Electromagnetic Induction Class 12 Notes | CBSE Physics Chapter 6

Chapter summary

Electromagnetic Induction covers magnetic flux, Faraday’s and Lenz’s laws of induction, motional EMF in moving conductors, eddy currents, and self and mutual inductance. It is an important NEET chapter where questions test the flux formula, the direction of induced current, the EMF of a moving rod, and the inductance of solenoids along with the energy stored in a magnetic field.

Chapter notes

Key Concepts

1. Magnetic Flux

Magnetic flux (Φ) through a surface is the total number of magnetic field lines passing through it.

Φ = B · A = BA cos θ

Unit: Weber (Wb) = T·m²


2. Faraday’s Laws of Electromagnetic Induction

First Law: Whenever the magnetic flux linked with a circuit changes, an EMF is induced in the circuit.

Second Law: The magnitude of induced EMF is equal to the rate of change of magnetic flux:

ε = −dΦ/dt

For a coil of N turns: ε = −N(dΦ/dt)

The negative sign represents Lenz’s law.

Ways to Change Flux (and Induce EMF)

  • Change the magnetic field strength (B)
  • Change the area of the loop (A)
  • Change the angle between B and the normal to the loop (θ)
  • Move the conductor in the field

3. Lenz’s Law

The direction of the induced current is such that it opposes the change in flux that produced it.

This is a consequence of the law of conservation of energy. If the induced current aided the change, it would create a perpetual motion machine - which is impossible.


4. Motional EMF

When a conductor of length l moves with velocity v perpendicular to a magnetic field B:

ε = Blv

This is because the magnetic force on the free electrons in the conductor creates a potential difference.


5. Self-Inductance

When current through a coil changes, the changing magnetic flux induces an EMF in the same coil that opposes the change.

ε = −L(dI/dt)

L = NΦ/I (self-inductance, unit: Henry, H)

For a solenoid: L = μ₀n²Al (n = turns per unit length, A = area, l = length)

Energy stored in an inductor: U = ½LI²


6. Mutual Inductance

When current in one coil changes, the changing flux induces an EMF in a nearby coil.

ε₂ = −M(dI₁/dt)

M = coefficient of mutual inductance (unit: Henry)

For two coaxial solenoids: M = μ₀n₁n₂Al


7. Eddy Currents

Eddy currents are loops of current induced in bulk conductors when exposed to changing magnetic fields.

Applications: Electromagnetic braking (trains), induction furnace, speedometers, electromagnetic damping

Disadvantage: Energy loss as heat in transformer cores. Minimised by using laminated cores.


Important Definitions

TermDefinition
Magnetic fluxΦ = BA cos θ - total field lines through a surface (unit: Weber)
Electromagnetic inductionProduction of EMF due to changing magnetic flux
Lenz’s lawInduced current opposes the change in flux that caused it
Self-inductance (L)Property of a coil to oppose change in its own current; ε = −LdI/dt
Mutual inductance (M)EMF induced in one coil due to changing current in another nearby coil
Eddy currentsCirculating currents induced in bulk conductors by changing magnetic fields

Read the rest of the chapter →Hide the rest ↑

Solved Examples

Example 1

A coil of 100 turns has flux changing from 0.02 Wb to 0.04 Wb in 0.1 s. Find the induced EMF.

Answer: ε = N(ΔΦ/Δt) = 100 × (0.04 − 0.02)/0.1 = 100 × 0.2 = 20 V

Example 2

A rod of length 50 cm moves at 4 m/s perpendicular to a field of 0.5 T. Find the motional EMF.

Answer: ε = Blv = 0.5 × 0.5 × 4 = 1 V

Example 3

A solenoid of self-inductance 2 H carries a current of 3 A. Find the energy stored.

Answer: U = ½LI² = ½ × 2 × 9 = 9 J

Example 4

The mutual inductance of two coils is 0.5 H. If the current in the first coil changes at 10 A/s, find the EMF in the second coil.

Answer: ε = M(dI/dt) = 0.5 × 10 = 5 V


Important Questions for Board Exams

1-Mark

  1. State Lenz’s law.
  2. What is the SI unit of self-inductance?

3-Mark

  1. State Faraday’s laws of electromagnetic induction. Give one application.
  2. Derive the expression for motional EMF.
  3. What are eddy currents? Give two applications and one disadvantage.

5-Mark

  1. Derive the expression for self-inductance of a long solenoid. Also find the energy stored.
  2. State and explain Faraday’s laws and Lenz’s law. Show that Lenz’s law is a consequence of conservation of energy.

Quick Revision Points

  • Φ = BA cos θ; ε = −NdΦ/dt (Faraday’s law)
  • Lenz’s law: induced current opposes the cause (conservation of energy)
  • Motional EMF: ε = Blv
  • Self-inductance: ε = −LdI/dt; L = μ₀n²Al (solenoid); U = ½LI²
  • Mutual inductance: ε₂ = −MdI₁/dt
  • Eddy currents: used in braking, induction furnace; minimised by lamination

Previous: Ch 5 - Magnetism and Matter
Next: Ch 7 - Alternating Current

🃏 Flash Cards: Electromagnetic Induction

Class 12 Physics · Chapter 6 – swipe through all 10 cards to understand the whole chapter.

🧲Start here1/10

Magnetic Flux

Flux counts how many field lines pass through a surface.

Φ_B = B A cosθ

θ is from the area normal, not the surface. Unit: weber (Wb) = T·m2

  • Max when B ⊥ surface (θ = 0); zero when B ∥ surface (θ = 90°)
  • Trap: angle with plane → angle with normal = 90° − that
  • For N turns, total linkage = N Φ_B
Core law2/10

Faraday’s Law

A changing flux induces an EMF; a steady flux induces nothing.

ε = −N (dΦ_B / dt)

Faster change → bigger EMF. N turns → N× the EMF.

  • Only a CHANGE in flux drives induction
  • Flux changes 3 ways: change B, change A, or rotate (change θ)
  • Magnitude = N · ΔΦ_B / Δt for uniform change
🔄Direction rule3/10

Lenz’s Law

The induced current opposes the very change that caused it.

minus sign in ε = −N dΦ_B/dt · opposes the change

This is just energy conservation — no free energy.

  • Flux increasing → current opposes; decreasing → current maintains it
  • Magnet pushed in is repelled; pulled out is attracted
  • You must do work; that work becomes the electrical energy
🔋NEET favourite4/10

Induced Charge

Total charge depends only on flux change and resistance, not speed.

q = N ΔΦ_B / R

Fast or slow gives the same CHARGE; only current/EMF differ.

  • Derived from ε = IR and q = ∫I dt
  • Path-independent — only ΔΦ_B and R matter
  • Common MCQ: same q whether magnet moves quick or slow
📏Key case5/10

Motional EMF

A rod sliding through a field becomes a tiny battery.

ε = B v l

Only the v-component ⊥ to both B and rod counts; if v ∥ B, ε = 0.

  • Same as Faraday’s law: rod sweeps area at rate l·v
  • On rails of resistance R, induced current I = B v l / R
  • It is a real EMF source even in a steady field
🛑Force & power6/10

Retarding Force on the Rod

The induced current makes the field push back on the moving rod.

F = B2l2v / R · P = B2l2v2 / R

F ∝ v but power dissipated ∝ v2. Mechanical power → heat in R.

  • Force opposes motion (Lenz’s law): F = B I l
  • Power supplied by the pusher equals ε2/R dissipated
  • Rotating rod about one end: ε = ½ B ω l2
🌀Bulk conductor7/10

Eddy Currents

Changing flux sets up looping currents inside solid metal.

Closed loops in metal → heat (I2R) · always oppose motion

Reduce losses with LAMINATED cores (thin insulated sheets).

  • Uses: electromagnetic braking, induction furnace/cooktop, dead-beat galvanometer
  • Magnet falls slowly through a copper pipe — eddy currents damp it
  • Lamination raises path resistance, cutting eddy heat loss
🔁Self opposition8/10

Self-Inductance

A coil opposes changes in its OWN current via a back-EMF.

ε = −L (dI/dt) · N Φ_B = L I

L depends on geometry/core only, not on current. Unit: henry (H).

  • Long solenoid: L = μ0 N2 A / l (so L ∝ N2 — double N → 4× L)
  • Energy stored: U = ½ L I2 (magnetic analogue of ½ C V2)
  • Inductor = electrical inertia; resists sudden current change
🔗Coupling9/10

Mutual Inductance

Changing current in one coil induces an EMF in a neighbour.

ε2 = −M (dI1/dt) · M = k √(L1 L2)

M is symmetric: M12 = M21. Unit: henry (H). 0 ≤ k ≤ 1.

  • Coaxial solenoids: M = μ0 N1 N2 A / l
  • k = 1 means perfect coupling (all flux links both coils)
  • Working principle of the transformer
🌐Application10/10

AC Generator

Rotating a coil in a field continuously changes flux → alternating EMF.

ε = N B A ω sin(ωt) · ε0 = N B A ω

Output is sinusoidal; peak EMF ε0 rises with N, B, A and speed ω.

  • Direct use of Faraday’s law with θ = ωt, so Φ = N B A cos(ωt)
  • Converts mechanical energy into electrical energy
  • EMF is zero when coil plane ⊥ B (flux max), peak when coil plane ∥ B
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📝 Practice Electromagnetic Induction — 10 NEET PYQs
Real previous-year questions · with answers & solutions
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Tap an option to check your answer and see the worked solution. Every question is a real NEET previous-year question.
Q1NEET 2021
Two conducting circular loops of radii R₁ and R₂ are placed in the same plane with their centres coinciding. If R₁ ≫ R₂, the mutual inductance M between them is directly proportional to:
Correct answer: D. Field at the centre of the large loop due to current I: B = (μ₀ I)/(2R₁). Flux through the small loop: Φ = Bπ R₂² = (μ₀ Iπ R₂²)/(2R₁). Since Φ = MI, M = (μ₀π R₂²)/(2R₁) ∝ (R₂²)/(R₁).
🔎 See the full step-by-step solution in the app →
Q2NEET 2020
The magnetic flux (in Wb) linked with a coil is given by Φ = 5t² + 3t + 16, where t is in seconds. The magnitude of the induced EMF in the coil at the end of the fourth second is:
Correct answer: D. The instantaneous EMF is ε = (dΦ)/(dt) = 10t + 3. The EMF ‘at the end of the fourth second’ here means the average over the fourth second: ε₄ – ε₃ = (10×4+3) – (10×3+3) = 43 – 33 = 10 V.
🔎 See the full step-by-step solution in the app →
Q3NEET 2020
A wheel with 20 metallic spokes, each 1 m long, is rotated at 120 rpm in a plane perpendicular to a magnetic field of 0.4 G (1 G = 10⁻⁴ T). The induced EMF between the axle and the rim of the wheel is:
Correct answer: A. All spokes are in parallel and develop the same EMF, so the number of spokes does not matter. For a rotating rod ε = (1)/(2)Bω l². ω = 2π(120)/(60) = 4π. ε = (1)/(2)×0.4×10⁻⁴×4π×1² = 2.51×10⁻⁴ V.
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Q4NEET 2020
A light bulb and an inductor coil are connected in series to an AC source through a key. The key is closed and after some time an iron rod is inserted into the interior of the inductor. The glow of the bulb:
Correct answer: A. Inserting an iron core increases the inductance L, hence the inductive reactance X_L = ω L rises. The current I = V/X_L therefore falls, so the bulb’s glow decreases.
🔎 See the full step-by-step solution in the app →
Q5NEET 2019
In which of the following devices is the eddy-current effect NOT used?
Correct answer: C. An electric heater works on the Joule heating (I²R) of a direct current through a resistive element, not on eddy currents. Magnetic braking, an electromagnet’s damping, and the induction furnace all involve eddy currents.
🔎 See the full step-by-step solution in the app →
Q6NEET 2019
A coil of 800 turns and effective area 0.05 m² is kept perpendicular to a magnetic field 5×10⁻⁵ T. When the plane of the coil is rotated by 90° about any of its coplanar axes in 0.1 s, the EMF induced in the coil is:
Correct answer: C. ε = N(ΔΦ)/(Δ t) = N B A(cos0° – cos90°)/(Δ t) = (800×5×10⁻⁵×0.05×(1-0))/(0.1) = (2×10⁻³)/(0.1) = 0.02 V.
🔎 See the full step-by-step solution in the app →
Q7NEET 2019
A cycle wheel of radius 0.5 m is rotated with a constant angular velocity of 10 rad/s in a region of magnetic field 0.1 T which is perpendicular to the plane of the wheel. The EMF generated between its centre and the rim is:
Correct answer: B. For a rotating spoke, ε = (1)/(2)Bω r² = (1)/(2)×0.1×10×(0.5)² = (1)/(2)×0.1×10×0.25 = 0.125 V.
🔎 See the full step-by-step solution in the app →
Q8NEET 2018
The magnetic potential energy stored in a certain inductor is 25 mJ when the current in the inductor is 60 mA. The self-inductance of this inductor is:
Correct answer: D. U = (1)/(2)LI² ⇒ L = (2U)/(I²) = (2×25×10⁻³)/((60×10⁻³)²) = (0.05)/(3.6×10⁻³) = 13.89 H.
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Q9NEET 2008
A circular disc of radius 0.2 m is placed in a uniform magnetic field of induction (1)/(π) Wb/m² in such a way that its axis makes an angle of 60° with B. The magnetic flux linked with the disc is:
Correct answer: A. The axis of the disc is its normal, so the angle between B and the area vector is 60°. Φ = BAcosθ = (1)/(π)×π(0.2)²×cos60° = (0.04)×0.5 = 0.02 Wb.
🔎 See the full step-by-step solution in the app →
Q10NEET 1988
Eddy currents are produced when:
Correct answer: A. Eddy currents are induced in the body of a bulk conductor only when the magnetic flux linked with it changes, i.e. when a metal is placed in a varying (changing) magnetic field. A steady field induces nothing.
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Frequently Asked Questions

What is magnetic flux and its SI unit?

Magnetic flux is a measure of how much magnetic field passes through a surface. For a uniform field B through area A, flux equals B times A times cosine of the angle between B and the area vector (the normal). Its SI unit is the weber (Wb), where one weber equals one tesla times one square metre.

State Faraday’s and Lenz’s laws.

Faraday’s law states that the induced EMF equals the negative rate of change of magnetic flux linkage, so a changing flux induces an EMF. Lenz’s law explains the negative sign: the induced current flows in a direction that opposes the change that produced it, which is a consequence of conservation of energy.

What is motional EMF and its formula?

Motional EMF is the EMF induced in a conductor moving through a magnetic field. For a rod of length l moving with velocity v perpendicular to a field B, the induced EMF equals B times v times l. It is just Faraday’s law applied to a conductor sweeping area in the field.

What is the difference between self-inductance and mutual inductance?

Self-inductance is the property of a coil to oppose a change in its own current by inducing a back-EMF, measured in henry and depending only on geometry and core material. Mutual inductance is the EMF induced in one coil due to a changing current in a neighbouring coil, and it is symmetric, meaning it is the same whichever coil carries the current.

What are eddy currents and where are they useful?

Eddy currents are loops of induced current set up in the body of a solid conductor when the magnetic flux through it changes, and by Lenz’s law they oppose the change and dissipate energy as heat. They are used in electromagnetic braking, induction furnaces, induction cooktops and dead-beat galvanometers, and are reduced by laminating cores.

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