CBSE Class 12 Physics Chapter 6 — Electromagnetic Induction: NCERT Notes, Faraday’s Law and Important Questions 2027

Last Updated: May 2026

CBSE Class 12 Physics Chapter 6 — Electromagnetic Induction is a 6-mark chapter in the Class 12 board exam (one short answer + one long answer typically). It is also high-yield in NEET (1–2 questions) and JEE Main (1 question). Faraday’s law, Lenz’s law and motional EMF questions appear in every CBSE 12 board paper since 2019. This 1,800-word CBSE Class 12 Physics Electromagnetic Induction guide covers all theorems, diagrams, derivations and 30 important questions with answer pointers.

1. Magnetic Flux (Φ)

Φ = B·A·cosθ, where B is magnetic field, A is area vector, θ is angle between B and area vector.
SI unit: Weber (Wb) = T·m². 1 Wb = 1 V·s.

2. Faraday’s Laws of Electromagnetic Induction

First Law: Whenever the magnetic flux linked with a closed circuit changes, an EMF is induced.
Second Law: The magnitude of the induced EMF equals the rate of change of magnetic flux:
ε = -dΦ/dt
The negative sign reflects Lenz’s law.

3. Lenz’s Law

The direction of induced current is such that it opposes the change in flux that produced it. This is a consequence of conservation of energy.

4. Motional EMF

A conductor of length L moving with velocity v perpendicular to a magnetic field B experiences EMF:
ε = BLv

The induced current direction follows from F = qv × B applied to charge carriers in the rod.

Derivation of Motional EMF

Consider a conducting rod PQ of length L sliding on parallel rails in a magnetic field B normal to the plane. As PQ moves with velocity v, area enclosed increases at rate L·v. So Φ = B·A → dΦ/dt = B·L·v. By Faraday’s law, ε = BLv.

5. Self-Induction

An induced EMF in a coil due to change in current through the same coil.
ε = -L·(dI/dt)
L is self-inductance (henry, H). For a solenoid: L = μ₀·n²·V, where n = turns per unit length, V = volume.

6. Mutual Induction

When current change in coil 1 induces EMF in coil 2 (and vice versa).
ε₂ = -M·(dI₁/dt)
M is mutual inductance. Coefficient of coupling k = M/√(L₁·L₂); 0 ≤ k ≤ 1.

7. Energy Stored in an Inductor

U = (1/2)·L·I²
Energy density: u = B²/(2μ₀)

8. Eddy Currents

Currents induced in bulk pieces of metal when flux changes. Cause heating (Joule’s law). Reduced by laminating cores. Applications: induction furnaces, electromagnetic braking, magnetic levitation, induction cookers.

9. AC Generator (Working Principle)

A coil rotated at constant angular velocity ω in a uniform magnetic field B. Flux Φ = NBA·cos(ωt). Induced EMF ε = NBA·ω·sin(ωt) = ε₀·sin(ωt).
Maximum EMF: ε₀ = NBAω.

10. Worked Numerical Examples

Example 1

A 100-turn coil of area 0.05 m² rotates at 1500 rpm in a uniform magnetic field of 0.5 T. Find peak EMF.
ω = 2π × 25 = 50π rad/s
ε₀ = NBAω = 100 × 0.5 × 0.05 × 50π = 125π ≈ 392.7 V

Example 2

A 0.5 m rod moves with v = 4 m/s perpendicular to B = 0.2 T. Find motional EMF.
ε = BLv = 0.2 × 0.5 × 4 = 0.4 V

Example 3

A solenoid of 1000 turns, length 0.5 m, area 10 cm². Find L.
n = 2000 turns/m. L = μ₀ · n² · V = (4π×10⁻⁷)·(2000)²·(10⁻³)·(0.5/0.5)… carefully: V = A·l = 10⁻³ × 0.5 = 5×10⁻⁴ m³.
L = 4π×10⁻⁷ × 4×10⁶ × 5×10⁻⁴ = 4π×10⁻⁷ × 2000 = 2.51×10⁻³ H ≈ 2.5 mH

11. 30 Important Questions (Sample 12)

1. State Faraday’s laws of electromagnetic induction. [2 marks]
2. Define magnetic flux. Give its SI unit. [1 mark]
3. State and explain Lenz’s law. Show that it is a consequence of conservation of energy. [3 marks]
4. Derive an expression for the EMF induced in a rod moving with velocity v perpendicular to a magnetic field B. [3 marks]
5. Define self-inductance. State its SI unit. [2 marks]
6. Derive the formula for self-inductance of a long solenoid. [3 marks]
7. What are eddy currents? Mention two applications and two methods to minimise their loss. [3 marks]
8. Describe the principle, construction, and working of an AC generator. [5 marks]
9. The number of turns in a solenoid is doubled and length halved. How does self-inductance change? [2 marks] — Answer: L = μ₀n²V; n doubles, V halves → L becomes 8L (since n² becomes 4×, V becomes 0.5×, net 2× × 4 = ?). Actually n=N/l. N→2N, l→0.5l. n=4n_old. L=μ₀(4n)²·(A·0.5l)=16μ₀n²·0.5·A·l=8L
10. A bar magnet falls through a vertical copper pipe. Will it accelerate at g? Explain. [3 marks]
11. State two factors on which mutual inductance of two coils depends. [2 marks]
12. Calculate the energy stored in an inductor of 0.5 H carrying 2 A. [2 marks]

12. Common Mistakes

• Forgetting the negative sign in Faraday’s law.
• Confusing motional EMF formula (BLv) with self-induction (-LdI/dt).
• Misapplying Lenz’s law direction — always check what ‘the change’ is.
• Forgetting unit conversions (cm² → m², gauss → tesla).

Frequently Asked Questions

Q1. CBSE Class 12 board weightage for this chapter?

5–6 marks (Unit IV — EMI and Alternating Currents = 8 marks combined, this chapter is 5–6).

Q2. Most repeated question in CBSE 12 boards?

Derivation of motional EMF (BLv) — appeared in 2019, 2021, 2023, 2024.

Q3. NCERT exercises enough?

For boards: yes. For NEET/JEE: solve HC Verma + DC Pandey additionally.

Internal Resources

• CBSE Class 12 Physics Chapter 1 — Electric Charges and Fields
• CBSE Class 12 Chemistry — Solutions
• Board Exam FAQ
• Board Exam Mock Test

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