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Electromagnetic Induction Class 12 Physics Chapter Test 2026-27 | CBSE
Electromagnetic Induction Class 12 Physics — Chapter Test
A syllabus-focused Electromagnetic Induction Class 12 Physics chapter test covering magnetic flux, Faraday’s laws, Lenz’s law, induced EMF/current, motional EMF, self-induction, mutual induction and energy stored in an inductor.
Current CBSE 2026–27 boundary: Chapter 6 covers electromagnetic induction, Faraday’s laws, induced EMF and current, Lenz’s law, self-induction and mutual induction. AC generator and transformer belong to Chapter 7: Alternating Current in the current curriculum and are not tested here as Chapter 6 core content. Unit IV carries 18 marks collectively for Chapters 6 and 7; CBSE does not prescribe a fixed standalone mark allocation for Chapter 6.
How to attempt: Try the complete paper before opening the answer key. Show formulas, substitutions and units in numerical questions. For direction questions, state the physical reasoning rather than writing only a final direction.
Section A — Objective Questions (8 × 1 = 8 marks)
Questions 1–6 are MCQs. Questions 7–8 are Assertion–Reason questions. Choose the most appropriate option.
Q1. A coil of N turns has magnetic flux Φ through each turn. If the flux changes from Φ1 to Φ2 in time Δt, the magnitude of average induced EMF is
Q2. A conducting loop is kept stationary in a uniform magnetic field. The field and the area of the loop remain constant, and the angle between B and the area vector does not change. The induced EMF is
Q3. The negative sign in Faraday’s law is associated with
Q4. A straight conducting rod of length 0.40 m moves at 5 m s−1 perpendicular to a magnetic field of 0.20 T in the standard motional-EMF arrangement. The induced EMF is
Q5. The self-inductance of a long air-core solenoid is proportional to
Q6. A current in a primary coil changes from 7 A to 3 A in 0.04 s. If mutual inductance is 0.5 H, the magnitude of induced EMF in the secondary is
Q7. Assertion (A): A changing magnetic flux can induce EMF in an open circuit.
Reason (R): Induced EMF is associated with the rate of change of magnetic flux, while a sustained conduction current additionally requires a closed conducting path.
Reason (R): Induced EMF is associated with the rate of change of magnetic flux, while a sustained conduction current additionally requires a closed conducting path.
Q8. Assertion (A): Lenz’s law opposes the change in magnetic flux that produces the induced current.
Reason (R): The induced current always opposes the magnetic field already present, irrespective of whether the flux is increasing or decreasing.
Reason (R): The induced current always opposes the magnetic field already present, irrespective of whether the flux is increasing or decreasing.
Section B — Short Answer Questions (4 × 2 = 8 marks)
Q9. A coil has 200 turns and area 5 × 10−3 m². Its plane is perpendicular to a uniform magnetic field of 0.40 T. The field becomes zero in 0.02 s. Calculate the magnitude of average induced EMF.
Q10. State Lenz’s law. Explain briefly how it is consistent with the law of conservation of energy.
Q11. Distinguish between self-induction and mutual induction. Give one defining equation for each.
Q12. A conducting rod of length ℓ moves with speed v in a magnetic field B in the standard arrangement. State the condition under which ε = Bℓv can be used and write the induced current if the closed circuit has resistance R.
Section C — Short Answer / Numerical Questions (3 × 3 = 9 marks)
Q13. A rectangular coil of 100 turns has area 2.0 × 10−2 m². A magnetic field perpendicular to the coil changes uniformly from 0.50 T to 0.10 T in 0.20 s. Find the magnitude of average induced EMF. If the total resistance of the circuit is 4 Ω, also find the induced current magnitude.
Q14. A long air-core solenoid has 1000 turns, length 0.50 m and cross-sectional area 4.0 × 10−3 m². Calculate its self-inductance. Take μ0 = 4π × 10−7 H m−1. Also state how L changes if the number of turns is doubled while all other quantities remain unchanged.
Q15. A coil of resistance 5 Ω experiences a change in magnetic flux per turn from 3.0 × 10−3 Wb to 1.0 × 10−3 Wb. The coil has 250 turns. Calculate the total charge transferred through the coil. State whether the answer depends on the time taken for the flux change, and explain why.
Section D — Case Study (1 × 4 = 4 marks)
Q16. Sliding Rod and Electromagnetic Induction 4 marks
A conducting rod of length 0.50 m slides on conducting rails that form a closed circuit of total resistance 2 Ω. The rod moves at constant speed 4 m s−1 perpendicular to a uniform magnetic field of 0.30 T. The magnetic field is perpendicular to the plane of the circuit.
- Write the expression for the induced EMF in the rod. (1)
- Calculate the induced EMF. (1)
- Calculate the induced current in the circuit. (1)
- State the direction principle that determines the induced current and explain what it opposes. (1)
Section E — Long Answer / Integrated Question (1 × 5 = 5 marks)
Q17. Self-Induction and Energy Storage
A coil of self-inductance 0.40 H carries a current that increases uniformly from 2 A to 5 A in 0.10 s.
- Calculate the magnitude of the self-induced EMF. (2)
- Explain the physical significance of the negative sign in ε = −L dI/dt. (1)
- Calculate the energy stored in the coil when the current reaches 5 A. (2)
Before checking: For Q13–Q17, write the governing formula first. A correct numerical answer without the correct physical relation is not a complete board-style solution.
Answer Key and Solutions
Section A — Answers
Q1 — B. |εavg| = N|ΔΦ|/Δt.
Q2 — B. Flux remains constant, so dΦ/dt = 0 and induced EMF is zero.
Q3 — B. The negative sign expresses Lenz’s law.
Q4 — C. ε = Bℓv = 0.20 × 0.40 × 5 = 0.40 V.
Q5 — B. For a long air-core solenoid, L = μ0N²A/ℓ.
Q6 — C. |ε| = M|ΔI|/Δt = 0.5 × 4/0.04 = 50 V.
Q7 — A. EMF can exist across an open circuit; a closed path is needed for sustained current. The reason explains the assertion.
Q8 — C. Assertion is true. Reason is false because Lenz’s law opposes the change in flux, not necessarily the existing field itself.
Q2 — B. Flux remains constant, so dΦ/dt = 0 and induced EMF is zero.
Q3 — B. The negative sign expresses Lenz’s law.
Q4 — C. ε = Bℓv = 0.20 × 0.40 × 5 = 0.40 V.
Q5 — B. For a long air-core solenoid, L = μ0N²A/ℓ.
Q6 — C. |ε| = M|ΔI|/Δt = 0.5 × 4/0.04 = 50 V.
Q7 — A. EMF can exist across an open circuit; a closed path is needed for sustained current. The reason explains the assertion.
Q8 — C. Assertion is true. Reason is false because Lenz’s law opposes the change in flux, not necessarily the existing field itself.
Section B — Solutions
Q9. Since the field is perpendicular to the plane, θ = 0° and Φ = BA. Initial flux per turn = 0.40 × 5 × 10−3 = 2 × 10−3 Wb. Thus |εavg| = N|ΔΦ|/Δt = 200 × 2 × 10−3/0.02 = 20 V.
Q10. Lenz’s law states that the induced current/EMF is directed so that its magnetic effect opposes the change in magnetic flux producing it. This opposition means external work is required to maintain the change; hence energy is not created from nothing.
Q11. Self-induction: changing current in a coil induces EMF in the same coil, ε = −L dI/dt. Mutual induction: changing current in one coil induces EMF in a coupled second coil, ε2 = −M dI1/dt.
Q12. ε = Bℓv applies to the standard arrangement with the relevant rod length, velocity and magnetic field mutually oriented so the full Bℓv result applies (commonly v ⟂ B and the rod is oriented appropriately). For a closed resistive circuit, I = ε/R = Bℓv/R.
Q10. Lenz’s law states that the induced current/EMF is directed so that its magnetic effect opposes the change in magnetic flux producing it. This opposition means external work is required to maintain the change; hence energy is not created from nothing.
Q11. Self-induction: changing current in a coil induces EMF in the same coil, ε = −L dI/dt. Mutual induction: changing current in one coil induces EMF in a coupled second coil, ε2 = −M dI1/dt.
Q12. ε = Bℓv applies to the standard arrangement with the relevant rod length, velocity and magnetic field mutually oriented so the full Bℓv result applies (commonly v ⟂ B and the rod is oriented appropriately). For a closed resistive circuit, I = ε/R = Bℓv/R.
Section C — Solutions
Q13. |ΔΦ| = A|ΔB| = 2.0 × 10−2 × 0.40 = 8.0 × 10−3 Wb per turn. Therefore |εavg| = N|ΔΦ|/Δt = 100 × 8.0 × 10−3/0.20 = 4.0 V. Current magnitude = ε/R = 4/4 = 1.0 A.
Q14. L = μ0N²A/ℓ = (4π × 10−7) × (1000)² × (4.0 × 10−3)/0.50 ≈ 1.01 × 10−2 H (about 10.1 mH). Since L ∝ N², doubling N makes L four times the original value.
Q15. Total charge q = N|ΔΦ|/R = 250 × (2.0 × 10−3)/5 = 0.10 C. It does not depend on the time taken for the flux change in this ideal constant-R case because q = ∫I dt = (N/R)∫|dΦ| = N|ΔΦ|/R.
Q14. L = μ0N²A/ℓ = (4π × 10−7) × (1000)² × (4.0 × 10−3)/0.50 ≈ 1.01 × 10−2 H (about 10.1 mH). Since L ∝ N², doubling N makes L four times the original value.
Q15. Total charge q = N|ΔΦ|/R = 250 × (2.0 × 10−3)/5 = 0.10 C. It does not depend on the time taken for the flux change in this ideal constant-R case because q = ∫I dt = (N/R)∫|dΦ| = N|ΔΦ|/R.
Section D — Case Study Solution
Q16.
1. ε = Bℓv.
2. ε = 0.30 × 0.50 × 4 = 0.60 V.
3. I = ε/R = 0.60/2 = 0.30 A.
4. The direction is determined by Lenz’s law: the induced current produces a magnetic effect that opposes the change in magnetic flux through the circuit.
1. ε = Bℓv.
2. ε = 0.30 × 0.50 × 4 = 0.60 V.
3. I = ε/R = 0.60/2 = 0.30 A.
4. The direction is determined by Lenz’s law: the induced current produces a magnetic effect that opposes the change in magnetic flux through the circuit.
Section E — Solution
Q17.
1. |ε| = L|ΔI|/Δt = 0.40 × (5 − 2)/0.10 = 12 V.
2. The negative sign shows that the self-induced EMF opposes the change in current that produces it, consistent with Lenz’s law.
3. U = ½LI² = ½ × 0.40 × 5² = 5 J.
1. |ε| = L|ΔI|/Δt = 0.40 × (5 − 2)/0.10 = 12 V.
2. The negative sign shows that the self-induced EMF opposes the change in current that produces it, consistent with Lenz’s law.
3. U = ½LI² = ½ × 0.40 × 5² = 5 J.
How to Use This Test for Revision
- First attempt: complete the test without notes and record your raw score out of 35.
- Then diagnose: classify every mistake as concept, formula selection, unit/conversion, calculation or direction/sign error.
- Target weak areas: use the relevant Chapter 6 resource below rather than rereading the whole chapter.
- Retest: after revision, repeat the numerical and reasoning questions without looking at the solutions.
Chapter 6 Resource Chain
Chapter 6 Notes
Important Questions
MCQs
Numericals
Case-Based Questions
Assertion–Reason
PYQs with Solutions
Formula Sheet + Quick Revision
Chapter 5 Notes — Previous Chapter
Official CBSE references:
Physics Curriculum 2026–27 ·
Class XII 2026–27 SQP & Marking Scheme
Student note: This is an original Learn Revise Hub chapter practice test, not an official CBSE question paper. The chapter boundary follows the official CBSE 2026–27 Physics curriculum, and the question mix is designed as a chapter-level practice assessment using the current CBSE-style objective, assertion–reason, short-answer, numerical and case-based formats.
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