Class 12 : Physics (English) – Chapter 6: Electromagnetic Induction
EXPLANATION & SUMMARY
🌿 Introduction
Electromagnetic induction is one of the greatest contributions of 19th-century physics. It was discovered independently by Michael Faraday in England and Joseph Henry in America. Both showed that whenever the magnetic flux linked with a circuit changes, an emf is induced, which may cause a current in the circuit.
This chapter not only unifies electricity and magnetism but also serves as the foundation of electrical power generation, transformers, and modern technology.
💡 Concept: A changing magnetic environment around a conductor produces an emf.
🟢 Magnetic Flux (ΦB)
✏️ Definition: Magnetic flux through a surface = measure of the number of magnetic field lines passing through it.
Formula:
ΦB = B · A = BA cosθ
➡️ B = magnetic field strength
➡️ A = area of the loop
➡️ θ = angle between field and normal to surface
✔️ Magnetic flux is analogous to electric flux.
🔴 Faraday’s Laws of Electromagnetic Induction
Through careful experiments, Faraday concluded two laws:
First Law: An emf is induced in a circuit whenever the magnetic flux linked with it changes.
Second Law: The magnitude of the induced emf is proportional to the rate of change of flux linkage.
✏️ Formula:
e = – dΦB/dt
✔️ The negative sign, introduced by Lenz, indicates opposition to the cause.
🟡 Lenz’s Law
Statement: The direction of induced current is always such that it opposes the change in flux producing it.
This ensures energy conservation.
Example:
Magnet approaches coil → induced current repels magnet.
Magnet moves away → induced current attracts magnet back.
💡 Without Lenz’s law, energy could be created freely, violating conservation.
🔵 Ways of Changing Flux
Flux linkage through a circuit can change due to:
✔️ Changing B (magnetic field strength).
✔️ Changing A (area of the circuit).
✔️ Changing θ (orientation).
✔️ Motion of conductor in magnetic field.
🟢 Motional emf (Derivation in Detail)
Consider a rod of length ℓ moving with velocity v perpendicular to a magnetic field B.
Force on a charge q inside rod: F = q(v × B).
Electrons accumulate at one end, creating emf.
✏️ emf induced:
e = Bℓv
If the conductor moves at an angle, emf = Bℓv sinθ.
✔️ Motional emf is central in generators and rail–gun experiments.
🔴 AC Generator (Derivation Expanded)
An AC generator converts mechanical rotation into alternating current.
Construction: Coil of N turns, area A, rotating with angular velocity ω in uniform B.
Flux linked at time t: Φ = NBA cos(ωt).
emf induced:
e = – dΦ/dt = NBAω sin(ωt).
✔️ Thus, emf varies sinusoidally with time. Peak emf = NBAω.
💡 Real-life: Every power station uses Faraday’s principle.
🟡 Transformer (Principle)
A transformer works on mutual induction.
Primary coil with turns N₁ connected to AC supply.
Secondary coil with N₂ turns linked magnetically.
emf induced in secondary: e₂ = –M dI₁/dt.
✏️ Ratio:
V₂/V₁ = N₂/N₁
✔️ Step-up or step-down AC voltage depending on ratio of turns.
🔵 Eddy Currents (Expanded)
When a solid conductor is subjected to changing flux, induced circular currents (eddy currents) flow.
Discovered by Foucault → sometimes called Foucault currents.
Cause unwanted heating but also useful applications.
✔️ Applications:
Electromagnetic brakes (eddy currents oppose wheel motion).
Damping in galvanometers.
Induction heating furnaces.
Smart electric meters.
💡 Example: Aluminium plate swinging in magnetic field slows down because of eddy current damping.
🟢 Self-Induction (Detailed)
When current through a coil changes, it changes magnetic flux linked with the coil, inducing emf in same coil.
Formula: e = –L dI/dt
L = coefficient of self-induction, depends on geometry and medium.
Unit: Henry (H).
💡 Analogy: L behaves like inertia — resists change in current just as mass resists change in velocity.
🔴 Mutual Induction (Detailed)
When current in coil 1 changes, flux linked with coil 2 also changes, inducing emf in coil 2.
Formula: e₂ = –M dI₁/dt
M = mutual inductance.
Symmetric: M₁₂ = M₂₁.
✔️ Basis of transformers, wireless charging, induction coils.
🟡 Energy in Inductor (Expanded)
When current builds in an inductor, work must be done against induced emf. This energy is stored in magnetic field.
Formula: U = ½ L I²
Energy density in field: u = B²/(2μ₀)
💡 Analogy: Capacitor stores energy in electric field, inductor stores in magnetic field.
🔵 Conservation of Energy (with Lenz’s Law)
Consider motional emf: If a conductor moves in field, induced current produces magnetic force opposing motion. Work done by external agent appears as:
Electrical energy in circuit, and/or
Heat in resistance.
✔️ Thus, conservation of energy is never violated.
🟢 Real-Life Applications (Expanded)
Generators: Power stations generate AC using electromagnetic induction.
Transformers: Step-up or step-down voltage in transmission.
Induction Cooker: Eddy currents heat vessel directly.
Magnetic Brakes: Trains stop smoothly with eddy current brakes.
Wireless charging: Based on mutual induction.
Choke coils: Limit AC current in circuits.
🔴 Common Misconceptions Cleared
❌ Steady magnetic field can induce emf → Wrong. ✔️ emf induced only if flux changes.
❌ Lenz’s law creates energy → Wrong. ✔️ It resists change, requires work input.
❌ Induction only by motion → Wrong. ✔️ Any change in flux (field strength, orientation, area) can cause induction.
🟢 Summary (~300 words)
Electromagnetic induction is the phenomenon of emf generation when magnetic flux changes through a circuit.
Flux: Φ = BA cosθ
Faraday’s Laws: Induced emf ∝ rate of change of flux; e = – dΦ/dt
Lenz’s Law: Direction opposes flux change → energy conservation.
Motional emf: e = Bℓv when conductor moves.
AC Generator: emf = NBAω sin(ωt).
Transformer: V₂/V₁ = N₂/N₁, works on mutual induction.
Eddy currents: Circular induced currents in bulk conductors → heating/damping.
Self-induction: e = –L dI/dt, with L as inductance.
Mutual induction: e₂ = –M dI₁/dt.
Energy in inductor: U = ½ L I², energy stored in field.
Applications:
Generators, transformers, induction cookers, brakes, wireless charging.
Every electric grid is built on electromagnetic induction.
💡 Core Idea: emf is induced only when flux changes — not by static fields.
📝 Quick Recap
✔️ Φ = BA cosθ
✔️ e = – dΦ/dt
✔️ Lenz’s law = opposition (energy conserved)
✔️ Motional emf = Bℓv
✔️ Generator emf = NBAω sin(ωt)
✔️ Transformer ratio = N₂/N₁
✔️ Self-induction = –L dI/dt
✔️ Mutual induction = –M dI/dt
✔️ Energy in inductor = ½ L I²
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QUESTIONS FROM TEXTBOOK
Question 6.1
Predict the direction of induced current in the situations described by the following Figs. 6.15 (a) to (f).
Answer
➡️ Apply Lenz’s Law: induced current opposes the change in flux.
(a) Magnet moves towards solenoid: Induced current opposes approach, so current flows anticlockwise (viewed from magnet side).
(b) Magnet moves away from solenoid: Induced current attracts magnet, so current flows clockwise (viewed from magnet side).
(c) Key just closed: Current in primary increases, flux increases through secondary, induced current opposes increase. If primary current is anticlockwise, induced is clockwise.
(d) Rheostat setting changed → current in primary changes. If resistance decreases (I increases), induced current in secondary opposes increase (opposite sense to primary). If resistance increases (I decreases), induced current supports primary’s direction.
(e) Key just released: Primary current falls, flux decreases. Induced current in secondary tries to support it (same direction as primary’s earlier current).
(f) Current in circular coil decreases steadily: Induced current supports existing current direction (so as to maintain flux).
Question 6.2
Use Lenz’s law to determine the direction of induced current in the situations described by Fig. 6.16:
(a) A wire of irregular shape turning into a circular shape.
(b) A circular loop being deformed into a narrow straight wire.
Answer
(a) When irregular loop reshapes into circle, magnetic flux enclosed increases (area increases). To oppose increase, induced current must produce opposite magnetic field. Hence, direction is such that it produces a field opposite to external.
(b) When circular loop is stretched into straight wire, area decreases. Flux decreases. Induced current will flow so as to support the lost flux (i.e., in such a direction that it maintains same magnetic field as before).
Question 6.3
A long solenoid with 15 turns per cm has a small loop of area 2.0 cm² placed inside the solenoid normal to its axis. If the current carried by the solenoid changes steadily from 2.0 A to 4.0 A in 0.1 s, what is the induced emf in the loop while the current is changing?
Answer
➡️ Data:
Turns per cm = 15 → n = 1500 turns/m.
Area = 2.0 cm² = 2.0 × 10⁻⁴ m².
Change in current ΔI = 2 A.
Time Δt = 0.1 s.
✏️ Field inside solenoid: B = μ₀ n I.
Change in flux = A ΔB = A μ₀ n ΔI.
Induced emf = (ΔΦ/Δt).
= (2.0 × 10⁻⁴ × 4π × 10⁻⁷ × 1500 × 2) / 0.1
= (2.0 × 10⁻⁴ × 3.77 × 10⁻³) / 0.1
≈ 7.5 × 10⁻⁶ V.
✔️ emf = 7.5 μV (microvolt).
Question 6.4
A rectangular wire loop of sides 8 cm and 2 cm with a small cut is moving out of a region of uniform magnetic field of 0.3 T directed normal to the loop. What is the emf developed across the cut if the velocity of the loop is 1 cm s⁻¹ in a direction normal to the (a) longer side; (b) shorter side of the loop? For how long does the induced voltage last in each case?
Answer
➡️ Data:
Sides: 8 cm = 0.08 m, 2 cm = 0.02 m.
B = 0.3 T, v = 0.01 m/s.
(a) If velocity normal to longer side → effective length = 0.08 m.
emf = Bℓv = 0.3 × 0.08 × 0.01 = 2.4 × 10⁻⁴ V.
Time for leaving = shorter side/v = 0.02 / 0.01 = 2 s.
(b) If velocity normal to shorter side → effective length = 0.02 m.
emf = 0.3 × 0.02 × 0.01 = 6 × 10⁻⁵ V.
Time = longer side/v = 0.08 / 0.01 = 8 s.
✔️ Answers:
(a) emf = 2.4 × 10⁻⁴ V, lasts 2 s.
(b) emf = 6 × 10⁻⁵ V, lasts 8 s.
Question 6.5
A 1.0 m long metallic rod is rotated with an angular frequency of 400 rad s⁻¹ about an axis normal to the rod passing through its one end. The other end of the rod is in contact with a circular metallic ring. A constant and uniform magnetic field of 0.5 T parallel to the axis exists everywhere. Calculate the emf developed between the centre and the ring.
Answer
➡️ Data: ℓ = 1 m, ω = 400 rad/s, B = 0.5 T.
✏️ Formula for rotational emf:
e = ½ B ω ℓ²
= 0.5 × 0.5 × 400 × 1²
= 100 V.
✔️ emf = 100 V.
Question 6.6
A horizontal straight wire 10 m long extending from east to west is falling with a speed of 5.0 m s⁻¹, at right angles to the horizontal component of the Earth’s magnetic field 0.3 × 10⁻⁴ Wb m⁻².
(a) What is the instantaneous value of the emf induced in the wire?
(b) Which end of the wire is at the higher electrical potential?
Answer
➡️ Data: ℓ = 10 m, v = 5 m/s, B = 0.3 × 10⁻⁴ T.
✏️ emf = B ℓ v
= (0.3 × 10⁻⁴) × 10 × 5
= 1.5 × 10⁻³ V.
(a) emf = 1.5 mV.
(b) Direction: Use Fleming’s right-hand rule. For east–west wire falling, induced emf makes east end at higher potential.
✔️ Answers: emf = 1.5 mV, east end higher potential.
Question 6.7
Current in a circuit falls from 5.0 A to 0.0 A in 0.1 s. If an average emf of 200 V is induced, give an estimate of the self-inductance of the circuit.
Answer
➡️ Formula: e = L (ΔI/Δt).
Given e = 200 V, ΔI = 5 A, Δt = 0.1 s.
L = e Δt / ΔI = 200 × 0.1 / 5 = 4 H.
✔️ Self-inductance = 4 Henry.
Question 6.8
A pair of adjacent coils has a mutual inductance of 1.5 H. If the current in one coil changes from 0 to 10 A in 0.5 s, what is the change of flux linkage with the other coil?
Answer
➡️ Formula: Φ₂ = M I₁.
M = 1.5 H, I₁ = 10 A.
Change in flux linkage = M ΔI = 1.5 × 10 = 15 Wb-turns.
✔️ Flux linkage change = 15 Wb-turns.
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OTHER IMPORTANT QUESTIONS
(CBSE MODEL QUESTIONS PAPER)
ESPECIALLY MADE FROM THIS LESSON ONLY
Section A (MCQs: Q1–Q18)
Question 1
The unit of magnetic flux is:
🔵 (A) Weber
🟢 (B) Tesla
🟠 (C) Gauss
🔴 (D) Ampere-turn
Answer: (A) Weber ✔️ (1 Weber = 1 Tesla·m²).
Question 2
Faraday’s second law of electromagnetic induction states that:
🔵 (A) emf is equal to magnetic flux
🟢 (B) emf is proportional to the rate of change of flux
🟠 (C) emf is constant for a coil
🔴 (D) emf is independent of flux
Answer: (B) emf ∝ dΦ/dt.
Question 3
Lenz’s law ensures conservation of:
🔵 (A) Momentum
🟢 (B) Energy
🟠 (C) Mass
🔴 (D) Flux
Answer: (B) Energy ✔️ Induced current resists flux change.
Question 4
Motional emf in a conductor of length ℓ moving with velocity v in magnetic field B is:
🔵 (A) e = Bv/ℓ
🟢 (B) e = Bℓv
🟠 (C) e = B/v
🔴 (D) e = ℓ/v
Answer: (B) e = Bℓv.
Question 5
Which phenomenon is used in AC generator?
🔵 (A) Self-induction
🟢 (B) Mutual induction
🟠 (C) Electromagnetic induction
🔴 (D) Eddy currents
Answer: (C) Electromagnetic induction.
Question 6
The SI unit of inductance is:
🔵 (A) Tesla
🟢 (B) Henry
🟠 (C) Weber
🔴 (D) Joule
Answer: (B) Henry (H).
Question 7
Which law gives the direction of induced emf?
🔵 (A) Ampere’s law
🟢 (B) Fleming’s right-hand rule
🟠 (C) Lenz’s law
🔴 (D) Biot-Savart law
Answer: (C) Lenz’s law.
Question 8
Energy stored in an inductor carrying current I is:
🔵 (A) LI
🟢 (B) ½ LI²
🟠 (C) I²/L
🔴 (D) L/I²
Answer: (B) ½ LI².
Question 9
Which current is produced in bulk metallic plates in varying field?
🔵 (A) Displacement current
🟢 (B) Eddy current
🟠 (C) Induced current
🔴 (D) Leakage current
Answer: (B) Eddy current.
Question 10
The direction of induced emf can be found using:
🔵 (A) Fleming’s left-hand rule
🟢 (B) Fleming’s right-hand rule
🟠 (C) Ampere’s rule
🔴 (D) Coulomb’s law
Answer: (B) Fleming’s right-hand rule.
Question 11
For a transformer: V₂/V₁ = ?
🔵 (A) N₂/N₁
🟢 (B) N₁/N₂
🟠 (C) √N₂/N₁
🔴 (D) V₁/N₁
Answer: (A) V₂/V₁ = N₂/N₁.
Question 12
Self-inductance is defined as:
🔵 (A) Flux per unit current
🟢 (B) Current per unit flux
🟠 (C) emf per unit resistance
🔴 (D) Flux per unit charge
Answer: (A) Flux linkage per unit current.
Question 13
Induced emf in coil depends on:
🔵 (A) Area of coil
🟢 (B) Number of turns
🟠 (C) Rate of change of flux
🔴 (D) All of these
Answer: (D) All of these.
Question 14
In eddy current damping, induced current:
🔵 (A) Supports motion
🟢 (B) Opposes motion
🟠 (C) Accelerates motion
🔴 (D) None
Answer: (B) Opposes motion.
Question 15
If current in a coil changes at 1 A/s and emf induced is 2 V, self-inductance is:
🔵 (A) 0.5 H
🟢 (B) 1 H
🟠 (C) 2 H
🔴 (D) 4 H
Answer: (C) 2 H (since e = L dI/dt).
Question 16
Which of these uses electromagnetic induction?
🔵 (A) Generator
🟢 (B) Transformer
🟠 (C) Induction cooker
🔴 (D) All of these
Answer: (D) All of these.
Question 17
Magnetic energy density in an inductor is:
🔵 (A) B²/2μ₀
🟢 (B) μ₀B²
🟠 (C) B²/μ₀
🔴 (D) μ₀/2B²
Answer: (A) B²/2μ₀.
Question 18
If mutual inductance between coils is M, emf induced in secondary =
🔵 (A) MI
🟢 (B) M dI/dt
🟠 (C) –M dI/dt
🔴 (D) –MI
Answer: (C) –M dI/dt.
Section B (Short Answer: Q19–Q23)
Question 19
State Faraday’s laws of electromagnetic induction.
Answer
✔️ First Law: Whenever magnetic flux linked with a circuit changes, an emf is induced in the circuit.
✔️ Second Law: The magnitude of induced emf is proportional to the rate of change of flux. Mathematically, e = – dΦ/dt.
Question 20
State Lenz’s law and its significance.
Answer
✏️ Lenz’s Law: The direction of induced current is always such that it opposes the cause of its production.
💡 Significance: Prevents violation of conservation of energy. If induced current aided the flux change, energy would be created freely, which is impossible.
Question 21
What is self-induction?
Answer
✔️ Phenomenon where changing current in a coil induces emf in the same coil.
Formula: e = –L dI/dt.
💡 L (inductance) is a measure of coil’s ability to oppose current change, analogous to inertia.
Question 22
What are eddy currents? Mention two uses.
Answer
✔️ Eddy currents are circulating induced currents in bulk metallic bodies when exposed to changing magnetic flux.
Uses:
In galvanometers to provide electromagnetic damping (needle comes to rest quickly).
In electromagnetic brakes of trains, where induced currents oppose wheel motion.
Question 23
What is energy stored in an inductor? Derive expression.
Answer
emf across inductor: e = L dI/dt.
Work done in small time: dW = e I dt = L I dI.
Total work W = ∫₀ᴵ L I dI = ½ L I².
✔️ Thus, energy stored = ½ L I².
Section C (Mid-length: Q24–Q28)
Question 24
A solenoid of 1000 turns/m carries current of 2 A. Find B inside solenoid.
Answer
B = μ₀ n I = 4π × 10⁻⁷ × 1000 × 2 = 2.51 × 10⁻³ T.
Question 25
Explain working of AC generator with expression for emf.
Answer
Rectangular coil of N turns, area A, rotated in B with angular velocity ω.
Magnetic flux: Φ = NBA cos(ωt).
Induced emf: e = – dΦ/dt = NBAω sin(ωt).
✔️ Peak emf = NBAω.
💡 Basis of electricity generation worldwide.
Question 26
A rod of length 0.5 m moves at 2 m/s perpendicular to B = 0.2 T. Find emf.
Answer
e = Bℓv = 0.2 × 0.5 × 2 = 0.2 V.
Question 27
Derive expression for energy density in inductor.
Answer
Energy stored in coil: U = ½ L I².
Magnetic field inside solenoid: B = μ₀ n I.
Energy per unit volume: u = U/volume = B²/2μ₀.
Question 28
State principle of transformer.
Answer
Works on mutual induction. AC in primary → varying flux → linked with secondary → emf induced.
Relation: V₂/V₁ = N₂/N₁.
Section D (Long Answer: Q29–Q31)
Question 29
Derive expression for motional emf.
Answer
Conductor of length ℓ moves with velocity v in field B.
Force on charge q: F = q(v × B).
Charges accumulate at ends → potential difference.
emf = Bℓv (maximum when v ⊥ B).
✔️ General case: e = Bℓv sinθ.
💡 This derivation directly explains operation of sliding conductor, rail–guns, and many measuring instruments.
Question 30
Explain transformer working with construction and principle.
Answer
Consists of two coils (primary, secondary) wound on soft iron core.
Primary connected to AC source → produces alternating flux in core.
This varying flux links secondary, inducing emf: V₂/V₁ = N₂/N₁.
If N₂ > N₁ → step-up; if N₂ < N₁ → step-down.
✔️ Used for power transmission with minimal loss.
💡 Efficiency is very high (~95–98%) in practical transformers due to laminated cores reducing eddy current losses.
Question 31
Explain eddy currents, their disadvantages and applications.
Answer
Eddy currents are induced circulating currents in bulk metals when exposed to changing flux.
They cause heating → energy loss (disadvantage in transformers, motors).
To reduce: cores laminated to minimize current paths.
Applications:
✔️ Eddy current damping in galvanometers (needle comes to rest quickly).
✔️ Magnetic brakes in trains (eddy currents oppose wheel motion).
✔️ Induction heating in furnaces.
✔️ Speedometers in vehicles (eddy currents oppose rotating disc).
Section E (Case/Application: Q32–Q33)
Question 32
A metallic rod moves with velocity 5 m/s in field B = 0.1 T, length 0.2 m. Resistance = 2 Ω. Find emf and induced current.
Answer
emf = Bℓv = 0.1 × 0.2 × 5 = 0.1 V.
Current = e/R = 0.1 / 2 = 0.05 A.
Question 33
A coil of inductance 2 H carries current 5 A. Current reduced to zero in 0.1 s. Find emf and energy lost.
Answer
emf = L ΔI/Δt = 2 × 5 / 0.1 = 100 V.
Energy lost = ½ LI² = ½ × 2 × 25 = 25 J.
💡 The energy stored in coil’s magnetic field is released as heat in resistance during decay of current.
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