Electromagnetic Induction – Class 12 Physics

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SECTION 6.1 – 6.2

Faraday's & Henry's Experiments

The Discovery

Around 1830, Michael Faraday (England) and Joseph Henry (USA) independently demonstrated that electric currents are induced in closed coils when subjected to changing magnetic fields. This is called Electromagnetic Induction.

Experiment 6.1 — Bar Magnet & Coil

North pole pushed towards coil → galvanometer deflects (current induced).
Magnet held stationary → no deflection (no current).
Magnet pulled away → deflection in opposite direction.
South pole moved → deflections opposite to north pole.
Faster motion → larger deflection (larger current).
Key: Relative motion between magnet and coil produces induced current.

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Fig 6.1 — Bar magnet pushed towards coil, galvanometer deflects (NCERT Figure 6.1)

Experiment 6.2 — Two Coils

Coil C2 (connected to battery) moved towards coil C1 (connected to galvanometer) → current induced in C1.
Again, it is the relative motion between coils that induces current.

Experiment 6.3 — Changing Current (No Motion Needed)

Two stationary coils C1 and C2. C2 connected to battery via tapping key K.
Key K pressed (current rises from 0 to max) → galvanometer deflects momentarily in C1.
Key held pressed (steady current) → no deflection.
Key released (current falls to 0) → deflection in opposite direction.
Conclusion: It is the changing magnetic flux (not motion) that induces EMF. Relative motion is NOT essential.

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Fig 6.2 — Two coils C1 and C2; current in C2 causes induction in C1 (NCERT Figure 6.2)

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Fig 6.3 — Experimental setup for Experiment 6.3 with tapping key K (NCERT Figure 6.3)

🎯 Practice MCQs — Faraday's Experiments

Q1. In Experiment 6.3, a galvanometer deflection is observed in coil C1 when the tapping key K is:

Deflection occurs only when current (and thus flux) is changing — i.e., when key is just pressed (current rising) or just released (current falling). Steady current = no change in flux = no deflection.

Q2. In Faraday's bar magnet experiment, when the magnet is moved towards the coil twice as fast, the induced EMF becomes:

ε = −dΦ/dt. Faster motion → faster rate of change of flux → larger induced EMF. Moving twice as fast makes EMF double.

Q3. Which of the following does NOT produce electromagnetic induction in a coil?

EMF is induced only when there is a change in magnetic flux. A stationary magnet produces a steady flux through the coil — no change, no EMF.

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SECTION 6.3

Magnetic Flux

Definition of Magnetic Flux

Magnetic flux through a surface of area A placed in uniform magnetic field B at angle θ:

Magnetic Flux Formula

Φ_B = B · A = BA cos θ

For non-uniform field:
Φ_B = Σ B_i · dA_i (sum over all area elements)

SI Unit: Weber (Wb) or T·m²
Nature: Scalar quantity

θ = angle between B and area vector A

Key Points about Magnetic Flux

Maximum flux when θ = 0° (B perpendicular to surface plane, i.e., parallel to area vector): Φ = BA
Minimum (zero) flux when θ = 90° (B parallel to surface plane): Φ = 0
Flux can be varied by changing B, A, or θ.
1 Wb = 1 T·m² = 1 V·s

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Fig 6.4 — Plane of area A in uniform magnetic field B at angle θ (NCERT Figure 6.4)

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Fig 6.5 — Non-uniform magnetic field B_i at area element dA_i (NCERT Figure 6.5)

🎯 Practice MCQs — Magnetic Flux

Q4. A square coil of side 10 cm is placed in a uniform magnetic field of 0.5 T. If the plane of the coil makes an angle of 60° with the field, the magnetic flux through the coil is:

Plane makes 60° with B, so area vector makes 90°−60° = 30° with B.
Φ = BA cos θ = 0.5 × (0.1)² × cos 30° = 0.5 × 0.01 × (√3/2) ≈ 0.5 × 0.01 × 0.866 ≈ 4.33 × 10⁻³ Wb
Wait — re-check: if plane makes 60° with B, area vector makes 30° with B: Φ = 0.5×0.01×cos30° ≈ 4.33×10⁻³ Wb.
Select C. [Note: Plane at 60° with B → area vector at 30° with B → Φ = BA cos 30°]

Q5. A rectangular coil of dimensions 8 cm × 2 cm is placed normally (θ = 0°) in a uniform field of 0.3 T. The magnetic flux through the coil is:

A = 8×2 = 16 cm² = 16×10⁻⁴ m²
Φ = BA cos 0° = 0.3 × 16×10⁻⁴ × 1 = 4.8 × 10⁻⁴ Wb

Q6. The SI unit of magnetic flux (Weber) is equivalent to:

1 Wb = 1 T·m² = 1 V·s (since ε = −dΦ/dt → Φ = ε×t = V·s = Volt·second)

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SECTION 6.4

Faraday's Law of Induction

Statement of Faraday's Law

The magnitude of the induced EMF in a circuit is equal to the time rate of change of magnetic flux through the circuit.

Faraday's Law — Mathematical Form

For single loop: ε = −dΦ_B / dt

For N-turn coil: ε = −N dΦ_B / dt

The negative sign indicates direction of induced EMF (Lenz's law).

Induced current: I = ε / R

Ways to Change Magnetic Flux (and thus induce EMF)

Changing the magnetic field B (as in Experiments 6.1 and 6.2)
Changing the area A of the coil (shrinking or stretching)
Changing the angle θ between B and A (rotating the coil)
Changing the number of turns N

Solved Example — Square Loop

Square loop (side 10 cm, R = 0.5 Ω) in B = 0.1 T at 45°. B decreases to zero in 0.7 s.

Initial Φ = BA cos 45° = 0.1 × 0.01 × (1/√2) = 10⁻³/√2 Wb
ΔΦ = Φ_final − Φ_initial = 0 − 10⁻³/√2
ε = |ΔΦ/Δt| = (10⁻³/√2) / 0.7 ≈ 1.0 mV
I = ε/R = 1.0×10⁻³ / 0.5 = 2 mA

🎯 Practice MCQs — Faraday's Law

Q7. A coil of 200 turns has a flux change of 5 × 10⁻³ Wb in 0.1 s. The induced EMF is:

ε = N × (ΔΦ/Δt) = 200 × (5×10⁻³ / 0.1) = 200 × 0.05 = 10 V

Q8. A circular coil (radius 10 cm, N = 500, R = 2Ω) is rotated through 180° in 0.25 s in Earth's horizontal field B_H = 3×10⁻⁵ T. The induced current is approximately:

ΔΦ = 2 × B_H × πr² = 2 × 3×10⁻⁵ × π×(0.1)² = 6π×10⁻⁷ Wb
ε = N(ΔΦ/Δt) = 500 × 6π×10⁻⁷ / 0.25 ≈ 3.77×10⁻³ V
I = ε/R = 3.77×10⁻³ / 2 ≈ 1.9 × 10⁻³ A

Q9. A long solenoid (15 turns/cm) has a small loop of area 2 cm² inside it. Current changes from 2A to 4A in 0.1 s. The induced EMF in the loop is:

n = 15 turns/cm = 1500 turns/m
dB/dt = μ₀n(dI/dt) = 4π×10⁻⁷ × 1500 × (2/0.1) = 4π×10⁻⁷ × 30000 = 12π×10⁻³ T/s
ε = A × dB/dt = 2×10⁻⁴ × 12π×10⁻³ ≈ 7.54 × 10⁻⁶ V

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SECTION 6.5 – 6.6

Lenz's Law & Motional EMF

Lenz's Law (1834)

"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."

Represented by the negative sign in ε = −dΦ/dt.
N pole approaching coil → flux increases → induced current creates N pole facing magnet (repulsion) to oppose increase.
N pole moving away → flux decreases → induced current creates S pole facing magnet (attraction) to oppose decrease.
Lenz's law is consistent with the law of conservation of energy.

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Fig 6.6 — Illustration of Lenz's law — direction of induced current in coil (NCERT Figure 6.6)

Lenz's Law and Energy Conservation

If the induced current aided the approaching magnet instead of opposing it, the magnet would accelerate indefinitely without energy input — a perpetual motion machine, which violates conservation of energy. The work done against the opposing force is converted to electrical energy (Joule heating).

Motional EMF (Section 6.6)

A conductor of length l moving with velocity v perpendicular to magnetic field B:

Free charges in the conductor experience Lorentz force qvB.
Work done moving charge from P to Q: W = qvBl
EMF = W/q = Blv
A time-varying magnetic field generates an electric field (even for stationary conductors).

Motional EMF Formula

ε = Blv

For a rotating rod (length R, angular velocity ω):
ε = (1/2) B ω R²

For N-turn rotating coil:
ε = NBAω sin ωt = ε₀ sin ωt

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Fig 6.10 — Rod PQ moving in uniform field B, motional EMF setup (NCERT Figure 6.10)

🎯 Practice MCQs — Lenz's Law & Motional EMF

Q10. A 1.0 m metallic rod is rotated at 400 rad/s about one end in a uniform magnetic field of 0.5 T (parallel to axis). The EMF developed between the centre and the ring is:

ε = (1/2) B ω R² = (1/2) × 0.5 × 400 × (1)² = 100 V

Q11. A wire of length 10 m falls horizontally with speed 5 m/s perpendicular to Earth's horizontal field B_H = 0.3×10⁻⁴ T. The instantaneous induced EMF is:

ε = Blv = 0.3×10⁻⁴ × 10 × 5 = 1.5 × 10⁻³ V

Q12. A wheel with 10 metallic spokes each 0.5 m long is rotated at 120 rev/min in Earth's field H_E = 0.4 G. The induced EMF between axle and rim is:

ω = 2π × (120/60) = 4π rad/s; B = 0.4×10⁻⁴ T; R = 0.5 m
ε = (1/2)ωBR² = (1/2)×4π×0.4×10⁻⁴×0.25 = 6.28 × 10⁻⁵ V
(Number of spokes doesn't matter — EMFs are in parallel)

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SECTION 6.7

Inductance (Mutual & Self)

Introduction to Inductance

Flux through a coil ∝ current through it: NΦ_B ∝ I → NΦ_B = LI
Inductance depends only on geometry of the coil and permeability of medium.
SI unit: Henry (H) — named after Joseph Henry.
Inductance is a scalar quantity. Dimensions: [ML²T⁻²A⁻²]

Mutual Inductance (M)

EMF induced in coil 1 due to changing current in coil 2: ε₁ = −M dI₂/dt
For two co-axial solenoids (inner radius r₁, n₁ turns/m; outer n₂): M = μ₀ n₁ n₂ π r₁² l
M₁₂ = M₂₁ = M (always equal — very useful!)
Two concentric circular coils (r₁ << r₂): M = μ₀πr₁² / 2r₂
Depends on separation and relative orientation of coils.

Self-Inductance (L)

EMF induced in a coil due to change in its own current: ε = −L dI/dt
Called back EMF — always opposes change in current.
L is electrical analogue of mass (inertia) in mechanics.
For a long solenoid: L = μ₀ n² A l
With magnetic core (relative permeability μᵣ): L = μᵣ μ₀ n² A l

Key Inductance Formulae

Mutual inductance (two coaxial solenoids):
  M = μ₀ n₁ n₂ π r₁² l

Self inductance of solenoid:
  L = μ₀ n² A l (air core)
  L = μᵣ μ₀ n² A l (magnetic core)

Magnetic energy stored:
  W = (1/2) L I²

Magnetic energy density:
  u_B = B² / 2μ₀

Energy Stored in Inductor

Work done against back EMF is stored as magnetic potential energy.
W = (1/2)LI² — analogous to kinetic energy (1/2)mv² in mechanics.
L plays the role of mass (inertia); I plays the role of velocity.

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Fig 6.12 — Two long co-axial solenoids of same length l (NCERT Figure 6.12)

🎯 Practice MCQs — Inductance

Q13. The current in a circuit falls from 5.0 A to 0.0 A in 0.1 s inducing an average EMF of 200 V. The self-inductance of the circuit is:

ε = L × |ΔI/Δt| → L = ε × Δt / ΔI = 200 × 0.1 / 5 = 4 H

Q14. A pair of adjacent coils has mutual inductance M = 1.5 H. If current in one coil changes from 0 to 20 A in 0.5 s, the change of flux linkage with the other coil is:

NΔΦ = M × ΔI = 1.5 × (20 − 0) = 30 Wb
(Flux linkage = M × I; change = M × ΔI = 1.5 × 20 = 30 Wb)

Q15. The energy stored in an inductor of self-inductance 2 H carrying a current of 10 A is:

W = (1/2)LI² = (1/2) × 2 × (10)² = 100 J

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SECTION 6.8

AC Generator

Principle

An AC generator converts mechanical energy to electrical energy using electromagnetic induction. Invented by Nicola Tesla.

A coil (armature) rotates in a uniform magnetic field B.
Rotating coil → changing flux → induced EMF.
Ends of coil connected to external circuit via slip rings and carbon brushes.

Working

Coil rotated at constant angular speed ω.
At time t, angle θ = ωt (θ = 0 at t = 0).
Flux: Φ_B = BA cos ωt
EMF: ε = NBAω sin ωt = ε₀ sin ωt
Maximum EMF: ε₀ = NBAω (when sin ωt = 1, i.e., θ = 90° or 270°)
EMF is zero when θ = 0° or 180° (maximum flux, minimum rate of change).

AC Generator Formulae

Instantaneous EMF: ε = ε₀ sin ωt = ε₀ sin 2πνt

Peak (max) EMF: ε₀ = NBAω = NBA × 2πν

ω = angular velocity (rad/s)
ν = frequency of rotation (Hz)

In India: frequency = 50 Hz
In USA: frequency = 60 Hz

Types of Commercial Generators

Hydro-electric generators: Mechanical energy from falling water (dams).
Thermal generators: Steam produced by burning coal rotates armature.
Nuclear generators: Nuclear fuel produces steam.
Modern generators produce up to 500 MW of electrical power.
In most commercial generators, coils are stationary; electromagnets rotate.

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Fig 6.13 — AC Generator — coil, slip rings, carbon brushes, magnetic poles N-S (NCERT Figure 6.13)

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Fig 6.14 — Sinusoidal EMF waveform generated by rotating coil (NCERT Figure 6.14)

🎯 Practice MCQs — AC Generator

Q16. A 100-turn coil (area 0.1 m²) rotates at 0.5 rev/s in a field of 0.01 T. The maximum EMF generated is:

ε₀ = NBA × 2πν = 100 × 0.01 × 0.1 × 2π × 0.5 = 100 × 0.01 × 0.1 × π = 0.314 V

Q17. In an AC generator, the induced EMF is maximum when the angle between the coil plane and magnetic field is:

ε = ε₀ sin θ. Maximum when sin θ = 1, i.e., θ = 90°. At this point, coil plane is parallel to B (area vector ⊥ to B) — rate of change of flux is greatest. Answer: 90°

Q18. An AC generator with N = 200 turns, A = 0.05 m², B = 0.5 T rotates at 50 Hz. The peak EMF ε₀ is:

ε₀ = NBAω = NBA × 2πν = 200 × 0.5 × 0.05 × 2π × 50 = 200 × 0.5 × 0.05 × 314.16 ≈ 3142 V

📋 Chapter Summary

⚡ Faraday's Law

ε = −N dΦ/dt. EMF induced by changing flux. More turns N → more EMF.

🔀 Magnetic Flux

Φ = BA cosθ. Unit: Weber (Wb) = V·s. Scalar quantity.

🔄 Lenz's Law

Induced current opposes flux change. Upholds conservation of energy. Negative sign in Faraday's law.

🏃 Motional EMF

ε = Blv for moving rod. ε = ½BωR² for rotating rod. Due to Lorentz force on charges.

🌀 Self Inductance

ε = −L dI/dt. L = μ₀n²Al. Energy W = ½LI². Back EMF = electrical inertia.

🔗 Mutual Inductance

ε₁ = −M dI₂/dt. M = μ₀n₁n₂πr₁²l. M₁₂ = M₂₁ always.

⚙️ AC Generator

ε = ε₀ sin ωt. ε₀ = NBAω. 50 Hz in India. Mechanical → electrical energy.

📊 Dimensions

Φ: Wb = [ML²T⁻²A⁻¹]. L, M: Henry = [ML²T⁻²A⁻²]. EMF: Volt.

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