b) Derive an expression for the mutual inductance of two long co-axial solenoids of same length wound one over the other. (c) In an experiment, two coils C₁ and C₂ are placed close to each other. Find out the expression for the emf induced in coil C₁ due to a change in the current through coil C₂.

Answer: Same as Question 45.

Question 78

Define mutual inductance between a pair of coils. Derive an expression for the mutual inductance of two long coaxial solenoids of same length wound one over the other.

Answer: Same as Question 44.

Question 79

(a) A metallic rod of length l and resistance R is rotated with a frequency v, with one end hinged at the centre and the other end at the circumference of a circular metallic ring of radius l, about an axis passing through the centre and perpendicular to the plane of the ring. A constant and uniform magnetic field B parallel to the axis is present everywhere. (i) Derive the expression for the induced emf and the current in the rod. (ii) Due to the presence of the current in the rod and of the magnetic field, find the expression for the magnitude and direction of the force acting on this rod. (iii) Hence obtain the expression for the power required to rotate the rod.

Answer: (i) \(\varepsilon = \dfrac{1}{2}B\omega l^2 = \pi B v l^2\). \(I = \dfrac{\varepsilon}{R} = \dfrac{\pi B v l^2}{R}\).

(ii) Force on the rod \(F = IlB = \dfrac{\pi B^2 v l^3}{R}\), directed opposite to the velocity (as per Lenz's law) — this opposes the rotation.

(iii) Power required to rotate = \(F \cdot v_{avg}\) or \(\dfrac{\varepsilon^2}{R}\) = \(\dfrac{\pi^2 B^2 v^2 l^4}{R}\).

Question 80

(a) A metallic rod of length l is moved perpendicular to its length with velocity v in a magnetic field \(\vec{B}\) acting perpendicular to the plane in which the rod moves. Derive the expression for the induced emf. (b) A wheel with 15 metallic spokes, each 60 cm long, is rotated at 360 rev/min in a plane normal to the horizontal component of Earth's magnetic field. The angle of dip at that place is 60°. If the emf induced between the rim of the wheel and the axle is 400 mV, calculate the horizontal component of Earth's magnetic field at the place. How will the induced emf change if the number of spokes is increased?

Answer: (a) Same derivation as Question 61: \(\varepsilon = Blv\).

(b) Given: \(\omega = \dfrac{360 \times 2\pi}{60} = 12\pi\ \text{rad/s}\), l = 0.6 m, \(\varepsilon = 0.4\ \text{V}\).

\(\varepsilon = \dfrac{1}{2}B_H \omega l^2 \Rightarrow 0.4 = \dfrac{1}{2} B_H \times 12\pi \times 0.36\).

\(0.4 = 2.16\pi B_H \Rightarrow B_H = \dfrac{0.4}{6.79} \approx\) \(5.89 \times 10^{-2}\ \text{T}\)

Increasing the number of spokes does not change the induced emf (all spokes are in parallel).

Key Formulas to Remember

Concept	Formula
Faraday's law	\(\varepsilon = -\dfrac{d\phi}{dt}\)
Motional emf (straight rod)	\(\varepsilon = Blv\)
Motional emf (rotating rod)	\(\varepsilon = \dfrac{1}{2}B\omega l^2\)
Self-inductance	\(\phi = LI\), \(\varepsilon = -L\dfrac{dI}{dt}\)
L of a solenoid	\(L = \mu_0 n^2 Al = \dfrac{\mu_0 N^2 A}{l}\)
Mutual inductance	\(\phi_2 = MI_1\), \(\varepsilon_2 = -M\dfrac{dI_1}{dt}\)
M of coaxial solenoids	\(M = \mu_0 n_1 n_2 Al\)
Magnetic energy stored in L	\(U = \dfrac{1}{2}LI^2\)
Energy density in B	\(u = \dfrac{B^2}{2\mu_0}\)
Force on movable arm	\(F = IlB = \dfrac{B^2 l^2 v}{R}\)
Power dissipated in moving arm	\(P = \dfrac{B^2 l^2 v^2}{R}\)
Maximum emf in rotating coil	\(\varepsilon_{max} = NBA\omega = 2\pi f NBA\)

FAQs

Are these questions enough for CBSE Class 12 Physics Chapter 6?

Yes. These questions cover every major topic — Faraday's laws, Lenz's law, self-inductance, mutual inductance, eddy currents, motional emf, energy stored in an inductor — that CBSE has asked in previous board exams. Combine them with NCERT exercises and you'll be well prepared.

Which topics are most important for board exams?

Faraday's and Lenz's laws, self-inductance of a solenoid and mutual inductance of coaxial solenoids (with derivations), motional emf for straight and rotating rods, eddy currents and their applications, energy stored in an inductor, and the consequence of Lenz's law via energy conservation are consistently high-weightage topics.

How should I prepare for numericals in this chapter?

Practise motional emf (Blv and ½Bωl²), induced emf calculations from changing flux, self-inductance of a solenoid, mutual inductance problems, energy stored in inductors, and problems involving eddy currents and energy conservation. Focus on understanding the direction of the induced current using Lenz's law and the right-hand rule.