NCERT Solutions for Class 12 Physics Chapter 6: Electromagnetic Induction

These Class 12 Physics Chapter 6 solutions cover Electromagnetic Induction for the 2026–27 session. Every NCERT Exercise (6.1–6.8) is reproduced word-for-word and solved step by step, with each numerical worked out fully and its result verified against the official NCERT answer key, complete with correct SI units.

Class: 12 Subject: Physics Chapter: 6 Name: Electromagnetic Induction Exercises: 6.1–6.8 Session: 2026–27
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Class 12 Physics Chapter 6 – Overview

Chapter 6, Electromagnetic Induction, shows that electricity and magnetism are two faces of the same phenomenon. While Oersted and Ampere proved that moving charges produce a magnetic field, the experiments of Faraday and Henry (around 1830) proved the converse — a changing magnetic flux through a circuit induces an emf in it. The chapter develops the idea of magnetic flux, states Faraday’s law (induced emf equals the rate of change of flux) and Lenz’s law (the induced current opposes the change that causes it, in keeping with conservation of energy), and then applies them to motional emf, mutual and self-inductance, energy stored in an inductor, and finally the AC generator. These principles power every generator and transformer in the modern world.

Key Concepts & Definitions

Magnetic flux (ΦB): the product of the magnetic field and the area it passes through, ΦB = B·A = BA cosθ, where θ is the angle between B and the area vector A. It is a scalar; SI unit weber (Wb) = T m2.

Faraday’s law of induction: the magnitude of the induced emf equals the time rate of change of magnetic flux; for a coil of N turns, ε = −N (dΦB/dt).

Lenz’s law: the induced current flows in a direction such that it opposes the very change in flux that produces it — this is what the negative sign in Faraday’s law expresses and it follows from the law of conservation of energy.

Motional emf: the emf ε = Blv induced across a rod of length l moving with speed v perpendicular to a field B; it can be derived both from flux change and from the Lorentz force on free charges.

Mutual inductance (M): links the emf induced in one coil to the rate of change of current in a neighbouring coil; ε1 = −M (dI2/dt), with M12 = M21.

Self-inductance (L): a coil’s opposition to a change in its own current (electrical inertia); ε = −L (dI/dt). SI unit of inductance is the henry (H).

AC generator: rotating a coil in a magnetic field changes the flux through it and produces a sinusoidal emf ε = ε0 sin ωt, converting mechanical energy into electrical energy.

Important Formulas

Magnetic flux: ΦB = BA cosθ  (Wb)

Faraday’s law: ε = −N (dΦB/dt)

Motional emf (straight rod): ε = Blv

Rotating rod / spoke about one end: ε = ½ B ω l2

Self-inductance: NΦB = LI  •  ε = −L (dI/dt)  •  solenoid L = μrμ0n2Al

Mutual inductance: N1Φ1 = MI2  •  ε1 = −M (dI2/dt)  •  co-axial solenoids M = μ0n1n2πr12l

Energy stored in an inductor: W = ½LI2  •  energy density uB = B2/(2μ0)

AC generator emf: ε = NBAω sinωt = ε0 sinωt, with ε0 = NBAω = NBA(2πν)

NCERT Exercises 6.1–6.8 — Solutions

Questions are reproduced verbatim from the NCERT textbook. Figure-based questions are answered from the data given in the figures, in words, since textbook images are not reproduced here.

6.1 Predict the direction of induced current in the situations described by the following Figs. 6.15(a) to (f).

ANSWER By Lenz’s law, the induced current always flows so as to oppose the change in magnetic flux. Tracing the loop letters as labelled in Fig. 6.15: (a) Along qrpq — the flux into the page increases, so the induced current is set up to oppose this increase. (b) Along prq, and along yzx — the induced currents oppose the respective flux changes in the two coils. (c) Along yzx. (d) Along zyx. (e) Along xry. (f) No induced current, since the magnetic field lines lie in the plane of the loop — the flux through the loop is zero and is not changing.

6.2 Use Lenz’s law to determine the direction of induced current in the situations described by Fig. 6.16: (a) A wire of irregular shape turning into a circular shape; (b) A circular loop being deformed into a narrow straight wire.

ANSWER (a) As an irregular shape becomes circular, the area enclosed increases, so the outward flux through the loop increases. By Lenz’s law the induced current opposes this increase, flowing along adcb (a′d′c′b′) so as to set up a magnetic field opposing the rising flux. (b) As a circular loop is squeezed into a narrow straight wire, the enclosed area decreases, so the flux decreases. The induced current opposes this decrease, flowing along a′d′c′b′ in the sense that maintains (tries to keep up) the flux through the loop. In both cases the rule is the same: increasing flux → current opposes (reduces) it; decreasing flux → current supports (maintains) it.

6.3 A long solenoid with 15 turns per cm has a small loop of area 2.0 cm2 placed inside the solenoid normal to its axis. If the current carried by the solenoid changes steadily from 2.0 A to 4.0 A in 0.1 s, what is the induced emf in the loop while the current is changing?

SOLUTION Given: n = 15 turns/cm = 1500 turns/m; A = 2.0 cm2 = 2.0 × 10−4 m2; current change ΔI = (4.0 − 2.0) = 2.0 A in Δt = 0.1 s. Field inside the solenoid: B = μ0nI, so the flux through the small loop is Φ = BA = μ0nIA. Induced emf: ε = dΦ/dt = μ0nA (dI/dt). dI/dt = 2.0 / 0.1 = 20 A s−1. ε = (4π × 10−7) × 1500 × (2.0 × 10−4) × 20 ε = (4π × 10−7) × 6.0 = 7.5 × 10−6 V (≈ 7.5 μV). (Matches NCERT key.)

6.4 A rectangular wire loop of sides 8 cm and 2 cm with a small cut is moving out of a region of uniform magnetic field of magnitude 0.3 T directed normal to the loop. What is the emf developed across the cut if the velocity of the loop is 1 cm s−1 in a direction normal to the (a) longer side, (b) shorter side of the loop? For how long does the induced voltage last in each case?

SOLUTION Given: sides = 8 cm and 2 cm; B = 0.3 T; v = 1 cm s−1 = 0.01 m s−1. Motional emf ε = Blv, where l is the side perpendicular to the motion (the side cutting field lines). (a) Moving normal to the longer side (8 cm): the effective length is l = 8 cm = 0.08 m. ε = Blv = 0.3 × 0.08 × 0.01 = 2.4 × 10−4 V. The loop must travel the width of the field equal to the other side (2 cm = 0.02 m): time = distance/v = 0.02 / 0.01 = 2 s. (b) Moving normal to the shorter side (2 cm): the effective length is l = 2 cm = 0.02 m. ε = Blv = 0.3 × 0.02 × 0.01 = 0.6 × 10−4 V (6 × 10−5 V). Distance to clear the field = 8 cm = 0.08 m: time = 0.08 / 0.01 = 8 s. (Matches NCERT key.)

6.5 A 1.0 m long metallic rod is rotated with an angular frequency of 400 rad s−1 about an axis normal to the rod passing through its one end. The other end of the rod is in contact with a circular metallic ring. A constant and uniform magnetic field of 0.5 T parallel to the axis exists everywhere. Calculate the emf developed between the centre and the ring.

SOLUTION Given: l = 1.0 m; ω = 400 rad s−1; B = 0.5 T. For a rod rotating about one end in a field parallel to the axis, ε = ½ B ω l2. ε = ½ × 0.5 × 400 × (1.0)2 = ½ × 200 = 100 V. (Matches NCERT key.)

6.6 A horizontal straight wire 10 m long extending from east to west is falling with a speed of 5.0 m s−1, at right angles to the horizontal component of the earth’s magnetic field, 0.30 × 10−4 Wb m−2. (a) What is the instantaneous value of the emf induced in the wire? (b) What is the direction of the emf? (c) Which end of the wire is at the higher electrical potential?

SOLUTION Given: l = 10 m; v = 5.0 m s−1; B = horizontal component = 0.30 × 10−4 Wb m−2. The wire, its velocity (downward) and B are mutually perpendicular, so ε = Blv. (a) ε = Blv = (0.30 × 10−4) × 10 × 5.0 = 1.5 × 10−3 V (1.5 mV). (b) Using the Lorentz force F = q(v × B) on the free electrons (v downward, B pointing north), the force drives positive charge from the western end to the eastern end. The emf is directed from West to East. (c) Since positive charge accumulates at the eastern end, the eastern end is at the higher electrical potential. (Matches NCERT key.)

6.7 Current in a circuit falls from 5.0 A to 0.0 A in 0.1 s. If an average emf of 200 V induced, give an estimate of the self-inductance of the circuit.

SOLUTION Given: ΔI = (5.0 − 0.0) = 5.0 A; Δt = 0.1 s; average emf ε = 200 V. From ε = L (dI/dt), the self-inductance is L = ε / (dI/dt) = ε Δt / ΔI. L = (200 × 0.1) / 5.0 = 20 / 5.0 = 4 H. (Matches NCERT key.)

6.8 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?

SOLUTION Given: M = 1.5 H; current change ΔI = (20 − 0) = 20 A in Δt = 0.5 s. The flux linkage with the second coil is N2Φ2 = M I1, so the change in flux linkage is Δ(NΦ) = M ΔI. Δ(NΦ) = 1.5 × 20 = 30 Wb (weber-turns). (Matches NCERT key.) Aside — the average induced emf, if needed, would be MΔI/Δt = 30/0.5 = 60 V.

Extra Practice Questions

Short Answer Type Questions

Q1. Define magnetic flux and give its SI unit.

ANSWERMagnetic flux through a surface is the total number of magnetic field lines passing through it, ΦB = BA cosθ. It is a scalar quantity and its SI unit is the weber (Wb), where 1 Wb = 1 T m2.

Q2. State Lenz’s law and explain how it is consistent with the conservation of energy.

ANSWERLenz’s law states that the induced current opposes the change in flux that causes it. If it did the opposite, the induced current would aid the magnet’s motion, the magnet would accelerate without any work being done, and energy would appear from nowhere. Because the current opposes the motion, an external agent must do work to keep the flux changing, and this work appears as electrical (Joule) energy — exactly conserving energy.

Q3. A 200-turn coil has its flux changing at 0.02 Wb s−1. Find the induced emf.

ANSWERε = N(dΦ/dt) = 200 × 0.02 = 4 V.

Q4. Why is the self-induced emf called “back emf”?

ANSWERWhen the current in a coil changes, the self-induced emf always opposes that change — it acts against the growth or decay of current, just like inertia opposes a change in motion. Because it works “back” against the applied source, it is called the back emf.

Q5. On what factors does the self-inductance of a long solenoid depend?

ANSWERFrom L = μrμ0n2Al, it depends on the geometry — the number of turns per unit length n, the cross-sectional area A and the length l — and on the relative permeability μr of the core material. It does not depend on the current.

Long Answer Type Questions

Q1. Derive the expression for the motional emf induced in a rod of length l moving with velocity v perpendicular to a uniform field B, using the Lorentz force.

ANSWERConsider a free charge q in a rod PQ moving with speed v perpendicular to field B. The magnetic (Lorentz) force on it is F = qvB, directed along the rod. This force pushes charges to one end of length l, doing work W = (qvB) × l in moving a charge from one end to the other. Since emf is the work done per unit charge, ε = W/q = (qvBl)/q = Blv. Thus a rod moving in a field acts like a source of emf ε = Blv, the same result obtained from the rate of change of flux Φ = Blx (since dx/dt = v). This shows motional emf can be understood both through Faraday’s law and through the Lorentz force.

Q2. Describe the principle, construction and working of an AC generator, and write the expression for the emf produced.

ANSWERPrinciple: electromagnetic induction — rotating a coil in a magnetic field changes the flux through it and induces an emf. Construction: a multi-turn coil (armature) mounted on a rotor shaft turns in a uniform magnetic field; its ends connect to the external circuit through slip rings and brushes. Working: as the coil rotates at angular speed ω, the angle between B and the area vector is θ = ωt, so the flux is ΦB = BA cosωt. By Faraday’s law, ε = −N(dΦB/dt) = NBAω sinωt = ε0 sinωt, where the peak emf ε0 = NBAω. The emf alternates in sign every half rotation, producing an alternating current. Mechanical energy (from water, steam or other sources) is thus converted to electrical energy.

Q3. Two long co-axial solenoids carry currents. Derive the expression for the mutual inductance per unit length of the arrangement.

ANSWERLet the inner solenoid S1 have radius r1 and n1 turns per unit length, and the outer solenoid S2 have n2 turns per unit length, each of length l. When current I2 flows in S2, it produces a near-uniform field B2 = μ0n2I2 over the inner region. The flux linkage with S1 is N1Φ1 = (n1l)(πr12)(μ0n2I2) = μ0n1n2πr12l I2. Since N1Φ1 = M12I2, we get M12 = μ0n1n2πr12l. By the reciprocity theorem M12 = M21 = M, so M = μ0n1n2πr12l (replace μ0 by μrμ0 for a magnetic core). The mutual inductance depends only on geometry and the core material.

MCQs & Assertion–Reason

1. The SI unit of magnetic flux is the:

(a) tesla    (b) weber    (c) henry    (d) volt

2. Faraday’s law states that the induced emf is proportional to the:

(a) magnetic flux    (b) magnetic field    (c) rate of change of magnetic flux    (d) area of the coil

3. Lenz’s law is a direct consequence of the conservation of:

(a) charge    (b) momentum    (c) energy    (d) mass

4. The motional emf across a rod of length l moving with speed v perpendicular to a field B is:

(a) Bvl2    (b) Blv    (c) ½Blv    (d) B/lv

5. The SI unit of inductance is the:

(a) weber    (b) tesla    (c) henry    (d) farad

6. The energy stored in an inductor carrying current I is:

(a) LI    (b) ½LI2    (c) LI2    (d) ½LI

7. A rod of length 1 m rotates at ω = 400 rad s−1 about one end in a field of 0.5 T parallel to the axis. The emf between centre and rim is:

(a) 50 V    (b) 100 V    (c) 200 V    (d) 400 V

8. For two coils, the mutual inductances always satisfy:

(a) M12 > M21    (b) M12 < M21    (c) M12 = M21    (d) M12 = −M21

9. The peak emf of an AC generator with N turns, area A, field B and angular speed ω is:

(a) NBA    (b) NBAω    (c) NBA/ω    (d) BAω

10. The magnetic energy density in a region where the field is B equals:

(a) B2/μ0    (b) B2/(2μ0)    (c) μ0B2/2    (d) 2B2/μ0

Answer key: 1-(b), 2-(c), 3-(c), 4-(b), 5-(c), 6-(b), 7-(b), 8-(c), 9-(b), 10-(b).

For each Assertion–Reason question, choose: (A) Both true and the Reason correctly explains the Assertion; (B) Both true but the Reason is not the correct explanation; (C) Assertion true, Reason false; (D) Assertion false, Reason true.

A-R 1. Assertion: An emf is induced in a coil only when the magnetic flux through it changes with time.

Reason: By Faraday’s law, the induced emf equals the time rate of change of magnetic flux.

A-R 2. Assertion: The induced current opposes the change in flux that produces it.

Reason: This is required by the law of conservation of energy and is the statement of Lenz’s law.

A-R 3. Assertion: A bar magnet held stationary near a coil induces a steady current in the coil.

Reason: A stationary magnet produces a constant flux, and a constant flux induces no emf.

A-R 4. Assertion: Self-inductance is called the electrical analogue of inertia.

Reason: Self-inductance opposes any change in the current through the coil, just as inertia opposes a change in motion.

A-R 5. Assertion: The mutual inductance of two coils depends on the current flowing through them.

Reason: Mutual inductance depends only on the geometry of the coils, their separation and orientation, and the medium.

Answer key: 1-(A), 2-(A), 3-(D), 4-(A), 5-(D).

Common Mistakes & Exam Tips

Watch out for these

  • Confusing flux with field — emf depends on the rate of change of flux, not on the flux or field value itself; a large but steady field induces no emf.
  • In motional emf ε = Blv, using the wrong length: l is the side perpendicular to the velocity, not the side along the motion (the cause of errors in Q6.4).
  • Forgetting the ½ in the rotating-rod formula ε = ½Bωl2, or using v at the tip instead of integrating along the rod.
  • Mixing up Φ = BA cosθ with θ as the angle from the plane — θ is between B and the area vector (normal), not the surface.
  • Dropping units when converting cm to m or cm2 to m2, which shifts the power of ten in the answer.
  • Treating the change of flux linkage (MΔI, in Wb) as an emf (MΔI/Δt, in V) — they differ by the time factor (Q6.8).

How to score full marks in this chapter

Always start a numerical by listing the given data in SI units. State the formula, substitute with units, and box the final answer with its correct unit (V, Wb, H). For direction questions, name Lenz’s law explicitly and say whether the flux is increasing or decreasing before deciding the current direction. Memorise the four core formulas — Φ = BA cosθ, ε = −N(dΦ/dt), ε = Blv and L = μrμ0n2Al — and the standard derivations (motional emf, AC generator, mutual inductance of co-axial solenoids), which are frequent board-exam questions.

Frequently Asked Questions

What is Class 12 Physics Chapter 6 Electromagnetic Induction about?

Chapter 6 explains how a changing magnetic flux induces an emf and current in a circuit. It covers magnetic flux, Faraday’s law, Lenz’s law, motional emf, mutual and self-inductance, energy stored in an inductor and the AC generator — the basis of all generators and transformers.

How many exercises are there in Chapter 6 and are all solved here?

There are 8 NCERT exercises (6.1 to 6.8). All eight are reproduced word-for-word and solved step by step on this page, with every numerical worked out in SI units and verified against the official NCERT answer key.

What is the difference between Faraday’s law and Lenz’s law?

Faraday’s law gives the magnitude of the induced emf (equal to the rate of change of flux), while Lenz’s law gives its direction (the induced current opposes the change in flux). The negative sign in ε = −N(dΦ/dt) combines both, and Lenz’s law follows from conservation of energy.

Are these Class 12 Physics Chapter 6 solutions free?

Yes. All solutions are free and follow the official NCERT Physics textbook for session 2026–27.

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