Electromagnetic Induction – Class 12 Physics Chapter 6 Notes & NCERT PDF

Introduction

Chapter 4 showed that an electric current produces a magnetic field. In 1830–31, Michael Faraday in England and Joseph Henry in the USA showed the reverse: a changing magnetic field produces an electric current. This phenomenon, electromagnetic induction, is the working principle of generators, transformers, induction cooktops, metal detectors and the wireless charging pad for your phone. Almost all the electrical energy used in India is produced by generators that rely on it. This chapter explains Faraday's law, Lenz's law, motional emf, mutual and self-inductance, and the ac generator.

Key Concepts

6.2 The Experiments of Faraday and Henry

6.3 Magnetic Flux

Φ_B = B·A = BA cos θ, where θ is the angle between B and the normal to the area. SI unit: weber (Wb) = T m².

6.4 Faraday's Law of Induction

The induced emf equals the rate of change of flux: ε = −dΦ_B/dt; for N turns, ε = −N dΦ_B/dt. The flux can change by changing B, the area A, or the angle θ.

6.5 Lenz's Law and Conservation of Energy

The induced current flows in a direction that opposes the change producing it. This is the meaning of the minus sign. If a north pole approaches a coil, the near face of the coil becomes a north pole and repels it; work must be done to push the magnet, and that work appears as electrical energy (Joule heat). If the induced current aided the change, we would get energy from nothing, violating conservation of energy.

6.6 Motional Electromotive Force

A rod of length l moving with velocity v perpendicular to B has ε = Blv. The same result follows from the Lorentz force qvB acting on free charges in the rod. A rod of length R rotating with angular speed ω about one end in a perpendicular field has ε = ½BωR².

6.7 Inductance

Flux linkage is proportional to current: NΦ ∝ I. The constant of proportionality is the inductance, unit henry (H).

6.8 AC Generator

A coil of N turns and area A rotates with angular speed ω in a field B. Flux Φ = BA cos ωt, so ε = NBAω sin ωt = ε₀ sin ωt, with peak ε₀ = NBAω. The emf is maximum when the plane of the coil is parallel to B (flux zero, changing fastest). Mains frequency is 50 Hz in India and 60 Hz in the USA. In hydroelectric plants falling water turns the coil; in thermal and nuclear plants, steam does.

Worked Examples

Example 1 (Book Example 6.2). A square loop of side 10 cm and resistance 0.5 Ω lies in a field of 0.10 T at 45° to its normal; the field falls to zero in 0.70 s. Φ₀ = BA cos 45° = 0.10 × 10⁻² × 0.707 = 7.07 × 10⁻⁴ Wb. ε = 7.07 × 10⁻⁴/0.70 ≈ 1.0 mV, I = ε/R = 2 mA.

Example 2 (Book Example 6.3). A coil of radius 10 cm, 500 turns and 2 Ω lies in the earth's field of 3.0 × 10⁻⁵ T and is turned through 180° in 0.25 s. ΔΦ = 2NBA = 2 × 500 × 3.0 × 10⁻⁵ × π(0.1)², so ε = ΔΦ/Δt ≈ 3.8 × 10⁻³ V and I ≈ 1.9 × 10⁻³ A.

Example 3 (Book Example 6.6). A 1 m rod rotates at 50 rev/s about one end in a 1 T field. ω = 2π × 50 rad/s, ε = ½BωR² = ½ × 1 × 314 × 1 = 157 V.

Example 4 (Book Example 6.7). A wheel with 10 spokes of length 0.5 m turns at 120 rev/min in a 0.4 G field. ε = ½ × 0.4 × 10⁻⁴ × 4π × 0.25 = 6.28 × 10⁻⁵ V. The number of spokes does not matter because the spokes are in parallel.

Example 5 (Book Example 6.10). Kamla's cycle generator: N = 100, A = 0.10 m², 0.5 rev/s, B = 0.01 T. ε₀ = NBA(2πν) = 100 × 0.01 × 0.10 × π = 0.314 V.

Formula and Key-Facts Sheet

Board Exam Focus

Frequent questions: state Faraday's and Lenz's laws; show that Lenz's law follows from energy conservation; derive ε = Blv; derive L or M of a solenoid; derive ½LI²; explain the ac generator; numericals from Exercises 6.3–6.8.

Answer outline: "Derive the expression for motional emf." Rod PQ of length l moves with v on rails in a field B normal to the plane. Area enclosed = lx, so Φ = Blx. ε = −dΦ/dt = −Bl dx/dt = Blv (magnitude). Alternatively, free charges feel qvB; work per unit charge over length l gives vBl.

Answer outline: "Show that the energy stored in an inductor is ½LI²." Power against back emf: dW/dt = |ε|I = LI dI/dt. Integrate from 0 to I: W = ∫LI dI = ½LI². Compare with ½mv² (L plays the role of mass).

Answer outline: Exercise 6.3. n = 1500 m⁻¹, A = 2.0 × 10⁻⁴ m², dI/dt = 20 A/s. ε = µ₀nA dI/dt = 4π × 10⁻⁷ × 1500 × 2 × 10⁻⁴ × 20 ≈ 7.5 × 10⁻⁶ V. Similarly, Exercise 6.5 gives ½BωR² = 100 V, Exercise 6.7 gives L = 200/(5/0.1) = 4 H and Exercise 6.8 gives ΔΦ = MΔI = 30 Wb.

Key Terms

Common Mistakes