Electromagnetic Induction — Class 12 MCQs with Answers

Electromagnetic Induction — important questions & MCQs with answers (Class 12 Physics)

14 solved questions from this chapter's difficulty levels, each with its answer and explanation. The other 76 are timed and scored when you sign in.

1. Q1Easy

   Induced EMF =

   A.dΦ/dt

   B.−dΦ/dt✓ Correct

   C.Φ·t

   D.Φ/t

   Answer: B. −dΦ/dt

   Explanation: Faraday's law gives EMF = −dΦ/dt; the minus sign (Lenz's law) shows the induced EMF opposes the flux change that caused it. Dropping the sign leaves only magnitude, not direction.

2. Q2Easy

   Lenz's law follows from:

   A.Charge quantisation

   B.Newton's first law

   C.Coulomb's law

   D.Conservation of energy✓ Correct

   Answer: D. Conservation of energy

   Explanation: Lenz's law follows from conservation of energy — if induced current aided the change instead of opposing it, the system would gain energy for free, which is forbidden.

3. Q3Easy

   SI unit of magnetic flux:

   A.Henry

   B.Tesla

   C.Weber✓ Correct

   D.Volt

   Answer: C. Weber

   Explanation: Magnetic flux Φ = B·A cos θ is measured in weber (Wb); tesla is the unit of B itself, and henry belongs to inductance, not flux.

4. Q4Easy

   SI unit of self-inductance:

   A.Weber

   B.Farad

   C.Tesla

   D.Henry✓ Correct

   Answer: D. Henry

   Explanation: Self-inductance relates flux linkage to current via ε = −L dI/dt, so its SI unit is the henry (H) = V·s/A, not weber or tesla which describe flux and field.

5. Q5Easy

   Magnetic flux through area A in uniform B is:

   A.BA sin θ

   B.B·A cos θ✓ Correct

   C.B/A

   D.B + A

   Answer: B. B·A cos θ

   Explanation: Flux through a flat area in a uniform field is Φ = B·A cos θ, θ being the angle between B and the area vector — not sin θ, which belongs to the motional-EMF formula.

6. Q6Easy

   Mutual inductance has unit:

   A.Henry✓ Correct

   B.Tesla

   C.Volt

   D.Weber

   Answer: A. Henry

   Explanation: Mutual inductance links the EMF induced in one coil to the changing current in another, ε₂ = −M dI₁/dt, so like self-inductance it is measured in henry.

7. Q7Medium

   Motional EMF: ε =

   A.B/v

   B.BLv✓ Correct

   C.L/B

   D.BLv²

   Answer: B. BLv

   Explanation: Motional EMF for a rod of length L sweeping speed v through field B is ε = BLv — linear in each quantity, so doubling v exactly doubles ε.

8. Q8Medium

   Self-inductance L:

   A.ε = −L(dI/dt)✓ Correct

   B.ε = L·I

   C.ε = I/L

   D.ε = L/t

   Answer: A. ε = −L(dI/dt)

   Explanation: An inductor's self-induced EMF is ε = −L(dI/dt): it always opposes the change in current through it, never the current's steady value itself.

9. Q9Medium

   Flux changes from 0.1 to 0.4 Wb in 0.2 s in a 10-turn coil. Induced EMF:

   A.30 V

   B.1.5 V

   C.150 V

   D.15 V✓ Correct

   Answer: D. 15 V

   Explanation: |ε| = N·ΔΦ/Δt = 10 × (0.4−0.1)/0.2 = 10 × 1.5 = 15 V; forgetting the 10 turns would give the wrong 1.5 V, a frequent slip on multi-turn coils.

10. Q10Medium

    L = 2 H, dI/dt = 5 A/s. EMF induced:

    A.−2.5 V

    B.10 V

    C.−10 V✓ Correct

    D.−5 V

    Answer: C. −10 V

    Explanation: ε = −L(dI/dt) = −2 H × 5 A/s = −10 V; the sign shows the EMF opposes the rising current, so its magnitude is 10 V, not 2.5 or 5 V from mixing up L and dI/dt.

11. Q11Medium

    Rod L = 0.5 m moves at 4 m/s perpendicular to B = 0.2 T. EMF:

    A.4 V

    B.0.4 V✓ Correct

    C.0.04 V

    D.2 V

    Answer: B. 0.4 V

    Explanation: ε = BLv = 0.2 T × 0.5 m × 4 m/s = 0.4 V — plug values straight into the motional-EMF formula, keeping units in metres and seconds.

12. Q12Hard

    Mutual inductance unit:

    A.Tesla

    B.Henry✓ Correct

    C.Weber

    D.Volt

    Answer: B. Henry

    Explanation: Mutual inductance, like self-inductance, is defined through ε = −M dI/dt, so it carries the same SI unit, the henry.

13. Q13Hard

    Energy stored in an inductor:

    A.½L²I

    B.LI

    C.½LI²✓ Correct

    D.I/L

    Answer: C. ½LI²

    Explanation: Building up current in an inductor stores energy in its magnetic field, U = ½LI², the magnetic analogue of ½CV² for a capacitor.

14. Q14Hard

    Conducting rod (mass m, length L, R = total resistance) slides on horizontal frictionless rails in B (vertical), pulled by weight Mg via pulley. Terminal velocity:

    A.MgR/(B²L²)✓ Correct

    B.MgL/(BR)

    C.Mg/(BL)

    D.BLR/(Mg)

    Answer: A. MgR/(B²L²)

    Explanation: At terminal velocity the magnetic braking force B²L²v/R exactly balances the driving weight Mg, so Mg = B²L²v_t/R, giving v_t = MgR/(B²L²) — it stops accelerating once these balance, not at maximum EMF.

Electromagnetic Induction — FAQs

More Class 12 Physics chapters

• 01Electric Charges and Fields
• 02Electrostatic Potential and Capacitance
• 03Current Electricity
• 04Moving Charges and Magnetism
• 05Magnetism and Matter
• 07Alternating Current

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