By the end of this chapter you'll be able to…

  • 1Compute magnetic flux and identify which of B, A or theta is changing in a given situation
  • 2Apply Faraday's law to find the magnitude of an induced emf, including for multi-turn coils
  • 3Use Lenz's law to determine the direction of an induced current and justify it by energy conservation
  • 4Derive and apply motional emf, including the half-B-omega-l-squared result for a rotating rod
  • 5Define mutual and self-inductance through flux linkage and use them to find induced emfs
  • 6Compute the inductance and stored energy of a solenoid
  • 7Explain the working of an AC generator and state when its emf is maximum
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Why this chapter matters
This is the chapter that makes electrical generation possible. Faraday's law and Lenz's law explain how every power station, transformer and induction motor works, and the inductance defined here is one of the three elements analysed throughout Chapter 7.

Electromagnetic Induction

1. Check this before you revise anything

The "Additional Exercises" section has been removed from this chapter, as from all 14 chapters of the current Class 12 Physics book. The questions run contiguously from 6.1 to 6.8 with no gaps.

Eddy currents have been removed. Older editions carried a full section on them, with magnetic braking in trains, the induction furnace, dead-beat galvanometers and electric power meters as applications. Searching this chapter for "eddy" returns zero hits, and so do "magnetic braking", "induction furnace" and "dead beat".

The section titled "Energy Consideration: A Quantitative Study" is also gone. The energy argument survives, but only qualitatively, inside section 6.5 on Lenz's law.

A cross-chapter gap worth knowing about. Example 6.3, Example 6.7 and Exercise 6.6 all use the horizontal component of the Earth's magnetic field as a given quantity. That component used to be defined in Chapter 5, but as noted on that chapter's page, the entire section on the Earth's magnetism has been removed — the word "Earth" no longer appears anywhere in Chapter 5.

So this chapter uses a quantity the book no longer teaches. Treat simply as "the component of the field parallel to the ground", which is all these questions require.

Textbook sectionTopic
6.1 to 6.2Introduction, and the experiments of Faraday and Henry
6.3Magnetic flux
6.4Faraday's law of induction
6.5Lenz's law and conservation of energy
6.6Motional electromotive force
6.7Inductance: mutual inductance and self-inductance
6.8AC generator

2. The Experiments, and Magnetic Flux (Textbook 6.2 to 6.3)

Faraday and Henry independently found that electricity can be produced from magnetism, but only under one condition.

The three experiments, and what each isolates:

  • A bar magnet moved towards or away from a closed coil deflects a galvanometer. Holding the magnet still, however close, gives nothing.
  • A coil carrying a steady current, moved relative to a second coil, induces a current in the second. Again, no relative motion means no effect.
  • With both coils stationary, merely switching the current on or off in the first still induces a current in the second.

The third experiment is decisive. Nothing moves, so motion cannot be the essential ingredient. What all three share is a changing magnetic flux through the circuit.

Magnetic flux measures how much field threads a surface:

where is the angle between and the normal to the area, not the plane. The unit is the weber, with 1 Wb = 1 T m.

Three ways to change the flux, and every induction question uses at least one:

  • Change , as when a magnet approaches or a current is switched.
  • Change , as when a loop is pulled out of a field region.
  • Change , as when a coil rotates — which is what a generator does.

3. Faraday's Law and Lenz's Law (Textbook 6.4 to 6.5)

Faraday's law. The induced emf equals the negative rate of change of flux:

and for a coil of turns, where the flux links each turn:

The emf lasts only while the flux is changing. In Exercise 6.3, current in a solenoid is switched off over a finite time, and the induced emf exists only during that interval. A steady current, however large, induces nothing.

Lenz's law is what the minus sign means: the induced current flows in the direction that opposes the change producing it.

Note the precise wording. The induced effect opposes the change, not the field itself. An approaching north pole is met by an induced north pole that repels it; a receding north pole is met by an induced south pole that attracts it, trying to hold it back. Both oppose the change, in opposite senses.

Why the sign must be negative. Suppose the induced current aided the change instead. It would strengthen the flux, which would drive more current, which would strengthen the flux further — energy from nothing. Lenz's law is conservation of energy written into the sign.

The mechanical counterpart: work must be done against the opposing force to keep a magnet moving, and that work is exactly the electrical energy dissipated in the circuit.

Applying it in practice. Exercises 6.1 and 6.2 are pure direction questions, over six and two configurations respectively. The reliable procedure is: decide whether flux is increasing or decreasing, decide which way an induced field must point to oppose that, then use the right-hand rule to get the current direction.

Exercise 6.1(f) is the instructive case — the field line lies in the plane of the loop, so the flux through it is zero and never changes, and no current is induced at all.


4. Motional emf (Textbook 6.6)

When a conducting rod of length moves with velocity perpendicular to a field :

Two derivations, and both are examinable. From flux: as the rod slides, the circuit area changes at rate , so . From the Lorentz force: free charges in the moving rod feel , which drives them to one end until the resulting electric field balances the magnetic force.

The second route explains which end is positive, which the flux argument alone does not. This is the whole content of Exercise 6.6(b) and (c).

The rotating rod. A rod rotating about one end sweeps out area at a rate that grows along its length, so is not constant and the field must be integrated:

The factor of is the single most-forgotten result in the chapter. Exercise 6.5 gives V.


5. Inductance (Textbook 6.7)

Inductance measures a circuit's opposition to a change in current, and both varieties are defined through flux linkage.

Mutual inductance (6.7.1). The flux linking coil 2 due to current in coil 1:

depends only on the geometry of the pair and the medium between them, never on the currents. It is symmetric: .

Exercise 6.8 asks only for the change in flux linkage, Wb. The time given is not needed for that part, though it would give the emf.

Self-inductance (6.7.2). A changing current in a coil changes its own flux, inducing an emf that opposes the change:

This is why Exercise 6.7 works: the average emf and the rate of current change give H.

For a long solenoid, computing the flux linkage gives:

with the turns per unit length. The dependence on — not — is worth noting: doubling the winding density quadruples the inductance.

Energy stored. Work done against the back emf while establishing the current is stored in the field:

Compare for a capacitor: the inductor stores energy in a magnetic field and does so through current, where the capacitor stores it electrically through voltage.


6. The AC Generator (Textbook 6.8)

A generator is the third route to changing flux, by changing : a coil is rotated at constant angular speed in a fixed magnetic field.

With the flux , Faraday's law gives:

Where the peaks fall is not where intuition puts them. The emf is maximum when the coil's plane is parallel to the field, where the flux is momentarily zero but changing fastest. It is zero when the plane is perpendicular, where the flux is maximum but momentarily stationary.

Peak emf grows with the number of turns, the area, the field and the rotation rate. Slip rings keep the connection to the external circuit while the coil turns, which is what makes the output alternating rather than direct.

This sinusoidal emf is exactly the supply that Chapter 7 analyses.


Summary

  • Induction requires a changing flux; a steady field or steady current induces nothing, however large.
  • , with measured from the normal; the unit is the weber, 1 Wb = 1 T m.
  • Flux can be changed through , through , or through — the last is how a generator works.
  • Faraday's law: , and for turns.
  • Lenz's law: the induced current opposes the change in flux, not the flux itself.
  • An approaching pole is repelled and a receding pole attracted — both oppose the change.
  • The minus sign is conservation of energy; the opposite sign would create energy from nothing.
  • Motional emf ; the Lorentz-force derivation also tells you which end goes positive.
  • A rod rotating about one end gives — the factor of one half is easily lost.
  • Mutual inductance: and , with set by geometry alone and symmetric between the coils.
  • Self-inductance: and .
  • Solenoid: , so inductance goes as the square of the turns per unit length.
  • Energy stored in an inductor is , the magnetic counterpart of .
  • AC generator: , peaking when the coil plane is parallel to the field.
  • Eddy currents and the quantitative energy section have been removed from this chapter.
  • The Earth's horizontal field component is used in Examples 6.3 and 6.7 and Exercise 6.6, although Chapter 5 no longer defines it.
  • The Additional Exercises block has been removed, leaving Exercises 6.1 to 6.8.

Key formulas & results

Everything you need to memorise, in one card. Screenshot this for revision.

Magnetic flux
Flux = B A cos(theta), measured in weber
theta is measured from the NORMAL to the area, not from its plane; 1 Wb = 1 T m squared
Faraday's law of induction
emf = -(rate of change of flux); for N turns, emf = -N (rate of change of flux)
The emf exists only while the flux is changing; a steady field induces nothing
Lenz's law
The induced current opposes the CHANGE in flux that produced it
An approaching pole is repelled and a receding pole attracted; both oppose the change
Motional emf
emf = B l v for a rod moving perpendicular to the field
Derivable both from the rate of area change and from the Lorentz force on the free charges
Rotating rod
emf = (1/2) B omega l squared
The speed varies along the rod, so the contribution must be integrated; the factor of one half is easily lost
Mutual inductance
N2 x Flux2 = M I1, and emf2 = -M (rate of change of I1)
M depends only on geometry and the medium, never on the currents, and is symmetric between the coils
Self-inductance
N x Flux = L I, and emf = -L (rate of change of I)
The induced emf opposes the change in the coil's own current, which is why it is called back emf
Inductance of a long solenoid
L = mu_0 n squared A l
Note the SQUARE of the turns per unit length: doubling the winding density quadruples L
Energy stored in an inductor
U = (1/2) L I squared
The magnetic counterpart of half C V squared for a capacitor; stored through current rather than voltage
AC generator emf
emf = N B A omega sin(omega t) = peak emf x sin(omega t)
Peak emf = N B A omega, reached when the plane of the coil is PARALLEL to the field
Flux linkage change
Change in flux linkage = M x change in current
This is what a question asks for when it wants webers rather than volts
Three ways to change flux
Change B, change A, or change theta
Changing theta by rotating a coil is exactly how a generator works
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Common mistakes & fixes

These are the exact errors that cost students marks in board exams. Read them once, save yourself the trouble.

WATCH OUT
Measuring the flux angle from the plane of the coil
In Flux = B A cos(theta), theta is the angle between B and the NORMAL to the surface. If the field lies in the plane of the loop, theta is 90 degrees and the flux is zero, which is why Exercise 6.1(f) induces no current at all.
WATCH OUT
Dropping the factor of one half for a rod rotating about one end
The speed v = omega r varies along the rod, so the emf must be integrated, giving half B omega l squared. Using B omega l squared doubles the answer.
WATCH OUT
Saying the induced current opposes the field rather than the change
Lenz's law opposes the CHANGE. A receding north pole induces a south pole that attracts it back, which is not opposition to the field itself but to its decrease.
WATCH OUT
Expecting an emf from a steady current or a stationary magnet
Only a changing flux induces an emf. In Exercise 6.3 the emf exists solely during the interval in which the solenoid current is switched off.
WATCH OUT
Using n as the total number of turns in L = mu_0 n squared A l
n is the turns per unit length. The dependence is on its square, so an error here is magnified.
WATCH OUT
Computing an emf when the question asks for the change in flux linkage
Exercise 6.8 wants M multiplied by the change in current, which is 30 Wb. The 0.5 s is not needed for that part, though it would give an emf of 60 V.
WATCH OUT
Thinking the generator emf peaks when the flux is maximum
The emf follows the RATE of change of flux. It is maximum where the flux passes through zero, with the coil plane parallel to the field, and zero where the flux is largest.
WATCH OUT
Giving only the magnitude in a motional emf question that asks about polarity
Use the Lorentz force on the free charges to see which end they accumulate at. That end is negative and the other is at the higher potential, which is what Exercise 6.6(c) requires.
WATCH OUT
Revising eddy currents, magnetic braking and the induction furnace
Eddy currents have been removed from this chapter. A search returns zero hits for eddy, magnetic braking, induction furnace and dead beat.
WATCH OUT
Being confused by the Earth's horizontal field component appearing without definition
Examples 6.3 and 6.7 and Exercise 6.6 all use it, but Chapter 5's section on the Earth's magnetism has been removed. Treat it simply as the component of the field parallel to the ground.

Practice problems

Work through this chapter's problems as a readiness check — reveal each solution, mark yourself honestly, and get your gap report at the end.

Readiness check

Are you exam-ready for Electromagnetic Induction?

9 problems from this chapter. Try each one, reveal the worked solution, mark yourself honestly — get your gap report at the end.

9 questions~6 min

5-minute revision

The whole chapter, distilled. Read this the night before the exam.

  • Induction needs a CHANGING flux; a steady field or current induces nothing
  • Flux = B A cos(theta), with theta from the normal; the unit is the weber
  • Flux changes through B, through A, or through theta
  • Faraday: emf = -(rate of change of flux), and -N times it for N turns
  • Lenz: the induced current opposes the CHANGE, not the field
  • An approaching pole is repelled, a receding pole attracted
  • The minus sign is conservation of energy; the opposite sign would create energy from nothing
  • Motional emf = B l v, with the polarity fixed by the Lorentz force on the carriers
  • A rod rotating about one end gives half B omega l squared
  • Mutual inductance: emf2 = -M times the rate of change of I1; M is geometric and symmetric
  • Self-inductance: emf = -L times the rate of change of I
  • Solenoid inductance L = mu_0 n squared A l, with n the turns per unit length
  • Energy stored in an inductor is half L I squared
  • AC generator emf = N B A omega sin(omega t), peaking when the coil plane is parallel to B
  • Eddy currents and the quantitative energy section have been removed
  • The Earth's horizontal field component is used here although Chapter 5 no longer defines it
  • The Additional Exercises block has been removed, leaving Exercises 6.1 to 6.8

CBSE marks blueprint

Where the marks come from in this chapter — so you can plan your prep.

Typical chapter weightage: Unit IV: Electromagnetic Induction and Alternating Currents, no chapter-wise split published by CBSE

Question typeMarks eachTypical countWhat it tests
Magnetic Flux, Faraday's Law and Lenz's Law2-31Flux calculation, emf magnitude, and the direction of the induced current
Motional emf and the Rotating Rod3-41Blv, integration for a rotating rod, and the polarity of the induced emf
Self-Inductance, Mutual Inductance, Energy Stored in an Inductor and the AC Generator4-51Flux linkage, solenoid inductance, stored energy, and the sinusoidal generator emf
Prep strategy
  • Identify which of B, A or theta is changing before writing any equation
  • Measure the flux angle from the normal, and say so explicitly in your working
  • For rotating rods, write the integral rather than reaching for Blv
  • Answer direction questions in two steps: is the flux rising or falling, and what opposes that
  • When a question asks for flux linkage, give webers; when it asks for emf, give volts
  • Do not spend time on eddy currents, which have been removed from this chapter

Where this shows up in the real world

This chapter isn't just an exam topic — it lives in the world around you.

Power generation

Every alternator in a power station is the AC generator of section 6.8, with turbines rotating coils in a magnetic field to give a sinusoidal emf.

Transformers

Mutual inductance between primary and secondary windings on a shared core is what allows voltages to be stepped up for transmission and down for use.

Induction cooktops

A rapidly changing field induces currents directly in the base of a metal pan, heating it without heating the hob surface.

Wireless charging

A coil in the charging pad and another in the device are coupled by mutual inductance, transferring energy across a small gap with no contacts.

Electric guitar pickups

A vibrating steel string changes the flux through a coil wound around a small magnet, inducing a voltage that reproduces the string's motion.

Exam strategy

Battle-tested tips from teachers and toppers for this chapter.

1
Name the changing quantity, B or A or theta, in the first line of every answer
2
Write the flux angle as measured from the normal and state that explicitly
3
Reach for the integral, not Blv, whenever anything rotates about one end
4
Split direction questions into rising-or-falling flux, then what opposes it, then the right-hand rule
5
Read carefully whether the answer wanted is a flux linkage in webers or an emf in volts
6
For a loop leaving a field, use the perpendicular edge for the emf and the parallel edge for the duration
7
Justify a Lenz's law direction with the energy argument if the question asks why, not just which way

Going beyond the textbook

For olympiad aspirants and curious learners — topics that build on this chapter.

STRETCH
Faraday's law in differential form as one of Maxwell's equations, with the curl of E equal to minus the rate of change of B
STRETCH
Eddy current braking and the skin effect, both consequences of induced currents in extended conductors
STRETCH
The betatron, in which a changing magnetic flux accelerates electrons around a fixed circular orbit
STRETCH
Inductive coupling and the transformer equations derived from mutual inductance including leakage flux
🚀

JEE Main & Advanced practice

Competitive-level problems on this chapter, above the board pattern. Try each one on paper before opening the solution.

JEE MainLoop leaving a field regionMotional emf and its duration

A rectangular loop of sides 8 cm and 2 cm with a small cut moves out of a region of uniform field T normal to the loop, at cm s. Find the induced voltage and how long it lasts if the motion is (a) along the long side, (b) along the short side.

Stuck? Show the approach

Only the edge perpendicular to the motion sweeps out area, so identify which side sets the emf and which sets the duration.

Show the full solution

The emf is , where is the length of the edge perpendicular to the velocity, since only that edge sweeps out new area. (a) Moving normal to the longer side, the cm edge cuts across the field, so V. The loop must travel its other dimension, cm, to leave the region, so at cm s the voltage lasts s. (b) Moving normal to the shorter side, the cm edge is now the active one, giving V, and the loop must now cover cm, so the voltage lasts s. The larger emf is the shorter-lived one.

Answer: (a) 2.4 x 10^-4 V lasting 2 s (b) 6 x 10^-5 V lasting 8 s
The trap

Using the same side for both the emf and the duration. The edge perpendicular to the motion sets the emf; the edge along the motion sets how long it lasts.

JEE MainRotating rodIntegrating a varying motional emf

A metallic rod m long is rotated at rad s about an axis through one end, perpendicular to a uniform field of T. Find the emf between the centre and the ring.

Stuck? Show the approach

Recognise that v = omega r varies along the rod, so the emf must be integrated rather than taken as Blv.

Show the full solution

Take an element at distance from the axis. Its speed is , so it contributes . Integrating from to : . Substituting, V. Equivalently, the rod sweeps out area at the rate per second, which gives the same result.

Answer: emf = 100 V
The trap

Applying Blv with v taken as the tip speed, which gives 200 V — exactly double. The speed is zero at the axis and maximum at the tip, so the average is half the tip value.

JEE AdvancedPolarity of a motional emfLorentz force determines which end is positive

A horizontal wire m long running east to west falls at m s, at right angles to the horizontal component of the Earth's field, T. Find the emf, its direction, and which end is at the higher potential.

Stuck? Show the approach

Use Blv for the magnitude, then apply the Lorentz force on the free charges inside the wire to settle the polarity.

Show the full solution

The magnitude is V. For the direction, consider a free positive charge in the wire: it moves downwards with the wire, and with downward and horizontal northward pushes it towards the east. Positive charge therefore accumulates at the eastern end, making that end the higher potential, and inside the wire the emf drives current from west to east.

Answer: 1.5 x 10^-3 V; emf drives current west to east inside the wire; the eastern end is at the higher potential
The trap

Answering only the magnitude. The flux argument gives the size of the emf but says nothing about polarity, which needs the Lorentz force on the carriers.

Where else this chapter is tested

CBSE board isn't the only one — other exams test this chapter too.

CBSE Class 12 BoardHigh
JEE MainHigh
NEETHigh
JEE AdvancedMedium

Questions students ask

The real ones — pulled from the Q&A community and tutor sessions.

Because the two are different, and the distinction decides half the direction questions. If a north pole approaches a coil the flux through it is increasing, so the induced current makes a north pole facing the magnet and repels it. If the same north pole recedes, the flux is decreasing, and the induced current now makes a south pole that attracts the magnet and tries to hold it back. In one case the induced field aids the original and in the other it opposes it, but in both cases it opposes the change.

It comes from integrating a varying speed. A rod rotating about one end has zero speed at the axis and maximum speed omega l at the tip, so no single value of v applies. Taking an element dr at distance r, its contribution is B omega r dr, and integrating from zero to l gives half B omega l squared. Using Blv with the tip speed treats every part of the rod as moving at the maximum speed, which doubles the answer.

Because Faraday's law involves the rate of change of flux, not the flux itself. When the plane of the coil is parallel to the field, the flux through it is momentarily zero, but it is passing through zero at its fastest rate, so the emf is at its peak. When the plane is perpendicular to the field the flux is maximum but momentarily stationary, so its rate of change and therefore the emf are zero. Sketching the cosine flux against its sine derivative makes this immediate.

Use the Lorentz force rather than the flux. Consider a free positive charge inside the rod: it shares the rod's velocity, so it feels a force q times v cross B. Work out which way that points and the positive charge accumulates at that end, making it the higher potential. The flux route gives only the magnitude of the emf. This is exactly what Exercise 6.6(c) is testing.

Self-inductance L relates a coil's own flux linkage to its own current, and produces a back emf opposing any change in that current. Mutual inductance M relates the flux linking a second coil to the current in the first, and produces an emf in the second coil when the first coil's current changes. Both are fixed by geometry and the medium, never by the currents themselves, and M is the same in either direction between a given pair of coils.

Not in this chapter as it now stands. Older editions had a full section on them, covering magnetic braking in trains, the induction furnace, dead-beat galvanometers and electric power meters. A search of the current chapter returns zero hits for eddy, magnetic braking, induction furnace and dead beat. The section titled Energy Consideration: A Quantitative Study has also gone, though the qualitative energy argument survives inside the treatment of Lenz's law.

Because the section on the Earth's magnetism was removed from Chapter 5 while the worked examples and exercise here were kept. Example 6.3, Example 6.7 and Exercise 6.6 all supply a value for it, so nothing is missing for solving them. Read it as the component of the Earth's magnetic field parallel to the ground, and use the supplied number directly. There is no need to learn declination or the angle of dip, which have both been removed.
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Last reviewed on 18 August 2026. Written and reviewed by subject-matter experts — read about our process.
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