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Electromagnetic Induction Class 12 Physics Case-Based Questions 2026-27 | Chapter 6
Class 12 Physics • Chapter 6 • CBSE 2026–27
Electromagnetic Induction Class 12 Physics Case-Based Questions
10 original CBSE-style case studies with 40 application-based questions and answers covering magnetic flux, Faraday’s law, Lenz’s law, motional EMF, induced charge, self-induction, magnetic energy and mutual induction.
Built for students searching for Class 12 Physics Chapter 6 case-based questions, Electromagnetic Induction case study questions with answers and CBSE 2026–27 Physics case-based practice.
Each case is designed as a 4-mark practice set to strengthen passage reading, formula selection, numerical reasoning and conceptual application.
How to use these case studies: Read the passage first and identify the physical process. Then answer each part independently. In numerical parts, write the formula, substitute SI units and check the final unit.
CBSE pattern note: The current Class XII Physics sample-paper structure includes a 4-mark case-study section. These are original practice sets written in that style; they are not official CBSE questions or predictions.
Current Chapter 6 scope: The official 2026–27 curriculum includes electromagnetic induction, Faraday’s laws, induced EMF/current, Lenz’s law, self-induction and mutual induction in Chapter 6. AC generator and transformer are listed under Chapter 7, so they are deliberately excluded from these core case studies.
Chapter 6 Case-Based Practice Set
Attempt the questions before reading the explanations. Each case contains four 1-mark parts, giving a 4-mark practice set.
Case Study 1 — Moving Magnet and Coil • 4 Marks
Case: A coil is connected to a sensitive galvanometer. A bar magnet is moved towards the coil and the galvanometer deflects. When the magnet is held stationary near the coil, the deflection disappears. When the magnet is moved away, the deflection reverses. The observations show that induction depends on a change in magnetic flux linked with the circuit.
(1) Which physical law directly gives the magnitude of the induced EMF?
Answer: Faraday’s law.
Explanation: Faraday’s law relates induced EMF to the rate of change of magnetic flux linkage.
Explanation: Faraday’s law relates induced EMF to the rate of change of magnetic flux linkage.
(2) Why does the galvanometer show no sustained deflection when the magnet is stationary?
Answer: There is no change in magnetic flux, so the induced EMF is zero.
Explanation: A magnetic field being present is not enough; the linked flux must change.
Explanation: A magnetic field being present is not enough; the linked flux must change.
(3) The magnet is moved twice as fast while producing the same total change in flux. What happens to the magnitude of average induced EMF?
Answer: It doubles.
Explanation: For a fixed flux change, |εavg| = N|ΔΦ|/Δt. Halving the time doubles the average EMF.
Explanation: For a fixed flux change, |εavg| = N|ΔΦ|/Δt. Halving the time doubles the average EMF.
(4) Why does the deflection reverse when the magnet is moved away?
Answer: The direction of the induced current reverses because the change in flux reverses.
Explanation: Lenz’s law determines the direction of the induced effect so that it opposes the change in flux.
Explanation: Lenz’s law determines the direction of the induced effect so that it opposes the change in flux.
Case Study 2 — Magnetic Flux and Orientation • 4 Marks
Case: A 100-turn coil of area 0.020 m² is placed in a uniform magnetic field of 0.50 T. Initially, the magnetic field is parallel to the area vector of the coil. The coil is then rotated so that the field makes 60° with the area vector.
(1) Write the expression for magnetic flux through one turn.
Answer: Φ = BA cosθ.
Explanation: Here θ is the angle between B and the area vector.
Explanation: Here θ is the angle between B and the area vector.
(2) Find the initial flux per turn.
Answer: 0.010 Wb.
Explanation: Φ₁ = BA cos0° = 0.50 × 0.020 = 0.010 Wb.
Explanation: Φ₁ = BA cos0° = 0.50 × 0.020 = 0.010 Wb.
(3) Find the final flux per turn.
Answer: 0.005 Wb.
Explanation: Φ₂ = BA cos60° = 0.50 × 0.020 × 0.5 = 0.005 Wb.
Explanation: Φ₂ = BA cos60° = 0.50 × 0.020 × 0.5 = 0.005 Wb.
(4) If the rotation occurs in 0.20 s, find the magnitude of average induced EMF.
Answer: 2.5 V.
Explanation: |εavg| = N|ΔΦ|/Δt = 100(0.005)/0.20 = 2.5 V.
Explanation: |εavg| = N|ΔΦ|/Δt = 100(0.005)/0.20 = 2.5 V.
Case Study 3 — Changing Magnetic Field • 4 Marks
Case: A single-turn conducting loop of area 0.25 m² is kept in a uniform magnetic field perpendicular to its plane. The magnetic field decreases from 0.80 T to 0.20 T in 0.15 s. The loop has resistance 2 Ω.
(1) What happens to the magnetic flux through the loop?
Answer: It decreases.
Explanation: Since the loop area and orientation remain constant, Φ = BA decreases with B.
Explanation: Since the loop area and orientation remain constant, Φ = BA decreases with B.
(2) Find the magnitude of average induced EMF.
Answer: 1.0 V.
Explanation: |ΔΦ| = A|ΔB| = 0.25×0.60 = 0.15 Wb. Thus |εavg| = 0.15/0.15 = 1.0 V.
Explanation: |ΔΦ| = A|ΔB| = 0.25×0.60 = 0.15 Wb. Thus |εavg| = 0.15/0.15 = 1.0 V.
(3) Find the magnitude of average induced current.
Answer: 0.50 A.
Explanation: I = ε/R = 1.0/2 = 0.50 A.
Explanation: I = ε/R = 1.0/2 = 0.50 A.
(4) Which law determines the direction of this induced current?
Answer: Lenz’s law.
Explanation: The induced current produces a magnetic effect that opposes the decrease in linked flux.
Explanation: The induced current produces a magnetic effect that opposes the decrease in linked flux.
Case Study 4 — Motional EMF • 4 Marks
Case: A conducting rod of length 0.50 m moves at 4 m s⁻¹ perpendicular to a uniform magnetic field of 0.30 T. The rod forms part of a closed circuit of total resistance 3 Ω. The geometry is such that ε = Bℓv applies.
(1) Find the motional EMF.
Answer: 0.60 V.
Explanation: ε = Bℓv = 0.30×0.50×4 = 0.60 V.
Explanation: ε = Bℓv = 0.30×0.50×4 = 0.60 V.
(2) Find the current in the circuit.
Answer: 0.20 A.
Explanation: I = ε/R = 0.60/3 = 0.20 A.
Explanation: I = ε/R = 0.60/3 = 0.20 A.
(3) If the speed becomes 8 m s⁻¹ with all else unchanged, what happens to the EMF?
Answer: It doubles to 1.20 V.
Explanation: For fixed B and ℓ, ε is directly proportional to v.
Explanation: For fixed B and ℓ, ε is directly proportional to v.
(4) What provides the energy associated with the electrical output?
Answer: Mechanical work done in moving the rod against the magnetic force.
Explanation: External mechanical work is converted into electrical energy and, in a resistive circuit, heat.
Explanation: External mechanical work is converted into electrical energy and, in a resistive circuit, heat.
Case Study 5 — Self-Induction • 4 Marks
Case: A coil of inductance 2 H carries a current of 3 A. The current is then reduced uniformly to zero in 0.20 s. The coil is treated as an ideal inductor.
(1) Write the relation for self-induced EMF.
Answer: |ε| = L|dI/dt|.
Explanation: The sign in ε = −L(dI/dt) gives the direction; the magnitude uses the absolute value.
Explanation: The sign in ε = −L(dI/dt) gives the direction; the magnitude uses the absolute value.
(2) Find the magnitude of average self-induced EMF.
Answer: 30 V.
Explanation: |εavg| = L|ΔI|/Δt = 2×3/0.20 = 30 V.
Explanation: |εavg| = L|ΔI|/Δt = 2×3/0.20 = 30 V.
(3) What does the negative sign in the law represent?
Answer: Lenz’s law: the induced EMF opposes the change in current.
Explanation: The induced effect resists the change that produced it.
Explanation: The induced effect resists the change that produced it.
(4) Why can opening an inductive circuit produce a large voltage?
Answer: A rapid current change gives a large |dI/dt|, hence a large induced EMF.
Explanation: Since |ε| = L|dI/dt|, a very short switching time can produce a large voltage.
Explanation: Since |ε| = L|dI/dt|, a very short switching time can produce a large voltage.
Case Study 6 — Energy Stored in an Inductor • 4 Marks
Case: An ideal inductor of inductance 4 H carries a steady current of 2 A. The current is increased to 4 A. The magnetic field stores energy in the inductor.
(1) Write the expression for energy stored in an inductor.
Answer: U = ½LI².
Explanation: The stored magnetic energy depends on the square of current.
Explanation: The stored magnetic energy depends on the square of current.
(2) Find the initial energy.
Answer: 8 J.
Explanation: U₁ = ½×4×2² = 8 J.
Explanation: U₁ = ½×4×2² = 8 J.
(3) Find the final energy.
Answer: 32 J.
Explanation: U₂ = ½×4×4² = 32 J.
Explanation: U₂ = ½×4×4² = 32 J.
(4) By what factor does the stored energy change when current doubles?
Answer: It becomes four times.
Explanation: Because U is proportional to I², doubling I multiplies U by 4.
Explanation: Because U is proportional to I², doubling I multiplies U by 4.
Case Study 7 — Mutual Induction • 4 Marks
Case: Two coils are placed close to each other. The current in the primary coil is changed, and an EMF appears across the secondary coil. The mutual inductance is 0.30 H. The primary current changes from 2 A to 8 A in 0.20 s.
(1) What phenomenon is illustrated?
Answer: Mutual induction.
Explanation: A changing current in one coil changes the magnetic flux linked with the other coil.
Explanation: A changing current in one coil changes the magnetic flux linked with the other coil.
(2) Find the magnitude of average induced EMF in the secondary.
Answer: 9 V.
Explanation: |ε₂,avg| = M|ΔI₁|/Δt = 0.30×6/0.20 = 9 V.
Explanation: |ε₂,avg| = M|ΔI₁|/Δt = 0.30×6/0.20 = 9 V.
(3) If the rate of change of primary current is doubled, what happens to induced EMF?
Answer: It doubles.
Explanation: For fixed M, |ε₂| = M|dI₁/dt|.
Explanation: For fixed M, |ε₂| = M|dI₁/dt|.
(4) Name one important factor that affects mutual inductance.
Answer: The degree of magnetic flux linkage between the two coils.
Explanation: Geometry, relative placement and magnetic medium affect flux linkage and therefore mutual inductance.
Explanation: Geometry, relative placement and magnetic medium affect flux linkage and therefore mutual inductance.
Case Study 8 — Induced Charge • 4 Marks
Case: A 200-turn coil of resistance 5 Ω experiences a change in magnetic flux per turn from 0.020 Wb to 0.005 Wb. The change takes 0.30 s. The circuit is closed throughout.
(1) Write the relation between induced charge and flux change for a simple constant-resistance coil.
Answer: q = N|ΔΦ|/R.
Explanation: It follows by integrating Faraday’s law with I = ε/R.
Explanation: It follows by integrating Faraday’s law with I = ε/R.
(2) Find the total induced charge.
Answer: 0.60 C.
Explanation: q = 200×0.015/5 = 0.60 C.
Explanation: q = 200×0.015/5 = 0.60 C.
(3) If the same flux change occurs in half the time, what happens to total induced charge?
Answer: It remains 0.60 C.
Explanation: For constant R, q depends on total flux change, not on the time taken.
Explanation: For constant R, q depends on total flux change, not on the time taken.
(4) Would the same total charge necessarily flow if the circuit were open?
Answer: No; there would be induced EMF, but no conducting path for current.
Explanation: Induced EMF and induced current are distinct ideas.
Explanation: Induced EMF and induced current are distinct ideas.
Case Study 9 — Flux-Time Graph Reasoning • 4 Marks
Case: For a single-turn loop, magnetic flux changes uniformly from 0.04 Wb to 0.00 Wb in 0.20 s. It then remains constant at zero for the next 0.30 s. The loop has resistance 2 Ω.
(1) During which interval is induced EMF present?
Answer: During the first 0.20 s.
Explanation: EMF is induced while flux is changing. During constant flux, dΦ/dt = 0.
Explanation: EMF is induced while flux is changing. During constant flux, dΦ/dt = 0.
(2) Find the magnitude of induced EMF during the first interval.
Answer: 0.20 V.
Explanation: |ε| = |ΔΦ|/Δt = 0.04/0.20 = 0.20 V.
Explanation: |ε| = |ΔΦ|/Δt = 0.04/0.20 = 0.20 V.
(3) Find the current during the first interval.
Answer: 0.10 A.
Explanation: I = ε/R = 0.20/2 = 0.10 A.
Explanation: I = ε/R = 0.20/2 = 0.10 A.
(4) What does the slope of a Φ–t graph represent in magnitude for one turn?
Answer: The magnitude of induced EMF.
Explanation: Faraday’s law gives |ε| = |dΦ/dt| for one turn.
Explanation: Faraday’s law gives |ε| = |dΦ/dt| for one turn.
Case Study 10 — Integrated Chapter 6 Application • 4 Marks
Case: A 300-turn coil of area 0.010 m² and resistance 6 Ω is in a uniform magnetic field. The field is perpendicular to the coil and decreases uniformly from 0.60 T to zero in 0.30 s. The coil is then kept in zero field. The same coil has self-inductance 1.5 H when considered as an isolated inductor.
(1) Find the magnitude of average induced EMF during the field change.
Answer: 6.0 V.
Explanation: |ΔΦ| = AΔB = 0.010×0.60 = 0.006 Wb. Thus |εavg| = 300×0.006/0.30 = 6 V.
Explanation: |ΔΦ| = AΔB = 0.010×0.60 = 0.006 Wb. Thus |εavg| = 300×0.006/0.30 = 6 V.
(2) Find the magnitude of average induced current.
Answer: 1.0 A.
Explanation: I = ε/R = 6/6 = 1 A.
Explanation: I = ε/R = 6/6 = 1 A.
(3) Find the total induced charge that passes through the coil.
Answer: 0.30 C.
Explanation: q = N|ΔΦ|/R = 300×0.006/6 = 0.30 C.
Explanation: q = N|ΔΦ|/R = 300×0.006/6 = 0.30 C.
(4) If the isolated 1.5 H inductor carries 2 A, what magnetic energy is stored?
Answer: 3 J.
Explanation: U = ½LI² = ½×1.5×4 = 3 J.
Explanation: U = ½LI² = ½×1.5×4 = 3 J.
What These Case-Based Questions Cover
Core Chapter 6: electromagnetic induction, magnetic flux, Faraday’s laws, induced EMF and current, Lenz’s law, motional EMF, self-induction and mutual induction.
Application skills: identifying what changes in flux, interpreting direction, connecting EMF with current, reading a flux–time relationship, solving short numerical situations and explaining energy transfer.
Revision depth: the set deliberately mixes direct concept questions with calculation, proportional reasoning, explanation and application instead of repeating definition-only case studies.
How These Case Studies Match CBSE-Style Practice
| Feature | How this page uses it |
|---|---|
| Case passage | A short physical situation supplies the context before the questions. |
| 4-mark structure | Each case contains four 1-mark sub-parts for a compact 4-mark practice set. |
| Concept + application | Questions move between definitions, interpretation, formula use, proportional reasoning and short calculations. |
| Step-wise explanation | Every answer includes a concise reason or calculation rather than an answer-only key. |
| Current syllabus boundary | Chapter 6 topics are prioritised; AC generator and transformer are kept under Chapter 7. |
Most Important Search Topics for Chapter 6 Case-Based Practice
Electromagnetic Induction Class 12 Case Study Questions · Chapter 6 Case-Based Questions · Faraday’s Law Case Study · Lenz’s Law Case Study · Magnetic Flux Case-Based Questions · Motional EMF Case Study · Self-Induction Case-Based Questions · Mutual Induction Case Study · Induced EMF and Current · CBSE Class 12 Physics 2026–27
Before You Start: 7-Step Case-Study Method
1. Read the whole passage once.
2. Identify the changing quantity: B, A, θ, current or flux linkage.
3. Decide whether the question asks magnitude, direction, current or energy.
4. Select the governing law before substituting values.
5. Keep the area-vector angle convention correct.
6. Use SI units and show the essential calculation.
7. Check whether the final answer is physically sensible.
High-Yield Skills Tested
| Skill | Cases |
|---|---|
| Recognising change in magnetic flux | 1, 3, 9, 10 |
| Using Φ = BA cosθ correctly | 2 |
| Faraday’s law and average EMF | 1, 2, 3, 9, 10 |
| Lenz’s law and direction | 1, 3, 5 |
| Motional EMF and energy conversion | 4 |
| Self-induction | 5 |
| Energy stored in an inductor | 6, 10 |
| Mutual induction | 7 |
| Induced charge | 8, 10 |
| Flux–time graph/slope reasoning | 9 |
Common Case-Study Traps
Trap 1: A magnetic field can exist without induced EMF. Induction requires a change in linked magnetic flux.
Trap 2: In Φ = BA cosθ, θ is measured between B and the area vector.
Trap 3: Lenz’s law opposes the change in flux, not necessarily the original field.
Trap 4: Induced EMF may exist in an open circuit, but induced current requires a closed conducting path.
Trap 5: For a simple constant-resistance circuit, q = N|ΔΦ|/R; the time taken for the same flux change does not alter total charge.
Trap 6: Do not mix Chapter 6 with Chapter 7 content merely because older resources combine the topics.
Case-Based Answer Strategy
✓ Identify the physical change first.
✓ Write the governing law before substituting.
✓ Use SI units.
✓ Check the angle convention for flux.
✓ Separate EMF from current.
✓ Use Lenz’s law for direction.
✓ Check average versus instantaneous EMF.
✓ Read all four parts before deciding what information matters.
Quick Self-Test
1. Why does a stationary magnet near a stationary coil not continuously induce EMF?
2. Why does reversing flux change reverse induced current?
3. Why does doubling the rate of flux change double induced EMF?
4. Why does doubling current make inductor energy four times larger?
5. Why does mutual induction require changing flux linkage?
Continue Chapter 6 Preparation
Notes: Electromagnetic Induction Class 12 Physics Notes
Important Questions: Chapter 6 Important Questions
MCQs: Chapter 6 MCQs with Answers
Numericals: Chapter 6 Solved Numericals
Previous chapter: Magnetism and Matter Notes
Official CBSE: Physics Curriculum 2026–27 · Class XII 2026–27 Sample Question Papers & Marking Schemes
Frequently Asked Questions
Are these official CBSE questions?
No. They are original CBSE-style practice questions designed around the current syllabus and sample-paper structure.
What is the most important idea in electromagnetic induction?
Induced EMF is associated with a change in magnetic flux linkage. The change may arise from B, area, orientation or relative motion.
Which formulas should I know for case-based questions?
Key relations include Φ = BA cosθ, ε = −N dΦ/dt, εavg = −NΔΦ/Δt, ε = Bℓv for the standard perpendicular rod arrangement, εself = −L dI/dt, ε2 = −M dI1/dt and U = ½LI².
Are AC generator and transformer included here?
No. For the current CBSE 2026–27 mapping, AC generator and transformer are under Chapter 7: Alternating Current.
Source discipline: These are original practice questions. Current CBSE sample-paper structure and syllabus scope were used for alignment; third-party pages were used only for search-intent and topic-coverage research. No third-party question is presented as an official CBSE question.
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