PhysicsCBSE Class 12 · Chapter 6

Electromagnetic Induction Class 12: Notes, Derivations and Numericals

Updated 14 October 2026 · Fact-checked

Electromagnetic Induction is Chapter 6 of CBSE Class 12 Physics. It explains how a changing magnetic flux induces an emf (Faraday's law), how Lenz's law gives the direction of the induced current, motional emf in a moving rod, self-inductance and mutual inductance with the solenoid results, and the working of an AC generator.

What Electromagnetic Induction is about

Chapter 4 showed that currents produce magnetic fields. Faraday and Henry asked the reverse question and found the answer in a series of experiments: a changing magnetic field produces a current. Move a magnet towards a coil, change the current in a neighbouring coil, or move a wire across a field, and an emf appears; hold everything still and it vanishes. The quantity that must change is the magnetic flux $Φ_{B} = B \cdot A$ .

Faraday's law says the induced emf equals the rate of change of flux linkage, and Lenz's law (the minus sign) says the induced current always opposes the change that causes it, which is a statement of energy conservation. You then derive the motional emf $B l v$ of a rod moving across a field, and meet inductance: a coil's own changing current induces an emf in it (self-inductance L), and in a neighbouring coil (mutual inductance M). The chapter ends with the AC generator, which turns mechanical rotation into an alternating emf $N B A ω sin ω t$ (the CBSE syllabus lists the generator with the next chapter, Alternating Current).

The quantitative energy treatment and eddy currents are not in the 2026-27 syllabus. In our question bank this chapter appears in all eight board papers from 2023 to 2026, mostly as 1-mark MCQs and assertion-reason items and 3-mark questions on mutual inductance, Faraday's and Lenz's laws and motional emf.

Weightage: This chapter is part of Units III and IV, Magnetic Effects of Current and Magnetism; Electromagnetic Induction and Alternating Currents, which carry 17 of the 70 theory marks together with Moving Charges and Magnetism, Magnetism and Matter and Alternating Current.Source: CBSE Physics syllabus 2026-27, Units III and IV Magnetic Effects of Current and Magnetism; Electromagnetic Induction and Alternating Currents

In real board papers: asked in 8 of the 8 board papers we hold (2023 to 2026), on average 5 marks per paper. See the questions

Key concepts

Magnetic flux

NCERT section 6.3

The magnetic flux through a plane area A in a uniform field B is

$Φ_{B} = B \cdot A = B A cos θ$

where θ is the angle between $B$ and the normal to the area. For a non-uniform field or curved surface, add $B \cdot d A$ over the surface. Flux is a scalar; its SI unit is the weber ( $1 W b = 1 T m^{2}$ ). For a coil of N turns, the flux linkage is $N Φ_{B}$ .

Flux can be changed in three ways: by changing B, by changing the area A inside the field, or by changing the angle θ (rotating the coil).

Worked example

A coil of 50 turns, each of area $0.02 m^{2}$ , is in a uniform field of 0.4 T. Find the flux linkage when the plane of the coil is (a) perpendicular to the field, (b) at $3 0^{\circ}$ to the field, (c) parallel to the field. [2 marks]

Try it first, then show the solutionSolution

(a) Plane perpendicular to B means the normal is along B, θ = 0: $N Φ = 50 \times 0.4 \times 0.02 = 0.4 W b$
(b) Plane at 30° to B means the normal is at 60°: $N Φ = 0.4 \times cos 6 0^{\circ} = 0.2 W b$
(c) Plane parallel to B means θ = 90°: flux linkage is zero

Answer: (a) 0.4 Wb (b) 0.2 Wb (c) zero

Faraday's law of induction

NCERT section 6.2 and 6.4

Faraday's experiments: (1) a magnet moved towards or away from a coil gives a current, larger for faster motion; (2) a current-carrying coil moved relative to another coil does the same; (3) switching the current in one coil on or off induces a brief current in a nearby stationary coil. The common factor is a change in magnetic flux.

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

$ε = - \frac{d Φ _{B}}{d t}$ , and for N turns $ε = - N \frac{d Φ _{B}}{d t}$

The emf depends on how fast the flux changes, not on how large the flux is. If the circuit has resistance R, the induced current is $I = \frac{ε}{R}$ , and the total charge that flows during a change $ΔΦ$ is $Q = \frac{N ΔΦ}{R}$ , independent of how quickly the change happens.

Worked example

A coil of 100 turns and area $0.01 m^{2}$ (resistance 10 Ω) is perpendicular to a field of 0.5 T. The field is switched off uniformly in 0.1 s. Find (a) the induced emf, (b) the current, (c) the charge that flows. [3 marks]

Try it first, then show the solutionSolution

Change in flux per turn: $ΔΦ = 0.5 \times 0.01 - 0 = 5 \times 1 0^{- 3} W b$
(a) $ε = N \frac{ΔΦ}{Δ t} = 100 \times \frac{5 \times 1 0 ^{- 3}}{0.1} = 5 V$
(b) $I = \frac{5}{10} = 0.5 A$
(c) $Q = I Δ t = 0.5 \times 0.1 = 0.05 C$ (same as $\frac{N ΔΦ}{R} = \frac{0.5}{10}$ )

Answer: (a) 5 V (b) 0.5 A (c) 0.05 C

Lenz's law and conservation of energy

NCERT section 6.5

Lenz's law: the polarity of the induced emf is such that it tends to produce a current which opposes the change in magnetic flux that produced it. This is the meaning of the minus sign in Faraday's law.

Example: push the north pole of a magnet towards a coil. The flux through the coil increases, so the induced current makes the near face of the coil a north pole, which repels the magnet. Pull the magnet away and the near face becomes a south pole, which attracts it back. Either way, you must do work against the opposition.

Why it must be so (energy). If the induced current helped the change instead, the magnet would be pulled in faster, inducing a bigger current, pulling it faster still: energy would be created from nothing. Lenz's law is therefore a consequence of the conservation of energy; the electrical energy of the induced current comes from the mechanical work you do. (The quantitative energy study is not in the 2026-27 syllabus.)

Worked example

A bar magnet is held vertically with its south pole downward and dropped through a horizontal circular coil connected to a galvanometer. Looking down on the coil from above, give the direction of the induced current (a) as the magnet approaches the coil, (b) as it leaves the coil below. [2 marks]

Try it first, then show the solutionSolution

(a) The south pole approaches, so the upper face of the coil must become a south pole to repel it. A face is a south pole when, seen from that side, the current flows clockwise.
So, seen from above, the current is clockwise.
(b) As the magnet leaves below, its north pole (upper end) moves away downward. The lower face must attract it, so the lower face becomes a south pole: seen from below clockwise, which is anticlockwise seen from above.

Answer: (a) Clockwise seen from above (b) anticlockwise seen from above

Motional electromotive force

NCERT section 6.6

A straight rod PQ of length l slides with velocity v on two parallel rails in a uniform field B perpendicular to the plane of the rails.

From the flux. If the enclosed part of the circuit has length x, the flux is $Φ = B l x$ . As the rod moves, x changes at rate $\frac{d x}{d t} = v$ (the area shrinks or grows), so

$ε = B l v$

From the Lorentz force. Each free charge q in the rod moves with the rod at speed v and feels $q v B$ along the rod. Moving charge q from one end to the other, this force does work $q v B l$ , so the emf (work per unit charge) is $B l v$ . This shows the source of the motional emf is the Lorentz force on the charges in the moving conductor.

For a rod of length l rotating about one end with angular speed ω in a field perpendicular to its plane, the speed at distance r is ωr, and adding $B (ω r) d r$ from 0 to l gives $ε = \frac{1}{2} B ω l^{2}$ .

Worked example

A rod 0.5 m long moves at $4 m s^{- 1}$ at right angles to a uniform field of 0.2 T, sliding on rails joined by a 2 Ω resistor (rod and rails have negligible resistance). Find the emf and the current. [2 marks]

Try it first, then show the solutionSolution

$ε = B l v = 0.2 \times 0.5 \times 4 = 0.4 V$
$I = \frac{ε}{R} = \frac{0.4}{2} = 0.2 A$

Answer: 0.4 V; 0.2 A

Mutual inductance

NCERT section 6.7.1

When the current $I_{1}$ in coil 1 changes, the flux it sends through coil 2 changes and an emf is induced in coil 2. The flux linkage with coil 2 is proportional to $I_{1}$ :

$N_{2} Φ_{2} = M I_{1}$ , so $ε_{2} = - M \frac{d I _{1}}{d t}$

M is the mutual inductance; unit henry ( $1 H = 1 W b A^{- 1} = 1 V s A^{- 1}$ ). It depends on the shapes, sizes, numbers of turns, separation and relative orientation of the coils and on the medium (a soft iron core raises it), not on the currents. And $M_{12} = M_{21}$ .

Two long coaxial solenoids of length l (inner radius $r_{1}$ with $n_{1}$ turns per metre, outer with $n_{2}$ turns per metre). The inner solenoid's current $I_{1}$ gives a field $μ_{0} n_{1} I_{1}$ inside it, and only the area $π r_{1}^{2}$ carries this field. Flux linkage with the outer: $(n_{2} l) (π r_{1}^{2}) (μ_{0} n_{1} I_{1})$ , so

$M = μ_{0} n_{1} n_{2} π r_{1}^{2} l$

With a core of relative permeability $μ_{r}$ , $M = μ_{r} μ_{0} n_{1} n_{2} π r_{1}^{2} l$ .

Worked example

The mutual inductance of two coils is 0.2 H. The current in the first coil rises steadily from 0 to 10 A in 0.25 s. Find the emf induced in the second coil. [1 mark]

Try it first, then show the solutionSolution

$\frac{d I _{1}}{d t} = \frac{10}{0.25} = 40 A s^{- 1}$
$∣ ε_{2} ∣ = M \frac{d I _{1}}{d t} = 0.2 \times 40 = 8 V$

Answer: 8 V

Self-inductance

NCERT section 6.7.2

A changing current in a coil changes the flux through the coil itself and induces an emf in it, opposing the change (a back emf). The flux linkage is proportional to the current:

$N Φ = L I$ , so $ε = - L \frac{d I}{d t}$

L is the self-inductance (unit henry). It plays the role of inertia for current: it opposes both a rise and a fall of current.

Long solenoid (n turns per metre, length l, cross-section A): $B = μ_{0} n I$ , flux linkage $N Φ = (n l) (μ_{0} n I) A$ , so

$L = μ_{0} n^{2} A l = \frac{μ _{0} N ^{2} A}{l}$

With a core, multiply by $μ_{r}$ . L grows as the square of the number of turns.

Energy stored. To build up the current from 0 to I, work must be done against the back emf: $W = \int_{0}^{I} L I d I = \frac{1}{2} L I^{2}$ . This energy is stored in the magnetic field of the coil, just as $\frac{1}{2} C V^{2}$ is stored in the electric field of a capacitor.

Worked example

An air-cored solenoid has 500 turns, length 0.5 m and radius 1 cm. (a) Find its self-inductance. (b) Find the average back emf when its current rises from 0 to 2 A in 0.01 s. ( $μ_{0} = 4 π \times 1 0^{- 7} T m A^{- 1}$ , $π^{2} \approx 9.87$ ) [3 marks]

Try it first, then show the solutionSolution

$A = π r^{2} = π \times 1 0^{- 4} m^{2}$
(a) $L = \frac{μ _{0} N ^{2} A}{l} = \frac{4 π \times 1 0 ^{- 7} \times 250 000 \times π \times 1 0 ^{- 4}}{0.5} = \frac{4 π ^{2} \times 2.5 \times 1 0 ^{- 6}}{0.5} \approx 1.97 \times 1 0^{- 4} H$
(b) $∣ ε ∣ = L \frac{Δ I}{Δ t} = 1.97 \times 1 0^{- 4} \times \frac{2}{0.01}$
$∣ ε ∣ \approx 3.9 \times 1 0^{- 2} V$

Answer: (a) about $1.97 \times 1 0^{- 4} H$ (0.2 mH) (b) about 0.039 V

AC generator

NCERT section 6.8

Principle: electromagnetic induction. A coil of N turns and area A is rotated with constant angular speed ω in a uniform field B. As it turns, the angle between the coil's normal and B is $θ = ω t$ , so the flux linkage is $N Φ = N B A cos ω t$ .

Emf: $ε = - N \frac{d Φ}{d t} = N B A ω sin ω t = ε_{0} sin ω t$ , with peak value $ε_{0} = N B A ω$ .

The emf is maximum when the plane of the coil is parallel to B (flux zero but changing fastest) and zero when the plane is perpendicular to B (flux maximum but momentarily not changing).
The emf reverses direction every half turn, so it is alternating, with frequency $ν = \frac{ω}{2 π}$ .

Parts: armature coil, field magnet, slip rings (which turn with the coil) and carbon brushes (which press on the rings and connect to the external circuit). The CBSE syllabus lists the AC generator in Unit IV with alternating current.

Worked example

A coil of 100 turns, each of area $0.01 m^{2}$ , rotates at 50 revolutions per second in a uniform field of 0.5 T. Find the peak emf, and the emf at the instant when the coil has turned $6 0^{\circ}$ ( $ω t = 6 0^{\circ}$ ) from the position of maximum flux. [2 marks]

Try it first, then show the solutionSolution

$ω = 2 π \times 50 = 100 π r a d s^{- 1}$
$ε_{0} = N B A ω = 100 \times 0.5 \times 0.01 \times 100 π = 50 π \approx 157 V$
$ε = ε_{0} sin 6 0^{\circ} = 157 \times 0.866 \approx 136 V$

Answer: Peak emf ≈ 157 V; about 136 V at that instant

Physics in real life: Electromagnetic Induction

Each concept, met in an everyday situation. Read the situation, try the problem, then check the solution. The one-line hook is what to remember.

AC generator

Suppose a bicycle dynamo spins a small coil between the poles of a magnet as the wheel turns, to light the front lamp.

Suppose the coil has 200 turns, each of area $4 c m^{2}$ , in a field of 0.1 T, and spins at 50 rad s⁻¹. Find the peak emf. What happens to it if the cyclist rides twice as fast?

Try it first, then show the solutionSolution

$A = 4 \times 1 0^{- 4} m^{2}$
$ε_{0} = N B A ω = 200 \times 0.1 \times 4 \times 1 0^{- 4} \times 50 = 0.4 V$
$ε_{0} \propto ω$ , so riding twice as fast doubles it to 0.8 V (and doubles the frequency).

Answer: 0.4 V peak; 0.8 V at double speed.

Memory hook: Spin faster, flux changes faster, lamp glows brighter.

Self-induction

Imagine switching off a circuit containing a large coil, such as an electromagnet. A small spark jumps across the switch contacts.

Suppose the coil has inductance 0.5 H and carries 2 A, and opening the switch makes the current fall to zero in 1 ms. Estimate the average induced emf.

Try it first, then show the solutionSolution

$∣ ε ∣ = L \frac{Δ I}{Δ t} = 0.5 \times \frac{2}{1 \times 1 0 ^{- 3}}$
$∣ ε ∣ = 1000 V$
This large back emf, opposing the fall in current, appears across the opening switch gap and can make the air conduct briefly: a spark.

Answer: About 1000 V, enough to cause a spark at the switch.

Memory hook: An inductor hates sudden change: break its current and it fights back with a big voltage.

Mutual induction

Suppose a phone is charged without a cable by placing it on a pad: a coil in the pad carries a rapidly changing current, and a coil in the phone picks up an emf.

Suppose the mutual inductance between the two coils is $2 \times 1 0^{- 5} H$ and the current in the pad's coil changes at $1 \times 1 0^{4} A s^{- 1}$ . Find the emf induced in the phone's coil. Why does the phone have to sit close to the pad?

Try it first, then show the solutionSolution

$∣ ε ∣ = M \frac{d I}{d t} = 2 \times 1 0^{- 5} \times 1 \times 1 0^{4} = 0.2 V$
M depends on the separation and alignment of the coils. Move the phone away and less of the pad's flux passes through its coil, so M and the emf fall sharply.

Answer: 0.2 V; M (and so the emf) drops quickly as the coils separate.

Memory hook: Mutual inductance is a handshake: it only works when the coils are close.

Topics in this chapter

Electromagnetic Induction formula sheet

Magnetic flux: $Φ_{B} = B \cdot A = B A cos θ$; θ from the normal; 1 Wb = 1 T m².
Faraday's law: $ε = - N \frac{d Φ _{B}}{d t}$
Charge through a circuit: $Q = \frac{N ΔΦ}{R}$; Independent of how fast the flux changes.
Motional emf: $ε = B l v$
Rod rotating about one end: $ε = \frac{1}{2} B ω l^{2}$
Mutual inductance: $N_{2} Φ_{2} = M I_{1}, ε_{2} = - M \frac{d I _{1}}{d t}$
Coaxial solenoids: $M = μ_{0} n_{1} n_{2} π r_{1}^{2} l$; $r_{1}$ = radius of the inner solenoid.
Self-inductance: $N Φ = L I, ε = - L \frac{d I}{d t}$
Solenoid inductance: $L = μ_{0} n^{2} A l = \frac{μ _{0} N ^{2} A}{l}$; With a core, multiply by $μ_{r}$ .
Energy in an inductor: $U = \frac{1}{2} L I^{2}$
AC generator: $ε = N B A ω sin ω t, ε_{0} = N B A ω$; $ω = 2 π ν$ .
Unit of inductance: $1 H = 1 W b A^{- 1} = 1 V s A^{- 1}$; Dimensions $[M L^{2} T^{- 2} A^{- 2}]$ .

Electromagnetic Induction: diagrams to practise

Practise each one on paper: draw it from the checklist, then tick off every label.

AC generator

Principle and working (3-5 marks).

Draw

Draw the N and S poles of a magnet facing each other.
Between them draw a rectangular coil (armature) on an axle, free to rotate.
Join the ends of the coil to two separate slip rings on the axle.
Draw two carbon brushes pressing on the rings, connected to an external resistor or lamp.

Label

Field magnet (N, S)
Armature coil
Axle and direction of rotation
Slip rings
Carbon brushes
External load

Watch out: Add a small sketch of ε against t (a sine curve) to show the alternating emf; it often earns the last mark.

Rod sliding on rails (motional emf)

Derivation of $ε = B l v$ (2-3 marks).

Draw

Draw two parallel horizontal rails joined at the left by a resistor R.
Draw a movable rod PQ across the rails, length l, with an arrow v to the right.
Mark B as crosses (into the page) or dots (out of the page) over the whole region.
Mark the distance x of the rod from the resistor end and the direction of the induced current.

Label

Rails
Rod PQ, length l
v
B (⊗ or ⊙)
x
R
Induced current I

Watch out: Show the current direction from Lenz's law: if the area is increasing, the current must reduce the flux.

How the board asks Electromagnetic Induction

The question patterns to expect, and the method that works for each.

Question type	How it is asked	Method
Flux and Faraday's law numericals	Emf when B, A or the angle changes; charge that flows; emf from a flux-time formula or graph.	Write Φ before and after (or as a function of t), find the rate of change, multiply by N; for charge use $N ΔΦ/ R$ .
Lenz's law direction	Direction of induced current for a magnet approaching or leaving a coil, a loop entering a field, or a current switched on in a nearby coil; falling magnets.	Ask: is the flux increasing or decreasing? The induced current makes a field that opposes that change; then use the right-hand rule.
Motional emf	Rod on rails, wire falling in a field, rotating rod; derivation of Blv.	Use $ε = B l v$ with l, v and B mutually perpendicular; for rotation use $\frac{1}{2} B ω l^{2}$ ; find the positive end with the Lorentz force.
Mutual inductance	Define M; derive M for coaxial solenoids; find emf in one coil from the rate of change of current in the other; flux-current graphs.	$M = N_{2} Φ_{2} / I_{1}$ ; for solenoids use the inner area only; slope of a $Φ$ -I graph gives M.
Self-inductance	Derive L for a solenoid; back emf; energy stored; effect of doubling turns or adding a core.	$L = μ_{0} N^{2} A / l$ , so L ∝ N²; energy $\frac{1}{2} L I^{2}$ .
AC generator	Principle, labelled diagram, derivation of $ε_{0} sin ω t$ , peak emf numericals, position of maximum emf.	Write $Φ = B A cos ω t$ , differentiate, and identify $ε_{0} = N B A ω$ .

Tricks to remember Electromagnetic Induction

Short memory aids. Say each one aloud once, then use it on a question the same day.

Change, not amount
Faraday cares about the CHANGE in flux. A huge steady flux gives zero emf; a small flux changing fast gives a big emf.
Use it when: Assertion-reason items on flux versus emf.
Lenz is lazy
Lenz's law is nature being lazy: the induced current always tries to keep things as they were.
Use it when: Fixing the direction of induced current.
Approach, repel; leave, attract
A magnet approaching a coil is repelled (same pole faces it); a magnet leaving is attracted (opposite pole faces it).
Use it when: Lenz direction for moving magnets.
BLV
'Blv' as in 'believe': B, l and v must be mutually perpendicular to use it directly.
Use it when: Motional emf.
L goes with N², M with N₁N₂
Double the turns of a solenoid: L becomes four times. M needs one factor from each coil.
Use it when: Ratio and design questions.
Parallel plane, peak emf
Generator emf is greatest when the coil's plane is parallel to B (flux zero) and zero when perpendicular (flux maximum).
Use it when: Generator MCQs.

Electromagnetic Induction: board-level solved questions

Ordered from board level to HOTS. Cover the solution, write your own, then compare step by step.

Example 1: Emf in a coil · 1 mark · MCQBoard level

A coil of 50 turns, each of area $0.02 m^{2}$ , has its plane perpendicular to a field that decreases uniformly at $0.5 T s^{- 1}$ . The induced emf is:
(a) 0.05 V
(b) 0.5 V
(c) 5 V
(d) 0.01 V

Try it first, then show the solutionSolution

$ε = N A \frac{d B}{d t} = 50 \times 0.02 \times 0.5 = 0.5 V$

Answer: (b) 0.5 V

Example 2: Magnet falling through a pipe · 1 mark · Assertion-reasonBoard level

Assertion (A): A small magnet dropped through a long vertical copper pipe falls with an acceleration less than g.
Reason (R): The currents induced in the pipe produce magnetic fields that oppose the motion of the magnet.
(a) Both true, R explains A. (b) Both true, R does not explain A. (c) A true, R false. (d) A false, R true.

Try it first, then show the solutionSolution

The falling magnet changes the flux through the pipe, inducing currents which, by Lenz's law, oppose the change and hence the motion. Both true; R explains A.

Answer: (a)

Example 3: Charge through a coil · 2 marks · Very short answerHarder board level

A coil of 100 turns and resistance 5 Ω has an area of $0.04 m^{2}$ and lies perpendicular to a field of 0.25 T. It is pulled out of the field completely. How much charge flows through it? Does the answer depend on how fast it is pulled out? [2 marks]

Try it first, then show the solutionSolution

$ΔΦ = B A = 0.25 \times 0.04 = 0.01 W b$ per turn
$Q = \frac{N ΔΦ}{R} = \frac{100 \times 0.01}{5} = 0.2 C$
No: faster pulling gives a bigger current for a shorter time; the product (charge) is the same.

Answer: 0.2 C; independent of the speed.

Example 4: Derivation: motional emf · 3 marks · Short answerHarder board level

A rod of length l slides with velocity v along two parallel rails in a uniform field B perpendicular to the plane of the rails. (a) Derive the emf from the change of flux. (b) Show how the Lorentz force on the free charges gives the same result. (c) If l = 0.5 m, v = 4 m s⁻¹, B = 0.2 T and the circuit resistance is 2 Ω, find the current. [3 marks]

Try it first, then show the solutionSolution

(a) With the rod at distance x, $Φ = B l x$ . $ε = - \frac{d Φ}{d t} = - B l \frac{d x}{d t}$ ; $∣ ε ∣ = B l v$ .
(b) A free charge q in the rod moves with it, so it feels $F = q v B$ along the rod. The work done carrying q along the rod is $q v B l$ , so emf = work per unit charge = $B l v$ .
(c) $ε = 0.2 \times 0.5 \times 4 = 0.4 V$ ; $I = \frac{0.4}{2} = 0.2 A$

Answer: $ε = B l v$ (both ways); 0.2 A

Example 5: Self-inductance of a solenoid and energy · 5 marks · Long answerHarder board level

(a) Define self-inductance and give its SI unit. (b) Derive the self-inductance of a long air-filled solenoid of length l, cross-section A and N turns. (c) Show that the energy stored when it carries current I is $\frac{1}{2} L I^{2}$ . (d) A solenoid of 500 turns, length 0.5 m and radius 1 cm carries 2 A. Find its inductance and the energy stored. [5 marks]

Try it first, then show the solutionSolution

(a) L is the flux linkage per unit current in the coil itself, $N Φ = L I$ , or the emf induced per unit rate of change of current, $ε = - L d I / d t$ ; unit henry.
(b) Field inside $B = μ_{0} \frac{N}{l} I$ ; flux linkage $N Φ = N \times μ_{0} \frac{N}{l} I \times A$ .
So $L = \frac{N Φ}{I} = \frac{μ _{0} N ^{2} A}{l}$ .
(c) While the current grows, the source does work against the back emf at rate $ε I = L I \frac{d I}{d t}$ ; total $W = \int_{0}^{I} L I d I = \frac{1}{2} L I^{2}$ , stored in the magnetic field.
(d) $L = \frac{4 π \times 1 0 ^{- 7} \times 250 000 \times π \times 1 0 ^{- 4}}{0.5} \approx 1.97 \times 1 0^{- 4} H$ ; $U = \frac{1}{2} \times 1.97 \times 1 0^{- 4} \times 4 \approx 3.9 \times 1 0^{- 4} J$

Answer: $L = μ_{0} N^{2} A / l$ ; energy $\frac{1}{2} L I^{2}$ ; $L \approx 1.97 \times 1 0^{- 4} H$ , $U \approx 3.9 \times 1 0^{- 4} J$

Example 6: Mutual inductance of coaxial solenoids · 3 marks · Short answerHarder board level

An inner solenoid of radius 1 cm has 1000 turns per metre; a coaxial outer solenoid of the same length 0.5 m has 2000 turns per metre. (a) Find their mutual inductance. (b) If the current in the inner solenoid changes at $50 A s^{- 1}$ , find the emf induced in the outer one. [3 marks]

Try it first, then show the solutionSolution

(a) $M = μ_{0} n_{1} n_{2} π r_{1}^{2} l = 4 π \times 1 0^{- 7} \times 1000 \times 2000 \times π \times 1 0^{- 4} \times 0.5$
$M = 4 π^{2} \times 1 0^{- 5} \approx 3.95 \times 1 0^{- 4} H$
(b) $∣ ε ∣ = M \frac{d I}{d t} = 3.95 \times 1 0^{- 4} \times 50 \approx 1.97 \times 1 0^{- 2} V$

Answer: (a) about $3.95 \times 1 0^{- 4} H$ (b) about 0.020 V

Example 7: Rotating rod: emf and polarity · 2 marks · Very short answerHOTS / competency

A copper rod 0.5 m long rotates at 40 rad s⁻¹ about one end O in a horizontal plane. A uniform field of 0.2 T points vertically upward. (a) Find the emf between O and the free end P. (b) Which end is at the higher potential if the rod turns anticlockwise seen from above? [2 marks]

Try it first, then show the solutionSolution

(a) $ε = \frac{1}{2} B ω l^{2} = \frac{1}{2} \times 0.2 \times 40 \times 0.25 = 1 V$
(b) Anticlockwise from above, a point on the rod moves along the tangent; take P on the +x axis moving along +y, with B along +z. Force on a positive charge: $q v \times B = q (v \hat{j} \times B \hat{k}) = q v B \hat{i}$ , outward towards P.
Positive charges pile up at P (electrons move towards O), so P is at the higher potential.

Answer: (a) 1 V (b) the free end P is at the higher potential.

Example 8: Emf from a flux formula · 3 marks · Short answerHOTS / competency

The flux through a 10-turn coil varies as $Φ = (2 t^{2} - 4 t + 1) \times 1 0^{- 3} W b$ , with t in seconds. (a) Find the emf at t = 3 s. (b) At what instant is the emf zero? (c) The coil's resistance is 4 Ω. Find the current at t = 3 s. [3 marks]

Try it first, then show the solutionSolution

$\frac{d Φ}{d t} = (4 t - 4) \times 1 0^{- 3} W b s^{- 1}$
(a) At t = 3 s: $\frac{d Φ}{d t} = 8 \times 1 0^{- 3}$ ; $∣ ε ∣ = N \frac{d Φ}{d t} = 10 \times 8 \times 1 0^{- 3} = 0.08 V$
(b) $4 t - 4 = 0 \Rightarrow t = 1 s$ (the flux is at its minimum then)
(c) $I = \frac{0.08}{4} = 0.02 A$

Answer: (a) 0.08 V (b) t = 1 s (c) 0.02 A

Case-study practice (Section D style)

Read the situation, then answer every part. In the board paper this is a 4-mark question split into 1 + 1 + 1 + 1 marks.

A magnet and a coil

Suppose a student connects a coil of many turns to a sensitive centre-zero galvanometer. She pushes the north pole of a bar magnet into the coil, holds it still inside, and then pulls it out quickly.

(i) · 1 mark
What does the galvanometer show while the magnet is held still inside the coil?
Answer
Answer: No deflection.
The flux is not changing, so no emf is induced.
(ii) · 1 mark
What pole appears on the face of the coil nearer the magnet as the magnet is pushed in?
Answer
Answer: A north pole.
By Lenz's law the coil opposes the approaching north pole by repelling it.
(iii) · 1 mark
How does the deflection when pulling out quickly compare with that when pushing in slowly?
Answer
Answer: Larger and in the opposite direction.
Faster change of flux gives a larger emf; removal reverses the change, so the current reverses.
(iv) · 1 mark
Which law of conservation does Lenz's law express?
Answer
Answer: Conservation of energy.
The student must do work against the opposing force; that work becomes the electrical energy of the induced current.

Inside a small generator

Suppose a simple AC generator has a coil of 100 turns, each of area $0.02 m^{2}$ , rotating at 50 revolutions per second in a uniform field of 0.5 T.

(i) · 1 mark
Find the peak emf.
Answer
Answer: About 314 V
$ε_{0} = N B A ω = 100 \times 0.5 \times 0.02 \times 100 π = 100 π \approx 314 V$
(ii) · 1 mark
What is the frequency of the alternating emf?
Answer
Answer: 50 Hz
One cycle of emf per revolution.
(iii) · 1 mark
In which position of the coil is the emf zero?
Answer
Answer: When the plane of the coil is perpendicular to the field.
The flux is then maximum and momentarily not changing.
(iv) · 1 mark
Find the emf 1/600 s after the coil passes the zero-emf position.
Answer
Answer: About 157 V
$ω t = 100 π \times \frac{1}{600} = \frac{π}{6}$ ; $ε = 314 \times sin 3 0^{\circ} = 157 V$

Where the marks are

How a typical answer is split into step marks. Show every step: a right final answer alone rarely earns full marks.

A coil of 200 turns and area $0.01 m^{2}$ is placed with its plane perpendicular to a field of 0.4 T. It is turned in 0.05 s so that its plane becomes parallel to the field. Find the average induced emf. [2 marks] (2 marks)

Step	Marks
$ΔΦ = B A - 0 = 0.4 \times 0.01 = 4 \times 1 0^{- 3} W b$ per turn	1
$ε = N \frac{ΔΦ}{Δ t} = \frac{200 \times 4 \times 1 0 ^{- 3}}{0.05} = 16 V$	1

Derive an expression for the mutual inductance of two long coaxial solenoids of the same length. [3 marks] (3 marks)

Step	Marks
Figure; field of inner solenoid $B_{1} = μ_{0} n_{1} I_{1}$ , existing only inside radius $r_{1}$	1
Flux linkage with outer: $N_{2} Φ_{2} = (n_{2} l) (π r_{1}^{2}) (μ_{0} n_{1} I_{1})$	1
$M = \frac{N _{2} Φ _{2}}{I _{1}} = μ_{0} n_{1} n_{2} π r_{1}^{2} l$ (and $M_{12} = M_{21}$ )	1

(a) State the principle of an AC generator. (b) With a labelled diagram, derive the expression for the emf induced. (c) A coil of 100 turns, area $0.01 m^{2}$ , rotates at 50 rev s⁻¹ in a field of 0.5 T. Find the peak emf. [5 marks] (5 marks)

Step	Marks
(a) Electromagnetic induction: a coil rotated in a magnetic field has a changing flux, so an emf is induced	1
(b) Diagram: armature coil, field magnet, slip rings, brushes	1
Flux linkage $N Φ = N B A cos ω t$	1
$ε = - N \frac{d Φ}{d t} = N B A ω sin ω t$ , $ε_{0} = N B A ω$	1
(c) $ε_{0} = 100 \times 0.5 \times 0.01 \times 2 π \times 50 \approx 157 V$	1

Where students lose marks in Electromagnetic Induction

The slips examiners see most often, and what to write instead.

Mistake: Taking θ from the plane of the coil in $B A cos θ$
Reading 'the plane makes 30° with B'.
Wrong working
Plane at 30° to B: $Φ = B A cos 3 0^{\circ}$
Right working
Normal at 60° to B: $Φ = B A cos 6 0^{\circ}$
Do this: Convert to the angle with the normal first.
Usually costs: 1 mark.
Mistake: Forgetting the number of turns N
Using the flux of one turn.
Do this: Write $ε = N d Φ/ d t$ and check that N appears in the substitution.
Usually costs: Final-answer mark.
Mistake: Explaining Lenz's law without the energy argument
Stating only the rule.
Do this: Add: if the induced current aided the change, energy would be created from nothing; so Lenz's law follows from conservation of energy.
Usually costs: 1 mark in 2-3 mark 'explain' questions.
Mistake: Using the outer solenoid's area for M
Not noticing where the field is.
Wrong working
$M = μ_{0} n_{1} n_{2} π r_{2}^{2} l$
Right working
$M = μ_{0} n_{1} n_{2} π r_{1}^{2} l$ : the inner solenoid's field exists only within $r_{1}$
Do this: Use the cross-section where the field actually is.
Usually costs: 1-2 marks.
Mistake: Saying the generator emf is maximum when the flux is maximum
Linking emf to flux instead of its rate of change.
Do this: Emf is maximum when the flux is zero and changing fastest (plane parallel to B).
Usually costs: 1 mark.
Mistake: Writing about eddy currents in answers
Older syllabi included them.
Do this: Eddy currents are not in the 2026-27 syllabus; explain falling-magnet questions with Lenz's law applied to the pipe or ring.
Usually costs: Time, and marks if the required idea is missed.

NCERT exercise guide: Electromagnetic Induction

What each exercise in the NCERT chapter tests, how to approach it, and a question of the same kind solved step by step. Questions here are written in our own words.

Exercises 6.1 and 6.2 (direction of induced current)

What it tests: Predicting the direction of induced current when magnets move, loops change shape or currents are switched.

How to approach it: Decide whether the flux through the loop rises or falls; the induced current makes its own field opposing that change.

Same kind as Exercise 6.2

A circular loop lies flat on a table in a uniform field pointing vertically upward. The loop is squeezed so that its area decreases. Seen from above, which way does the induced current flow?

Show the solutionSolution

Upward flux through the loop decreases as the area shrinks.
The induced current must add upward flux to oppose the decrease.
A current that is anticlockwise seen from above gives an upward field inside the loop.

Answer: Anticlockwise, seen from above

Exercises 6.3 and 6.4 (emf from changing flux)

What it tests: Emf in a small loop inside a solenoid whose current changes, and emf in a loop moving out of a field region.

How to approach it: For the solenoid, $Φ = μ_{0} n I A$ , so $ε = μ_{0} n A d I / d t$ . For a moving loop, $ε = B l v$ while it is crossing the boundary.

Same kind as Exercise 6.3

A long solenoid with 1000 turns per metre has a small loop of area $2 \times 1 0^{- 4} m^{2}$ inside it, coaxial with it. The solenoid current changes steadily from 3 A to 1 A in 0.2 s. Find the emf in the loop. ( $μ_{0} = 4 π \times 1 0^{- 7} T m A^{- 1}$ )

Show the solutionSolution

$\frac{d I}{d t} = \frac{2}{0.2} = 10 A s^{- 1}$
$ε = μ_{0} n A \frac{d I}{d t} = 4 π \times 1 0^{- 7} \times 1000 \times 2 \times 1 0^{- 4} \times 10$
$ε \approx 2.5 \times 1 0^{- 6} V$

Answer: About $2.5 \times 1 0^{- 6} V$

Exercises 6.5 and 6.6 (motional emf)

What it tests: Emf between the centre and the end of a rotating rod, and emf across a falling horizontal wire.

How to approach it: Rotating: $\frac{1}{2} B ω l^{2}$ . Falling wire: $B l v$ with the field component perpendicular to both the wire and its velocity.

Same kind as Exercise 6.5

A metal rod 1.0 m long rotates at 20 rad s⁻¹ about one end, in a plane perpendicular to a uniform field of 0.5 T. Find the emf between its ends.

Show the solutionSolution

$ε = \frac{1}{2} B ω l^{2} = \frac{1}{2} \times 0.5 \times 20 \times (1.0)^{2}$
$= 5 V$

Answer: 5 V

Exercises 6.7 and 6.8 (self- and mutual inductance)

What it tests: Self-inductance from the emf and the rate of change of current, and flux linkage and emf from a given mutual inductance.

How to approach it: $L = ε / (d I / d t)$ ; flux linkage $= M I_{1}$ ; emf $= M d I_{1} / d t$ .

Same kind as Exercise 6.7

The current in a coil rises steadily from 2 A to 6 A in 0.05 s and an average emf of 40 V is induced. Find the self-inductance and the energy stored at 6 A.

Show the solutionSolution

$\frac{d I}{d t} = \frac{4}{0.05} = 80 A s^{- 1}$
$L = \frac{40}{80} = 0.5 H$
$U = \frac{1}{2} L I^{2} = \frac{1}{2} \times 0.5 \times 36 = 9 J$

Answer: 0.5 H; 9 J

Electromagnetic Induction: previous year questions

Asked in 8 of the 8 board papers we hold (2023 to 2026), on average 5 marks per paper; 22 questions in all, counting OR alternatives. In 3 of 4 sample papers.

Counted from the real papers on our previous year papers page, not estimated. Average marks are over every board paper we hold, with an OR pair counted once. Chapter tags are ours.

Solved board questions from Electromagnetic Induction

Real questions from recent papers. Try each one first, then open our step-by-step solution with marks per step.

2026, set 55/1/1, Q33(a) · 5 marks · Long answer

(a) State Faraday’s law of electromagnetic induction.
(b) Derive an expression for the self-inductance of an air-filled long solenoid of length $l$ and cross-sectional area A having N turns.
(c) A conducting rod of length 50 cm, with one end pivoted, is rotated with angular speed of 60 rpm in a uniform magnetic field of 4.0 mT directed perpendicular to the plane of rotation of rod. Find the emf induced in the rod.

Try it first, then show our solutionSolution

(a) Faraday's law. The magnitude of the emf induced in a circuit equals the time rate of change of magnetic flux linked with it. Its direction is such that it opposes that change (Lenz's law).
$ε = - N \frac{d Φ _{B}}{d t}$
1 mark
(b) Field and flux linkage. Let a current $I$ flow in the solenoid. The number of turns per unit length is $n = N / l$ .
For a long solenoid, the field inside is uniform:
$B = μ_{0} n I = \frac{μ _{0} N I}{l}$
The flux through each turn is $Φ = B A$ . So the total flux linkage is:
$N Φ = N \cdot \frac{μ _{0} N I}{l} \cdot A = \frac{μ _{0} N ^{2} A}{l} I$
1 mark
Self-inductance. Since $N Φ = L I$ :
$L = \frac{μ _{0} N ^{2} A}{l} (or μ_{0} n^{2} A l)$
1 mark
(c) Convert the data to SI units.
- $l = 50$ cm $= 0.5$ m
- $B = 4.0$ mT $= 4.0 \times 1 0^{- 3}$ T
- $ω = 60$ rpm $= \frac{60 \times 2 π}{60} = 2 π$ rad s $^{- 1}$
For a rod rotating about one end, the emf is:
$ε = \frac{1}{2} B ω l^{2}$
1 mark
Substitute.
$ε = \frac{1}{2} \times 4.0 \times 1 0^{- 3} \times 2 π \times (0.5)^{2} = π \times 1 0^{- 3} V$
$\approx 3.14 \times 1 0^{- 3} V = 3.14 mV$
1 mark

Answer: (a) $ε = - N d Φ/ d t$ ; (b) $L = μ_{0} N^{2} A / l$ ; (c) $ε \approx 3.14 \times 1 0^{- 3}$ V

Where the marks are

Correct statement of Faraday's law with $ε = - d Φ/ d t$ (1 mark)
Writes $B = μ_{0} N I / l$ and the flux linkage $N Φ = μ_{0} N^{2} A I / l$ (1 mark)
Obtains $L = μ_{0} N^{2} A / l$ (1 mark)
Converts 60 rpm to $2 π$ rad/s and uses $ε = \frac{1}{2} B ω l^{2}$ (1 mark)
Correct result $\approx 3.14$ mV (1 mark)

See it in the full 2026 board paper, set 55/1/1

2026, set 55/5/1, Q4 · 1 mark · MCQ

The magnetic flux through a loop placed in a magnetic field can be changed by changing :

(A)area of the loop only
(B)the value of magnetic field only
(C)orientation of the loop in the magnetic field only
(D)any one or more of the factors given in (A), (B) and (C)

Try it first, then show our solutionSolution

Flux is $Φ = B A cos θ$ . Changing B, A or θ, or any combination of them, changes the flux.
1 mark

Answer: Option (D)

Where the marks are

Uses $Φ = B A cos θ$ to choose (D) (1 mark)

See it in the full 2026 board paper, set 55/5/1

2026, set 55/5/1, Q18 · 2 marks · Very short answer

A plane circular coil is rotated about its vertical diameter with a constant angular speed $ω$ in a uniform horizontal magnetic field. Initially the plane of the coil is parallel to the magnetic field. Draw plots showing the variation of the following physical quantities as the function of $ω t$ , where t represents time elapsed :
(a) Magnetic flux $ϕ$ linked with the coil, and
(b) emf induced in the coil.

Try it first, then show our solutionSolution

At $t = 0$ the plane of the coil is parallel to $B$ , so its normal is perpendicular to $B$ and the flux is zero. After time t the normal has turned through $ω t$ from this position, so the angle between normal and $B$ is $(9 0^{\circ} - ω t)$ :
$ϕ = N B A cos (9 0^{\circ} - ω t) = N B A sin ω t$
(a) Plot of $ϕ$ vs $ω t$ : a sine curve starting at zero, rising to maximum $+ N B A$ at $ω t = π /2$ , zero at $π$ , minimum $- N B A$ at $3 π /2$ , zero at $2 π$ .
1 mark
(b) $ε = - \frac{d ϕ}{d t} = - N B A ω cos ω t$ .
Plot of $ε$ vs $ω t$ : a (negative) cosine curve: $- N B A ω$ at $ω t = 0$ , zero at $π /2$ , $+ N B A ω$ at $π$ , zero at $3 π /2$ , $- N B A ω$ at $2 π$ . The emf is maximum in magnitude when the flux is zero, and zero when the flux is maximum (90° phase difference).
1 mark

Answer: φ = NBA sin ωt (sine curve from zero); ε = −NBAω cos ωt (cosine curve, maximum magnitude at ωt = 0)

Where the marks are

Correct sinusoidal plot of flux starting from zero at ωt = 0 (1 mark)
Correct plot of emf as (negative) cosine, maximum magnitude at ωt = 0, 90° out of phase with flux (1 mark)

See it in the full 2026 board paper, set 55/5/1

Every Electromagnetic Induction question, by year

For 2027

Sample paper 2026-27: Q3 (1 mark, solved), Q28 (3 marks, solved)

2026

Set 55/1/1: Q33(a) (5 marks, solved)

Set 55/5/1: Q4 (1 mark, solved), Q13 (1 mark, solved), Q18 (2 marks, solved), Q24 (3 marks, solved)

Sample paper 2025-26: Q9 (1 mark, solved), Q28 (3 marks, solved)

2025

Set 55/1/1: Q4 (1 mark, solved), Q5 (1 mark, solved), Q13 (1 mark, solved), Q24 (3 marks, solved)

Set 55/7/1: Q3 (1 mark, solved), Q6 (1 mark, solved), Q7 (1 mark, solved), Q24 (3 marks, solved)

2024

Set 55/1/1: Q14 (1 mark, solved)

Set 55/5/1: Q5 (1 mark, solved), Q24(a) (3 marks, solved)

Sample paper 2023-24: Q11 (1 mark, solved), Q28(a) (3 marks, solved)

2023

Set 55/1/1: Q6 (1 mark, solved), Q7 (1 mark, solved), Q28 (3 marks, solved)

Set 55/5/1: Q4 (1 mark, solved), Q25 (2 marks, solved), Q28 (3 marks, solved)

Practice questions on Electromagnetic Induction

Write each answer on paper first. Open the hint only if you are stuck, and the answer only when you have finished.

Q1 · 1 mark · MCQBoard level
The SI unit of inductance, the henry, is equal to:
(a) Wb A
(b) V s A⁻¹
(c) V A s⁻¹
(d) T m²
Hint
$ε = L d I / d t$ .
Answer
(b) V s A⁻¹ (also Wb A⁻¹)
Q2 · 1 mark · MCQBoard level
The dimensions of self-inductance are:
(a) $[M L^{2} T^{- 2} A^{- 2}]$
(b) $[M L^{2} T^{- 2} A^{- 1}]$
(c) $[M L T^{- 2} A^{- 2}]$
(d) $[M L^{2} T^{- 1} A^{- 2}]$
Hint
Energy $= \frac{1}{2} L I^{2}$ .
Answer
(a), since $L = \frac{2 U}{I ^{2}}$ and energy has dimensions $[M L^{2} T^{- 2}]$ .
Q3 · 1 mark · Very short answerBoard level
State Lenz's law.
Hint
Opposes the change.
Answer
The direction of the induced current is such that it opposes the change in magnetic flux that produces it.
Q4 · 1 mark · Assertion-reasonHarder board level
Assertion (A): The mutual inductance of two coils depends on the currents flowing in them.
Reason (R): The flux linked with one coil is proportional to the current in the other.
(a) Both true, R explains A. (b) Both true, R does not explain A. (c) A true, R false. (d) A false, R true.
Hint
What does M depend on?
Answer
(d). M depends only on geometry and the medium; R is true (that proportionality defines M).
Q5 · 2 marks · Very short answerHarder board level
A coil of 150 turns, area $0.02 m^{2}$ and resistance 6 Ω has its plane perpendicular to a uniform field of 0.3 T. It is flipped over through $18 0^{\circ}$ in 0.1 s. Find (a) the average induced emf, (b) the average current, (c) the charge that flows through the coil.
Hint
After a half turn the flux through each turn has reversed, so the change is 2BA, not BA.
Answer
(a) $ΔΦ = B A - (- B A) = 2 \times 0.3 \times 0.02 = 0.012 W b$ per turn; $ε = \frac{150 \times 0.012}{0.1} = 18 V$ . (b) $I = \frac{18}{6} = 3 A$ . (c) $Q = I Δ t = 0.3 C$ (check: $\frac{N ΔΦ}{R} = \frac{1.8}{6} = 0.3 C$ ).
Q6 · 2 marks · Very short answerHarder board level
Suppose a horizontal rod 2 m long, lying east-west, falls with speed 5 m s⁻¹ at a place where the horizontal component of the earth's field is $3 \times 1 0^{- 5} T$ (northward). Find the emf across its ends.
Hint
Rod, velocity and the horizontal field are mutually perpendicular.
Answer
$ε = B l v = 3 \times 1 0^{- 5} \times 2 \times 5 = 3 \times 1 0^{- 4} V$ .
Q7 · 2 marks · Very short answerHarder board level
The current through a coil of self-inductance 0.4 H varies as $i = (3 t^{2} - 2 t) A$ , with t in seconds. Find (a) the magnitude of the induced emf at t = 2 s, (b) the energy stored in the coil at t = 2 s, (c) the instant at which the induced emf is zero.
Hint
$ε = - L \frac{d i}{d t}$ and $U = \frac{1}{2} L i^{2}$ .
Answer
(a) $\frac{d i}{d t} = 6 t - 2 = 10 A s^{- 1}$ at t = 2 s, so $∣ ε ∣ = 0.4 \times 10 = 4 V$ . (b) $i = 3 (4) - 2 (2) = 8 A$ ; $U = \frac{1}{2} \times 0.4 \times 64 = 12.8 J$ . (c) $6 t - 2 = 0 \Rightarrow t = \frac{1}{3} s$ .
Q8 · 2 marks · Very short answerHarder board level
A bar magnet's north pole is moved towards a closed loop lying on a table (the magnet is above the loop). Seen from above, which way does the induced current flow? What happens if the magnet is moved away?
Hint
The upper face must repel an approaching N pole.
Answer
Approaching: the upper face becomes a north pole, so the current is anticlockwise seen from above. Moving away: the upper face becomes a south pole, clockwise seen from above.
Q9 · 5 marks · Long answerHarder board level
(a) Explain, with a labelled diagram, the principle and working of an AC generator. (b) Derive $ε = N B A ω sin ω t$ . (c) A coil of 50 turns and area $0.05 m^{2}$ rotates at 100 rad s⁻¹ in a field of 0.4 T. Find the peak emf and say in which position it occurs.
Hint
Flux linkage $N B A cos ω t$ .
Answer
(a) Principle: EMI; a coil rotating in a field has a changing flux; parts: armature, magnet, slip rings, brushes. (b) $N Φ = N B A cos ω t$ , so $ε = - \frac{d ( N Φ )}{d t} = N B A ω sin ω t$ . (c) $ε_{0} = 50 \times 0.4 \times 0.05 \times 100 = 100 V$ , when the plane of the coil is parallel to B.
Q10 · 3 marks · Short answerHOTS / competency
A small circular loop of radius 1 cm is placed at the centre of a large circular loop of radius 20 cm, coplanar and concentric. Find their mutual inductance. ( $μ_{0} = 4 π \times 1 0^{- 7}$ , $π^{2} \approx 9.87$ )
Hint
The large loop's field is nearly uniform over the small loop.
Answer
Field at the centre of the large loop: $\frac{μ _{0} I}{2 R}$ . Flux through the small loop: $\frac{μ _{0} I}{2 R} π r^{2}$ . So $M = \frac{μ _{0} π r ^{2}}{2 R} = \frac{4 π \times 1 0 ^{- 7} \times π \times 1 0 ^{- 4}}{0.4} \approx 9.9 \times 1 0^{- 10} H$ .
Q11 · 3 marks · Short answerHOTS / competency
A rectangular loop moves at constant speed into, through and out of a region of uniform magnetic field that is wider than the loop. Sketch how the induced emf varies with time and explain each part.
Hint
Emf appears only while the flux is changing.
Answer
Entering: flux rises steadily, so a constant emf (say positive) of $B l v$ . Fully inside: flux constant, emf zero. Leaving: flux falls steadily, constant emf of the same size but opposite sign. The graph is a positive step, a gap at zero, then an equal negative step.
Q12 · 2 marks · Very short answerHarder board level
What happens to the self-inductance of a solenoid if (a) the number of turns is doubled with the length unchanged, (b) a soft iron core of relative permeability 1000 is inserted?
Hint
$L = μ_{r} μ_{0} N^{2} A / l$ .
Answer
(a) It becomes four times. (b) It becomes 1000 times.

Test your recall

Answer in your head or on paper first, then open the card. Pulling an answer out of memory makes it stick far better than re-reading.

1.State Faraday's law of electromagnetic induction.Show answer
The induced emf in a circuit equals the negative of the rate of change of magnetic flux linked with it: $ε = - N d Φ/ d t$ .
2.State Lenz's law and the principle it follows from.Show answer
The induced current opposes the change in flux that produced it; it follows from conservation of energy.
3.Write the motional emf of a rod and of a rod rotating about one end.Show answer
$B l v$ ; $\frac{1}{2} B ω l^{2}$ .
4.Define mutual inductance and give its SI unit.Show answer
The flux linkage with one coil per unit current in the other, $M = N_{2} Φ_{2} / I_{1}$ (or emf per unit rate of change of current); henry.
5.Write the self-inductance of a long solenoid.Show answer
$L = μ_{0} n^{2} A l = μ_{0} N^{2} A / l$ (times $μ_{r}$ with a core).
6.Energy stored in an inductor?Show answer
$\frac{1}{2} L I^{2}$ .
7.Peak emf of a generator with N = 100, A = 0.02 m², B = 0.5 T, ω = 100 rad s⁻¹?Show answer
$ε_{0} = N B A ω = 100 V$ .
8.Name three ways to change magnetic flux through a coil.Show answer
Change B, change the area in the field, or change the orientation (angle).

Best order to study it

Start with Faraday's experiments and the definition of magnetic flux.
Learn Faraday's law with N turns, then the charge-flow result $Q = N ΔΦ/ R$ .
Practise Lenz's law directions with magnets and coils, and the energy argument.
Derive motional emf by both the flux and the Lorentz-force methods; add the rotating rod.
Study mutual inductance with the coaxial-solenoid derivation, then self-inductance and energy stored.
Finish with the AC generator: principle, diagram, derivation and numericals.

Quick revision checklist

The night before the exam: can you do each of these without looking?

Can you calculate flux with the angle measured from the normal?
Can you state Faraday's law and find the emf from a flux-time formula?
Can you find the charge that flows in a coil when the flux changes?
Can you give the direction of an induced current using Lenz's law in four standard cases?
Can you explain why Lenz's law follows from energy conservation?
Can you derive $ε = B l v$ in two ways?
Can you derive M for two coaxial solenoids and L for one solenoid?
Can you say what happens to L when turns are doubled or a core is inserted?
Can you write the energy stored in an inductor?
Can you derive the AC generator emf and say when it is maximum?

Electromagnetic Induction: frequently asked questions

Are eddy currents in the 2026-27 syllabus?

No. Eddy currents and the quantitative study of energy in electromagnetic induction are not in the 2026-27 syllabus.

Is the AC generator part of this chapter or the next?

NCERT explains it at the end of this chapter (section 6.8), while the CBSE syllabus lists 'AC generator' in Unit IV along with alternating current. Either way it is in the syllabus: learn its principle, diagram and the derivation of \varepsilon = NBA\omega\sin\omega t.

What does the minus sign in Faraday's law mean?

It expresses Lenz's law: the induced emf drives a current that opposes the change in flux causing it.

Which derivations are important?

Motional emf (Blv), mutual inductance of two coaxial solenoids, self-inductance of a solenoid, and the emf of an AC generator. Mutual inductance was the most frequent topic for this chapter in the board papers in our question bank.

Why does a magnet fall slowly through a metal pipe or ring?

As it falls, the flux through each part of the pipe changes, inducing currents. By Lenz's law these currents produce fields that oppose the magnet's motion, so its acceleration is less than g.

Does mutual inductance depend on the current?

No. It depends only on the geometry of the coils (turns, sizes, separation, orientation) and the medium between them.

Based on: CBSE Physics (042) syllabus 2026-27; NCERT Physics Part I, Textbook for Class XII, Chapter 6 (Reprint 2026-27)