Electromagnetic Induction — Class 12 Physics NCERT Solutions (Free)
Free step-by-step NCERT solutions for Class 12 Physics chapter "Electromagnetic Induction" — 7 important questions with detailed answers for CBSE board exam preparation.
TL;DR: Free step-by-step NCERT solutions for Class 12 Physics chapter "Electromagnetic Induction" — 7 important questions with detailed answers for CBSE boar…
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NCERT Solutions for Class 12 Physics — Electromagnetic Induction. Step-by-step answers to all 7 textbook questions from this chapter, written for CBSE board preparation and free to use.
Electromagnetic Induction — All 7 Questions Solved
Q1. State Faraday's law of electromagnetic induction. Write mathematical expression.
Faraday's Law of Electromagnetic Induction: Rate of change of magnetic flux through a circuit induces emf in the circuit.
Mathematical Expression:
ε = -dΦ_B/dt
where ε = induced emf (volts)
Φ_B = magnetic flux (Weber, Wb)
t = time (seconds)
Alternatively, for N turns:
ε = -N(dΦ_B/dt)
Integral form over time interval t₁ to t₂:
ε = -N × (Φ₂ - Φ₁)/(t₂ - t₁)
Magnetic Flux:
Φ_B = ∫B⋅dA = BA cos θ (for uniform field)
where B = magnetic flux density
A = area of loop
θ = angle between B and normal to A
SI Unit: Weber (Wb) or V⋅s
Lenz's Law (negative sign):
Induced emf acts to oppose the change producing it.
- If flux increases, induced current creates field opposing increase
- If flux decreases, induced current creates field opposing decrease
Physical Interpretation:
- Moving magnetic field through coil → induced emf
- Moving coil in magnetic field → induced emf
- Changing current in nearby coil → induced emf (mutual inductance)
- Changing current in same coil → induced emf (self-inductance)
Q2. A square loop of side 10 cm is placed in magnetic field 0.5 T. If field decreases to 0.2 T in 0.1 s, find induced emf.
Given: Side of square loop = 10 cm = 0.1 m
Initial magnetic field B₁ = 0.5 T
Final magnetic field B₂ = 0.2 T
Time interval Δt = 0.1 s
Field perpendicular to loop (θ = 0°)
Area of loop:
A = side² = (0.1)² = 0.01 m²
Initial flux:
Φ₁ = B₁A cos 0° = 0.5 × 0.01 = 0.005 Wb
Final flux:
Φ₂ = B₂A cos 0° = 0.2 × 0.01 = 0.002 Wb
Change in flux:
ΔΦ = Φ₂ - Φ₁ = 0.002 - 0.005 = -0.003 Wb
Rate of change of flux:
dΦ/dt = ΔΦ/Δt = -0.003 / 0.1 = -0.03 Wb/s
Induced emf (using Faraday's law):
ε = -dΦ/dt = -(-0.03) = 0.03 V
|ε| = 0.03 V = 30 mV
Final Answer: Induced emf = 0.03 V or 30 mV
The emf is induced to oppose the decrease in magnetic flux (Lenz's law).
Q3. Define self-inductance (L). Derive expression for self-inductance of solenoid.
Self-Inductance (L): Property of coil to oppose change in its own current.
Definition: L = |ε|/|dI/dt| = Φ_B/I
where ε = induced emf
dI/dt = rate of change of current
Φ_B = magnetic flux through coil
I = current in coil
SI Unit: Henry (H) = Wb/A = V⋅s/A
Smaller units: mH (millihenry), μH (microhenry)
For Solenoid:
Consider solenoid with N turns, length L, cross-sectional area A.
Magnetic field inside solenoid:
B = μ₀nI = μ₀(N/L)I
Magnetic flux through each turn:
Φ = BA = μ₀(N/L) × I × A
Total flux linkage (flux × number of turns):
Λ = NΦ = N × μ₀(N/L) × I × A = μ₀N²A I / L
Self-inductance:
L = Λ/I = μ₀N²A/L
Alternatively: L = μ₀n²AL (in terms of turns per unit length n = N/L)
where μ₀ = 4π × 10⁻⁷ H/m
N = total number of turns
A = cross-sectional area (m²)
L = length of solenoid (m)
Properties:
- L independent of current I
- L proportional to N²
- L inversely proportional to length
- L proportional to cross-sectional area
- Depends only on geometry and material (μ₀)
Q4. What is mutual inductance (M)? Explain with two coils.
Mutual Inductance (M): Property describing how change in current in one coil induces emf in nearby coil.
Definition:
M₁₂ = Φ₂₁/I₁ = |ε₂|/|dI₁/dt|
where Φ₂₁ = flux through coil 2 due to current I₁ in coil 1
ε₂ = induced emf in coil 2
I₁ = current in coil 1
SI Unit: Henry (H)
For two coils:
Coil 1 (primary): Current I₁, N₁ turns, area A₁
Coil 2 (secondary): N₂ turns, area A₂, separation d
Mutual inductance:
M₁₂ = M₂₁ = M (reciprocal property)
Expression (ideal case, coils wound on same core):
M = k√(L₁L₂)
where k = coupling coefficient (0 ≤ k ≤ 1)
L₁, L₂ = self-inductances of coils
For perfect coupling (k = 1):
M = √(L₁L₂)
Induced emf in secondary:
ε₂ = -M(dI₁/dt)
Physical Mechanism:
Step 1: Current I₁ in coil 1 creates magnetic field B₁
Step 2: This field passes through coil 2 (partial or complete)
Step 3: Flux linkage with coil 2: Λ = M × I₁
Step 4: Change in I₁ induces emf in coil 2
Coupling Coefficient:
k = M/√(L₁L₂) = Φ/Φ_max
- k = 1: Perfect coupling (all flux links both coils)
- k < 1: Partial coupling
- k = 0: No coupling
Applications:
- Transformers (k ≈ 1)
- Wireless power transmission
- Inductive coupling circuits
Q5. Define AC generator. Derive expression for induced emf in rotating rectangular coil.
AC Generator: Device that converts mechanical energy into electrical energy using electromagnetic induction.
Principle: Rotating coil in uniform magnetic field generates alternating emf.
Derivation for Rectangular Coil:
Setup:
- Rectangular coil with N turns, area A
- Rotates in uniform magnetic field B
- Angular velocity ω (rad/s)
- At t = 0, coil plane parallel to field (flux = 0)
Step 1: Angle θ between normal and field at time t
θ = ωt (starting from θ = 0)
Step 2: Magnetic flux through coil
Φ = BA cos θ = BA cos(ωt)
Step 3: For N turns
Λ = NΦ = NBA cos(ωt)
Step 4: Induced emf (Faraday's law)
ε = -dΛ/dt = -d[NBA cos(ωt)]/dt
ε = NBA ω sin(ωt)
Step 5: Let ε₀ = NBaω (peak emf)
ε = ε₀ sin(ωt)
Alternatively, if starting with coil perpendicular to field:
ε = ε₀ cos(ωt)
Final Expression:
ε = NBaω sin(ωt) = ε₀ sin(ωt)
where ε₀ = NBA ω = peak (maximum) emf
N = number of turns
B = magnetic flux density
A = area of coil
ω = angular velocity (rad/s)
Frequency relation:
ω = 2πf (where f = frequency in Hz)
ε₀ = 2πfNBA
RMS value:
ε_rms = ε₀/√2 = NBA ω/√2 = 2πfNBA/√2
Q6. A rectangular coil of 100 turns, area 0.05 m², rotates in magnetic field of 0.2 T at 50 Hz. Find peak and RMS values of induced emf.
Given: N = 100 turns
A = 0.05 m²
B = 0.2 T
f = 50 Hz (frequency)
Angular velocity:
ω = 2πf = 2π × 50 = 100π rad/s
Peak (Maximum) EMF:
ε₀ = NBAω
Step 1: ε₀ = 100 × 0.2 × 0.05 × 100π
Step 2: ε₀ = 100 × 0.2 × 0.05 × 100π
Step 3: ε₀ = 100 × 1 × π = 100π V
Step 4: ε₀ = 314.16 V ≈ 314 V
RMS Value:
ε_rms = ε₀/√2
Step 1: ε_rms = 100π/√2
Step 2: ε_rms = 100π/1.414
Step 3: ε_rms = 314.16/1.414
Step 4: ε_rms = 222.14 V ≈ 222 V
Alternatively:
ε_rms = 2πfNBA/√2 = (2π × 50 × 100 × 0.2 × 0.05)/√2
ε_rms = (100π)/√2 ≈ 222 V
Final Answer:
Peak emf = 100π V ≈ 314 V
RMS emf = 100π/√2 V ≈ 222 V
Q7. Define eddy currents. Explain how they are produced and their applications.
Eddy Currents: Induced currents flowing in closed loops within conductor when magnetic flux through it changes.
Also called: Foucault currents
How Produced:
Step 1: Changing magnetic flux passes through conductor
Step 2: By Faraday's law, emf is induced
Step 3: Since conductor is continuous, current flows in closed loops
Step 4: These closed loops are eddy currents
Step 5: Lenz's law: Direction opposes change in flux
Characteristics:
- Flow in closed paths (no external circuit needed)
- Perpendicular to direction of changing flux
- Magnitude depends on: rate of flux change, conductivity, geometry
- Create magnetic field opposing the change (Lenz's law)
- Dissipate energy as heat (I²R losses)
Applications:
1. Electromagnetic Damping:
- Moving conductor in magnetic field: eddy currents oppose motion
- Used in moving coil meters for damping
- Shock absorbers in mechanical systems
- Eddy current brakes in trains
2. Metal Detection:
- Detector coil creates changing field
- Metal object develops eddy currents
- Eddy currents create opposing field
- Detects change in field strength
- Airport security, mine detection
3. Eddy Current Testing:
- Nondestructive testing of metals
- Detects cracks, defects, material thickness
- Quality control in manufacturing
4. Induction Heating:
- High-frequency changing field in conductor
- Eddy currents heat metal rapidly
- Induction cooktops, industrial heating
5. Energy Dissipation:
- Eddy current brake on dental drills
- Stopping mechanisms
Disadvantages:
- Causes energy loss in transformers
- Reduced efficiency in electric motors
- Heat generation in power transmission
Mitigation:
- Laminated cores (thin layers separated by insulation)
- Reduces eddy current paths
- Decreases heat loss in transformers and motors
Electromagnetic Induction — Practice MCQs with Answers
Attempt these 20 multiple-choice questions after working through the solutions above. Each carries the correct option and the reasoning behind it, so a wrong answer tells you which idea to revisit.
MCQ 1. What is the SI unit of magnetic flux?
- A. Tesla
- B. Weber
- C. Henry
- D. Ampere
Answer: B. Weber — The SI unit of magnetic flux (Φ) is the Weber (Wb), defined as Tesla-meter squared (T·m²).
MCQ 2. According to Faraday's law of electromagnetic induction, the magnitude of the induced electromotive force (EMF) is directly proportional to:
- A. The magnetic field strength
- B. The rate of change of magnetic flux
- C. The area of the coil
- D. The resistance of the coil
Answer: B. The rate of change of magnetic flux — Faraday's second law states that the magnitude of the induced EMF is directly proportional to the rate of change of magnetic flux linked with the coil.
MCQ 3. A bar magnet is dropped through a long vertical copper tube. The acceleration of the magnet inside the tube will be:
- A. Equal to 'g'
- B. Greater than 'g'
- C. Less than 'g'
- D. Zero
Answer: C. Less than 'g' — As the magnet falls, the changing magnetic flux induces eddy currents in the copper tube. By Lenz's law, these currents create a magnetic field that opposes the magnet's motion, leading to an upward force that reduces its acceleration.
MCQ 4. The motional EMF induced in a conductor of length 'L' moving with velocity 'v' perpendicular to a uniform magnetic field 'B' is given by:
- A. E = BL/v
- B. E = Bv/L
- C. E = BLv
- D. E = B/Lv
Answer: C. E = BLv — When a conductor moves perpendicular to a magnetic field, the induced motional EMF across its ends is directly proportional to the magnetic field strength (B), the length of the conductor (L), and its velocity (v).
MCQ 5. The SI unit of self-inductance is:
- A. Farad
- B. Weber
- C. Henry
- D. Tesla
Answer: C. Henry — The SI unit of self-inductance (L) is the Henry (H), which is equivalent to Weber per Ampere (Wb/A).
MCQ 6. The self-inductance of a long solenoid is independent of:
- A. The number of turns per unit length
- B. The area of cross-section
- C. The current flowing through it
- D. The permeability of the core material
Answer: C. The current flowing through it — The self-inductance of a solenoid depends on its geometry (number of turns, length, area) and the permeability of the core material, but it is a constant of the coil and does not depend on the current flowing through it.
MCQ 7. The energy stored in an inductor of inductance 'L' carrying a current 'I' is given by:
- A. (1/2)LI²
- B. LI
- C. (1/2)L²I
- D. LI²
Answer: A. (1/2)LI² — The energy stored in the magnetic field of an inductor is proportional to the square of the current flowing through it and its inductance, given by the formula U = (1/2)LI².
MCQ 8. The principle on which an AC generator works is:
- A. Heating effect of current
- B. Chemical effect of current
- C. Electromagnetic induction
- D. Magnetic effect of current
Answer: C. Electromagnetic induction — An AC generator works on the principle of electromagnetic induction, where a rotating coil in a magnetic field experiences a changing magnetic flux, inducing an EMF.
MCQ 9. A current-carrying wire is placed near a conducting loop as shown. If the current in the wire is increasing, what is the direction of the induced current in the loop? (Assume wire is above the loop, parallel)
- A. Clockwise
- B. Anti-clockwise
- C. No current is induced
- D. Depends on the material of the loop
Answer: B. Anti-clockwise — If the current in the straight wire is increasing, the magnetic field (into the page below the wire) through the loop is increasing. By Lenz's law, the induced current in the loop will create a magnetic field out of the page to oppose this increase, which corresponds to an anti-clockwise current.
MCQ 10. A circular coil of 10 turns and radius 10 cm is placed in a uniform magnetic field of 0.5 T such that its plane is perpendicular to the magnetic field. The magnetic flux linked with the coil is:
- A. 0.5 Wb
- B. 0.05π Wb
- C. 5π Wb
- D. 0.005π Wb
Answer: B. 0.05π Wb — Magnetic flux Φ = N * B * A * cosθ. Here N=10, B=0.5 T, A = πr² = π(0.1 m)² = 0.01π m². Since the plane is perpendicular, θ=0°, so cosθ=1. Φ = 10 * 0.5 * 0.01π = 0.05π Wb.
MCQ 11. A metallic rod of length 'L' is moved with a velocity 'v' in a uniform magnetic field 'B' perpendicular to the rod and the field. If the rod has a resistance 'R', the power dissipated in the rod is:
- A. (BLv)²/R
- B. (BLv)R
- C. BLv/R
- D. (BLv)²R
Answer: A. (BLv)²/R — The induced EMF in the rod is E = BLv. The current induced in the rod is I = E/R = BLv/R. The power dissipated is P = I²R = (BLv/R)²R = (BLv)²/R.
MCQ 12. Two coils have a mutual inductance of 0.005 H. If the current in the primary coil changes from 5 A to 1 A in 0.02 s, the induced EMF in the secondary coil is:
- A. 1 V
- B. 0.1 V
- C. 0.5 V
- D. 0.05 V
Answer: A. 1 V — Induced EMF in secondary E = -M (dI/dt). Here M = 0.005 H, dI = (1 A - 5 A) = -4 A, dt = 0.02 s. So, E = -0.005 * (-4 / 0.02) = 0.005 * 200 = 1 V.
MCQ 13. Eddy currents are produced when:
- A. A conductor is placed in a static magnetic field
- B. A conductor moves perpendicular to a static electric field
- C. A conductor is subjected to a changing magnetic flux
- D. A conductor carries a steady current
Answer: C. A conductor is subjected to a changing magnetic flux — Eddy currents are loops of electric current induced within bulk conductors by a changing magnetic field in the conductor, in accordance with Faraday's law of induction.
MCQ 14. Which of the following is NOT an application of eddy currents?
- A. Magnetic braking in trains
- B. Induction furnace
- C. Speedometer
- D. DC motor
Answer: D. DC motor — Magnetic braking, induction furnaces, and speedometers all utilize eddy currents. A DC motor primarily works on the principle of the force experienced by a current-carrying conductor in a magnetic field.
MCQ 15. A 10 H inductor carries a current of 2 A. The energy stored in the inductor is:
- A. 10 J
- B. 20 J
- C. 40 J
- D. 5 J
Answer: B. 20 J — The energy stored in an inductor is given by U = (1/2)LI². Substituting L = 10 H and I = 2 A, U = (1/2) * 10 * (2)² = (1/2) * 10 * 4 = 20 J.
MCQ 16. Lenz's law is a consequence of the law of conservation of:
- A. Charge
- B. Momentum
- C. Energy
- D. Mass
Answer: C. Energy — Lenz's law states that the direction of the induced current opposes the cause producing it. If it aided the cause, it would violate the law of conservation of energy by leading to perpetual motion.
MCQ 17. The time constant of an RL circuit is given by:
- A. R/L
- B. L/R
- C. LR
- D. 1/(LR)
Answer: B. L/R — The time constant (τ) of an RL circuit, which represents the time required for the current to reach approximately 63.2% of its maximum value, is given by the ratio of inductance (L) to resistance (R).
MCQ 18. A rectangular loop is moving out of a uniform magnetic field directed into the page. The induced current in the loop will be:
- A. Clockwise
- B. Anti-clockwise
- C. Zero
- D. Depends on the speed of the loop
Answer: A. Clockwise — As the loop moves out of the field, the magnetic flux into the page through the loop decreases. By Lenz's law, the induced current will create a magnetic field into the page to oppose this decrease, which corresponds to a clockwise current.
MCQ 19. Which of the following does NOT affect the magnitude of induced EMF in a coil?
- A. Number of turns in the coil
- B. Strength of the magnetic field
- C. Resistance of the coil
- D. Rate of change of magnetic flux
Answer: C. Resistance of the coil — The magnitude of the induced EMF depends on the number of turns, the strength of the magnetic field, and the rate of change of magnetic flux. The resistance of the coil affects the induced current, but not the induced EMF itself.
MCQ 20. If an iron core is inserted into a solenoid, its self-inductance will:
- A. Increase
- B. Decrease
- C. Remain the same
- D. Become zero
Answer: A. Increase — The self-inductance of a solenoid is proportional to the permeability of the core material (L ∝ μ). Since iron is a ferromagnetic material with high relative permeability, inserting it significantly increases the solenoid's self-inductance.
How to Use These Solutions
Attempt each question yourself first and only then compare with the worked answer. Marks in the CBSE board exam are awarded for the METHOD as much as the final result, so reproduce the steps rather than memorising the last line. Where a solution states a law, a formula or a definition, learn that wording — examiners look for it.
Related practice for this chapter: chapter MCQs, previous-year questions, revision notes and sample papers.