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CBSE · Class 12 · Physics

Electromagnetic Induction

Standard difficulty · 20 questions · 40 marks

Name: Date: Score: ____ / 40

Show your working for Sections B and C. Answers are on the last page — try every question first!

Section A — Choose the correct answer8 × 1 = 8 marks
1.
A loop of area 210 cm2 is in a field of 2 × 10−1 T. If the normal to the loop makes 90° with the field, the flux through it is:
  1. 2.1 × 10−3 Wb
  2. 4.2 × 10−3 Wb
  3. 1 Wb
  4. 0 Wb
2.
The flux through each turn of a 402-turn coil changes by 37 mWb in 10 × 10−2 s. The average induced emf is:
  1. 148.74 V
  2. 0.37 V
  3. 1.4874 V
  4. 1.4874 × 104 V
3.
A rod 5 cm long moves at 20 m/s at right angles to a field of 4 × 10−1 T. The emf across its ends is:
  1. 1 × 10−3 V
  2. 40 V
  3. 0.8 V
  4. 0.4 V
4.
The current in a 419 mH coil changes by 2 A in 5 × 10−2 s. The self-induced emf is:
  1. 1,676 V
  2. 0.1676 V
  3. 0.0419 V
  4. 16.76 V
5.
The SI unit of magnetic flux is the:
  1. gauss
  2. tesla
  3. weber
  4. henry
6.
The energy stored in a 129 mH inductor carrying 6 A is:
  1. 2.322 × 10−3 J
  2. 2.322 J
  3. 4.644 J
  4. 0.387 J
7.
Two coils have mutual inductance 371 mH. If the current in one changes by 6 A in 1 × 10−2 s, the emf in the other is:
  1. 222.6 V
  2. 445.2 V
  3. 2.226 V
  4. 0.02226 V
8.
Assertion (A): A rod 87 cm long moving at 19 m/s at right angles to a 3 × 10−1 T field has an emf of 4.959 V across it.
Reason (R): The emf across a rod moving at right angles to a field is e = Bl/v.
  1. Both A and R are true, and R is the correct explanation of A.
  2. Both A and R are true, but R is not the correct explanation of A.
  3. A is true, but R is false.
  4. A is false, but R is true.
Section B — Short answer8 × 2 = 16 marks
9.
A metal rod 60 cm long spins at 120 rad/s about one end, in a plane at right angles to a field of 1 × 10−1 T. Find the emf between its ends.
10.
A coil of 128 turns and area 60 cm2 rotates at 260 rad/s in a field of 7 × 10−1 T. Find the peak emf.
11.
The flux through each turn of a 466-turn coil of resistance 36 Ω falls by 38 mWb. Find the charge that flows round the coil.
12.
A rod 19 cm long slides at 14 m/s on rails joined by a 17 Ω resistor, at right angles to a field of 8 × 10−1 T. Find the emf and the current.
13.
A long solenoid has 1743 turns, cross-section 49 cm2 and length 86 cm. Find its self-inductance. (μ0 = 4π × 10−7 T m/A)
14.
An emf of 32 V is induced in a coil when the current in a nearby coil changes by 7 A in 5 × 10−2 s. Find their mutual inductance.
15.
A coil of 442 turns and area 370 cm2 lies at right angles to a field of 7 × 10−1 T. It is pulled out of the field in 5 × 10−1 s. Find the average emf.
16.
The south pole of a magnet is moved away from one face of a coil, changing the flux by 4 × 10−3 Wb in 2 × 10−1 s. What polarity does that face of the coil take, and what is the average emf in one turn?
Section C — Word problems4 × 4 = 16 marks
17.
A rod 100 cm long slides at a steady 17 m/s along rails joined by a 4 Ω resistor, at right angles to a field of 1 × 10−1 T. Find the emf, the current, the force needed to keep it moving, and the power supplied.
18.
A generator coil of 193 turns and area 310 cm2 turns at 100 rad/s in a field of 7 × 10−1 T and is joined to a 31 Ω load. Find the peak emf, the peak current and the rms emf.
19.
A coil of 462 turns, area 80 cm2 and resistance 44 Ω lies at right angles to a field of 4 × 10−1 T. The field is reversed in 5 × 10−1 s. Find the change in flux linkage, the average emf, the average current and the charge that flows.
20.
The current in a 241 mH inductor rises from 9 A to 14 A in 10 × 10−2 s. Find the emf induced and the extra energy stored.
AmbifulWorksheets
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CBSE · Class 12 · Physics

Answer key

Electromagnetic Induction · AW-12P-EMI-2026-V1 · For parents — fold back or tear off before handing the sheet to your child.

QAnswerWorking
1(d) 0 WbΦ = BA cos θ
2(a) 148.74 Ve = NΔΦ/Δt
3(d) 0.4 Ve = Blv
4(d) 16.76 Ve = L ΔI/Δt
5(c) weber—
6(b) 2.322 JU = ½LI2
7(a) 222.6 Ve = M ΔI/Δt
8(c) A is true, but R is false.A is true: Blv = 3 × 10−1 × 87 × 10−2 × 19 = 4.959 V. R is false.
92.16 Ve = ½Bωl2
10139.776 Ve0 = NBAω
110.492 Cq = NΔΦ/R (it does not depend on how fast the flux changes)
122.128 V; 0.125 Ae = Blv; I = e/R
136.92π × 10−3 HL = μ0N2A/l = 4π × 10−7 × 17432 × 49 × 10−4 ÷ (86 × 10−2)
140.229 HM = e/(ΔI/Δt)
1522.8956 VΔΦ = NBA, e = NBA/t
16the face becomes a north pole; 0.02 VLenz's law: the induced current opposes the change, so it attracts the receding pole
171.7 V; 0.425 A; 0.0425 N; 0.7225 We = Blv; I = e/R; F = BIl; P = Fv = e2/R
18418.81 V; 13.51 A; 418.81/√2 Ve0 = NBAω; I0 = e0/R; e_rms = e0/√2
192.9568 Wb; 5.9136 V; 0.1344 A; 0.0672 CΔΦ = 2NBA
2012.05 V; 13.8575 Je = L ΔI/Δt; ΔU = ½L(I22 − I12)