Electromagnetic InductionNEET MCQs with solutions
Electromagnetic Induction covers Faraday's laws, Lenz's law, motional EMF, eddy currents, self and mutual inductance. NEET tests EMF calculations using Faraday's law, Lenz's law direction, motional EMF in rods and inductance formulas.
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- 12 Physics
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Q1Electromagnetic Induction: Introduction & Faraday's Experiments
A coil is connected to a galvanometer. A bar magnet is moved towards the coil and the galvanometer shows a deflection. When the magnet is held stationary inside the coil, the galvanometer returns to zero. What is the essential cause of the observed induced current?
Not quite — the answer is B.
A changing magnetic flux is the essential condition for induced emf by Faraday's law. A stationary magnet produces no flux change, so no emf results. Option A is the classic trap: presence of B alone is insufficient without a rate of change.
Q2Electromagnetic Induction: Introduction & Faraday's Experiments
A bar magnet is moved towards a conducting coil. In a second experiment, the magnet is kept stationary while the coil is moved towards it with the same speed. In both cases galvanometer deflection is observed. Which conclusion is most appropriate?
Not quite — the answer is C.
Both experiments show that it is the relative motion that matters, not which object moves. Relative motion changes flux linked with the coil. Option D is wrong: emf depends on rate of flux change, not on final position alone.
Q3Electromagnetic Induction: Introduction & Faraday's Experiments
A conducting coil is placed in a region of uniform magnetic field. The coil and the field source are both stationary, and the field remains constant with time. Which observation is expected?
Not quite — the answer is D.
With constant B and stationary coil, magnetic flux Φ = BAcosθ is constant, so dΦ/dt = 0 and induced emf = 0. Options A and B are wrong because no flux change exists. Option C incorrectly attributes deflection to thermal effects.
Q4Electromagnetic Induction: Introduction & Faraday's Experiments
A bar magnet is first pushed towards a coil connected to a galvanometer, causing deflection in one direction. The same magnet is then pulled away from the coil. Which of the following correctly describes what happens to the galvanometer deflection?
Not quite — the answer is A.
Withdrawing the magnet reduces the linked flux, giving dΦ/dt of opposite sign compared to approach. By Faraday's law, the induced emf reverses polarity, reversing the galvanometer deflection. Option D is wrong: direction reversal depends on the direction of flux change, not on the comparative speed.
Q5Faraday's Laws of Electromagnetic Induction
A coil of 100 turns experiences a change in magnetic flux through each turn from 4.0×10⁻³ Wb to 1.0×10⁻³ Wb in 0.05 s. The magnitude of the average induced emf is:
Not quite — the answer is B.
|ΔΦ| = 3.0×10⁻³ Wb, Δt = 0.05 s. |ε| = N|ΔΦ|/Δt = 100 × 3.0×10⁻³ / 0.05 = 6 V. Option A (0.06 V) is the trap for omitting N and computing ΔΦ/Δt alone. Every turn contributes equally, so N is an essential multiplicative factor.
Q6Faraday's Laws of Electromagnetic Induction
Which statement correctly distinguishes Faraday's first and second laws of electromagnetic induction?
Not quite — the answer is C.
Faraday's first law states that an emf is induced whenever magnetic flux through a circuit changes. Faraday's second law gives |ε| = N|dΦ/dt|, quantifying the magnitude. Option D is the standard trap — it reverses the roles of the two laws entirely.
Q7Faraday's Laws of Electromagnetic Induction
Consider the following statements about Faraday's law: I. The negative sign in ε = −NdΦ/dt represents the direction of induced emf consistent with Lenz's law. II. The magnitude of induced emf depends on the rate of change of magnetic flux. III. For N turns, induced emf is N times the single-turn value when each turn has the same flux change. IV. A larger magnetic flux always produces a larger induced emf even when the flux is constant. How many of the above statements are correct?
Not quite — the answer is C.
Statement I is true: the negative sign encodes Lenz's law and gives the direction of ε. Statement II is true by Faraday's second law. Statement III follows from ε = N|dΦ/dt|. Statement IV is false: constant flux gives dΦ/dt = 0 and zero induced emf regardless of magnitude of Φ. Hence three statements are correct.
Q8Faraday's Laws of Electromagnetic Induction
All of the following statements about Faraday's law (ε = −NdΦ/dt) are correct EXCEPT:
Not quite — the answer is D.
Faraday's law shows that induced emf depends on dΦ/dt, the rate of change of flux, not on the instantaneous value of Φ itself. A large constant flux produces zero emf. Options A, B and C are all direct, correct consequences of ε = −NdΦ/dt.
Q9Lenz's Law
A north pole of a bar magnet is moved towards a conducting coil. According to Lenz's law, the face of the coil nearest the magnet behaves as:
Not quite — the answer is B.
The approaching N-pole increases inward flux. Lenz's law requires the induced effect to oppose this increase, so the near face becomes a north pole and repels the approaching magnet. Option A (south pole) is the classic trap — a south face would attract and assist the motion, violating energy conservation.
Q10Lenz's Law
A conducting loop is placed in a uniform magnetic field directed into the page. The field is then switched off completely. The induced current in the loop is:
Not quite — the answer is A.
When B is switched off, inward flux decreases to zero. The induced current must oppose this decrease by producing inward flux. By the right-hand rule, a clockwise current (viewed from front) produces a magnetic field into the page. Option B (anticlockwise) is the trap — it would produce outward flux, further reducing inward flux instead of opposing the decrease.
Q11Lenz's Law
A bar magnet is moved away from a conducting coil with its north pole facing the coil. The face of the coil nearest the receding magnet becomes:
Not quite — the answer is C.
When the N-pole recedes, inward flux decreases. Lenz's law requires opposition to this decrease, so the near face becomes a south pole to attract the receding north pole and oppose the separation. Option A is the trap — a north pole would repel the magnet and assist its motion, violating Lenz's law.
Q12Lenz's Law
A conducting loop lies in the plane of the page. A magnetic field directed into the page is increasing with time. The induced current in the loop is:
Not quite — the answer is D.
The external inward flux is increasing. Lenz's law requires the induced field to oppose this increase, so the induced field must point out of the page. By the right-hand rule, an anticlockwise current produces a field out of the page. Option A is the standard trap — clockwise current would produce inward flux, reinforcing rather than opposing the increase.
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Get RankUp on Google PlayQ13Motional EMF
A straight conducting rod of length 0.50 m moves with speed 4 m/s perpendicular to a uniform magnetic field of 0.20 T, with the rod, velocity and magnetic field mutually perpendicular. The induced emf across the rod is:
Not quite — the answer is B.
Motional emf is ε = Blv = 0.20 × 0.50 × 4 = 0.40 V. Option A arises from omitting the factor of 10 in the field value. Options C and D result from successive factor-of-10 errors.
Q14Motional EMF
A conducting rod moves parallel to a uniform magnetic field. Which statement correctly explains why the motional emf due to that field is zero?
Not quite — the answer is C.
Motional emf arises from F = q(v×B) on free charges. When v is parallel to B, sin 0° = 0, so the force is zero and no charge separation occurs. Option B confuses flux-based (Faraday) reasoning with the microscopic force origin of motional emf.
Q15Motional EMF
A 10 m long horizontal wire extending east to west falls vertically downward at 5 m/s. The horizontal component of Earth's magnetic field is 0.30×10⁻⁴ T directed north. The instantaneous motional emf is:
Not quite — the answer is A.
The wire, velocity and horizontal field component are mutually perpendicular, so ε = Blv = 0.30×10⁻⁴ × 10 × 5 = 1.5×10⁻³ V. Option B results from omitting the length factor; option C from doubling the field value in error.
Q16Motional EMF
A conducting rod of length l moves with velocity v in a uniform magnetic field B. The angle between v and B is θ. Which expression gives the magnitude of the motional emf?
Not quite — the answer is D.
The force on charge carriers is qvB sinθ, driving charge separation along the rod, so ε = Blv sinθ. Option A is the trap — cosθ would apply if θ were measured from the perpendicular, not from B. Option C assumes perpendicular geometry and ignores the angle.
Q17Magnetic Flux
The magnetic flux through a plane surface is maximum when the angle between the magnetic field and the area vector is:
Not quite — the answer is A.
Φ = BAcosθ is maximum when cosθ = 1, i.e., at θ = 0°. The standard trap is using the angle between B and the plane (which would be 90° for maximum flux) instead of the angle between B and the area vector.
Q18Magnetic Flux
A rectangular loop of area 0.50 m² is placed in a uniform magnetic field of 0.40 T. The magnetic field makes an angle of 30° with the plane of the loop. What is the magnetic flux through the loop?
Not quite — the answer is A.
Angle with plane = 30°, so angle with area vector = 90° − 30° = 60°. Therefore Φ = BAcos60° = 0.40 × 0.50 × 0.5 = 0.10 Wb. Option B (0.17 Wb) is the trap for using cos30° directly without converting the angle to the area vector.
Q19Magnetic Flux
A magnetic field of 2 T passes through a rectangular surface of dimensions 20 cm × 30 cm. The field makes an angle of 60° with the area vector. What is the magnetic flux?
Not quite — the answer is C.
Area = 0.20 × 0.30 = 0.06 m². Φ = BAcos60° = 2 × 0.06 × 0.5 = 0.06 Wb. Option B (0.10 Wb) is the trap for using sin60° ≈ 0.866 instead of cos60° = 0.5. Option D (0.12 Wb) is the trap for using the full product BA without any cosine factor.
Q20Magnetic Flux
All of the following statements about magnetic flux through a surface are correct EXCEPT:
Not quite — the answer is D.
Magnetic flux Φ = BAcosθ is a scalar dot product. Its sign depends on the direction of the chosen area vector: if the angle between B and the area vector exceeds 90°, cosθ < 0 and flux is negative. Options A, B, and C are all correct statements about magnetic flux.
Q21Electromagnetic Induction: Introduction & Faraday's Experiments
A current is induced in a closed stationary coil. Which of the following correctly explains how this is possible without any mechanical motion?
Not quite — the answer is A.
Faraday's law states emf = -NdΦ/dt. If B changes with time, flux changes even without motion. Options C and D are physically incorrect. Option B confuses resistance with the cause of induction.
Q22Electromagnetic Induction: Introduction & Faraday's Experiments
An electron moves along a straight path past a nearby conducting coil. What happens to the induced current as the electron approaches the coil and then moves away from it?
Not quite — the answer is D.
As the electron approaches, the magnetic flux linked with the coil increases; as it recedes, flux decreases. The sign of dΦ/dt reverses as the electron passes, so the induced emf and current reverse direction. Option B is wrong: a moving charge produces a magnetic field and changing flux.
Q23Electromagnetic Induction: Introduction & Faraday's Experiments
Two identical coils are connected separately to galvanometers. In experiment I, a bar magnet is moved rapidly into the coil. In experiment II, the same magnet is moved slowly into the coil through the same final position. Which statement is correct?
Not quite — the answer is B.
Induced emf = N|dΦ/dt|. Faster motion gives a larger dΦ/dt and hence a larger instantaneous emf. The total flux change may be equal in both, but the rate is higher in experiment I. Option C is the classic trap: confusing total flux change with rate of flux change.
Q24Electromagnetic Induction: Introduction & Faraday's Experiments
A coil has 200 turns. The magnetic flux through each turn changes uniformly from 2.0 × 10⁻³ Wb to 5.0 × 10⁻³ Wb in 0.10 s. What is the magnitude of the average induced emf?
Not quite — the answer is C.
ε = N|ΔΦ|/Δt = 200 × (3.0 × 10⁻³) / 0.10 = 6 V. Option B (0.6 V) is the trap: students who forget to multiply by N get ΔΦ/Δt = 0.03 V. Option A uses only ΔΦ without dividing by time. Option D divides by wrong Δt.
ELITE question · AIR under 50 level
This chapter has 78 ELITE questions for students aiming at the very top. They are only in the app.
Unlock ELITE questions in the appKey EMI Concepts
Quick revision: most questions in this chapter test these facts.
| Concept | Key Formula |
|---|---|
| Faraday's law | EMF = −dΦ/dt; induced EMF proportional to rate of change of flux |
| Lenz's law | Induced current opposes the cause; determines direction of EMF |
| Motional EMF | ε = Bvl; rod moving in magnetic field perpendicular to both B and l |
| Self inductance | ε = −L(dI/dt); solenoid: L = μ₀n²Al |
| Mutual inductance | ε₂ = −M(dI₁/dt); M = μ₀n₁n₂Al (coaxial solenoids) |
| Energy in inductor | U = ½LI² |
What the app covers in this chapter
199 questions in total, each with a detailed explanation.
| Electromagnetic Induction: Introduction & Faraday's Experiments | 40 |
| Faraday's Laws of Electromagnetic Induction | 40 |
| Lenz's Law | 40 |
| Motional EMF | 40 |
| Magnetic Flux | 39 |
Questions students ask
Is Electromagnetic Induction important for NEET?
Yes — Faraday's law, Lenz's law and motional EMF are tested every year. Both conceptual and numerical questions appear.
Which topics should I revise first?
Focus on Faraday's law EMF calculations, Lenz's law for direction, motional EMF (ε = Bvl), self and mutual inductance formulas, and energy stored in an inductor.
How many questions from this chapter are on RankUp?
The RankUp app has 199 questions on Electromagnetic Induction, including 78 ELITE questions. Every question has a detailed explanation.
