A Level Physics 20.5 Electromagnetic Induction MCQs – 30 Practice Questions with Answers (Cambridge 9702)
These 30 original practice questions cover Cambridge AS & A Level Physics (9702) Section 20.5, Electromagnetic Induction, an A Level (A2) topic. They build from magnetic flux and flux linkage, through Faraday’s law and Lenz’s law, to induced EMF calculations including motional EMF and the simple AC generator.
Formula recap: Magnetic flux Φ = B A cos θ, where θ is the angle between the field B and the normal to the area. Flux linkage = NΦ. Faraday’s law: E = −Δ(NΦ)/Δt. Motional EMF: E = B L v. Peak generator EMF: E0 = B A N ω. See the full A Level Physics Formula Sheet for more.
Flux & Flux Linkage
Q1. The magnetic flux Φ through a plane area A in a uniform field of flux density B is given by
- (A) Φ = B A sin θ
- (B) Φ = B A cos θ, where θ is the angle between B and the normal to the area
- (C) Φ = B / A
- (D) Φ = B A tan θ
Show Answer
Answer: B – Flux counts only the component of the field passing through the area, which is B cos θ when θ is measured from the normal. Multiplying by the area gives Φ = B A cos θ.
Q2. The magnetic flux through a flat loop is zero when
- (A) the field is perpendicular to the plane of the loop
- (B) the normal to the loop is parallel to the field
- (C) the field lies in the plane of the loop
- (D) the area of the loop is made as large as possible
Show Answer
Answer: C – If the field lies in the plane of the loop, the angle between B and the normal is 90°, so cos θ = 0 and Φ = B A cos θ = 0. No field lines pass through the loop.
Q3. A uniform field of flux density 0.40 T passes perpendicularly through a rectangular area of 2.0 × 10−3 m2. The magnetic flux through the area is
- (A) 8.0 × 10−4 Wb
- (B) 2.0 × 10−3 Wb
- (C) 8.0 × 10−2 Wb
- (D) 2.0 × 102 Wb
Show Answer
Answer: A – The field is perpendicular to the area, so θ = 0 and cos θ = 1. Φ = B A = 0.40 × 2.0 × 10−3 = 8.0 × 10−4 Wb.
Q4. A coil has 200 turns and the flux through each turn is 3.0 × 10−3 Wb. The flux linkage of the coil is
- (A) 1.5 × 10−5 Wb
- (B) 3.0 × 10−3 Wb
- (C) 6.0 Wb turns
- (D) 0.60 Wb turns
Show Answer
Answer: D – Flux linkage = NΦ = 200 × 3.0 × 10−3 = 0.60 Wb turns. Each turn links the same flux, so the linkage is the flux multiplied by the number of turns.
Q5. The unit of magnetic flux is the
- (A) tesla
- (B) weber, where 1 Wb = 1 T m2
- (C) henry
- (D) volt
Show Answer
Answer: B – Flux is measured in webers (Wb). Since Φ = B A, one weber equals one tesla times one square metre. The tesla is the unit of flux density, not flux.
Q6. A flat coil lies in a uniform magnetic field. The flux through the coil can be changed by rotating the coil because rotation changes
- (A) the flux density B
- (B) the area A of the coil
- (C) the angle θ between the field and the normal to the coil, so B A cos θ changes even though B and A stay the same
- (D) the number of turns on the coil
Show Answer
Answer: C – Rotating the coil changes θ in Φ = B A cos θ. This is exactly how a generator works: B and A are constant, but the angle changes continuously, so the flux changes continuously.
Q7. Magnetic flux density B is best described as
- (A) the magnetic flux through unit area held perpendicular to the field
- (B) the total magnetic flux leaving a magnet
- (C) the magnetic flux linkage per turn of a coil
- (D) the magnetic flux multiplied by the area
Show Answer
Answer: A – B is flux per unit area, which is why it is called flux density. A small area in a strong field can have the same flux as a large area in a weak field, because Φ = B A.
Q8. A uniform field of flux density 0.50 T makes an angle of 60° with the normal to a flat area of 4.0 × 10−3 m2. The flux through the area is
- (A) 2.0 × 10−3 Wb
- (B) 1.0 × 10−3 Wb
- (C) 1.7 × 10−3 Wb
- (D) 4.0 × 10−3 Wb
Show Answer
Answer: B – Φ = B A cos θ = 0.50 × 4.0 × 10−3 × cos 60° = 0.50 × 4.0 × 10−3 × 0.50 = 1.0 × 10−3 Wb. (Option C uses sin 60°, which would be correct only if the angle were measured from the plane.)
Q9. A rectangular coil is fully inside a uniform magnetic field. Which action does NOT change the flux through the coil?
- (A) Rotating the coil through 30°
- (B) Pulling the coil partly out of the field
- (C) Increasing the strength of the field
- (D) Sliding the coil sideways while it stays fully inside the uniform field
Show Answer
Answer: D – Sliding the coil sideways changes nothing: B, A and θ all stay the same, so Φ = B A cos θ is unchanged. The other actions change θ, the area in the field, or B.
Q10. A coil of 500 turns has a flux of 2.0 × 10−4 Wb through each turn. Its flux linkage is
- (A) 4.0 × 10−7 Wb turns
- (B) 2.0 × 10−4 Wb turns
- (C) 0.10 Wb turns
- (D) 2.5 × 106 Wb turns
Show Answer
Answer: C – Flux linkage = NΦ = 500 × 2.0 × 10−4 = 0.10 Wb turns.
Faraday’s Law & Lenz’s Law
Q11. Faraday’s law of electromagnetic induction states that the magnitude of the induced EMF in a circuit is proportional to
- (A) the rate of change of magnetic flux linkage
- (B) the magnetic flux linkage itself
- (C) the magnetic flux density
- (D) the resistance of the circuit
Show Answer
Answer: A – E = −Δ(NΦ)/Δt. It is the rate at which the flux linkage changes, not the size of the flux linkage, that sets the EMF.
Q12. A strong bar magnet is held stationary inside a coil connected to a sensitive voltmeter. The voltmeter reads
- (A) a large steady value, because the flux through the coil is large
- (B) a small steady value
- (C) a value that slowly increases
- (D) zero, because the flux is not changing
Show Answer
Answer: D – An EMF is induced only while the flux linkage is changing. A stationary magnet gives a large but constant flux, so Δ(NΦ)/Δt = 0 and no EMF is induced.
Q13. A magnet is moved towards a coil and an EMF is induced. The size of this EMF can be increased by
- (A) moving the magnet more slowly
- (B) moving the magnet faster towards the coil
- (C) using a coil of larger electrical resistance
- (D) holding the magnet still for longer near the coil
Show Answer
Answer: B – Moving the magnet faster changes the flux linkage more quickly, so Δ(NΦ)/Δt is larger and the EMF is larger. Resistance affects the induced current, not the EMF.
Q14. Lenz’s law, which says the induced current opposes the change producing it, is a consequence of
- (A) conservation of charge
- (B) conservation of momentum
- (C) conservation of energy
- (D) conservation of mass
Show Answer
Answer: C – If the induced current helped the change instead of opposing it, energy would be created from nothing. Opposition means work must be done to produce the electrical energy, as energy conservation requires.
Q15. The north pole of a bar magnet is pushed towards one end of a coil. As the magnet approaches, the end of the coil nearest the magnet behaves as
- (A) a south pole, attracting the magnet
- (B) a north pole, repelling the approaching magnet
- (C) an unmagnetised end, with no current induced
- (D) a north pole only after the magnet stops moving
Show Answer
Answer: B – By Lenz’s law the induced current opposes the approaching north pole, so the near end of the coil becomes a north pole and repels it. Work done against this repulsion becomes electrical energy.
Q16. The north pole of a bar magnet, already inside a coil, is now pulled away from the coil. The induced current flows so that the end of the coil nearest the magnet becomes
- (A) a north pole, pushing the magnet away faster
- (B) unmagnetised, as no flux passes through the coil
- (C) a north pole, attracting the magnet
- (D) a south pole, attracting the magnet and opposing its removal
Show Answer
Answer: D – The flux through the coil is decreasing, so the induced current acts to oppose the decrease: the near end becomes a south pole, which attracts the retreating north pole and opposes the removal.
Q17. Work must be done to push the north pole of a magnet into a coil connected to a resistor because
- (A) the induced current makes the coil repel the magnet, and the mechanical work done against this repulsion is converted into electrical energy
- (B) the coil becomes heavier as current flows in it
- (C) the magnet loses mass as it enters the coil
- (D) the resistor pushes back on the magnet directly
Show Answer
Answer: A – Lenz’s law makes the coil oppose the magnet’s entry. The work done against this opposing force is the source of the electrical energy dissipated in the resistor.
Q18. Two coils are placed close together. An EMF is induced in the second coil when
- (A) the first coil carries a large steady current
- (B) both coils are kept completely still with a steady current in the first
- (C) the current in the first coil is changed, for example switched on, switched off or increased
- (D) a magnet is held stationary inside the second coil
Show Answer
Answer: C – A changing current in the first coil produces a changing magnetic field, so the flux linkage of the second coil changes and an EMF is induced. A steady current gives steady flux and no induction. This is the principle of the transformer.
Q19. A coil is wound with twice as many turns, and the flux through each turn is changed at the same rate as before. The induced EMF
- (A) stays the same
- (B) doubles, because the rate of change of flux linkage doubles
- (C) halves
- (D) becomes four times larger
Show Answer
Answer: B – Flux linkage is NΦ. Doubling N doubles the flux linkage and so doubles its rate of change: E = Δ(NΦ)/Δt doubles.
Q20. A small magnet dropped through a vertical copper tube takes noticeably longer to fall than the same magnet dropped through a plastic tube. This is because
- (A) copper is denser than plastic, so air resistance is greater
- (B) the copper tube attracts the magnet and holds it up
- (C) the magnet sticks to the inside of the copper tube
- (D) currents induced in the copper produce effects that oppose the magnet’s motion, as Lenz’s law requires
Show Answer
Answer: D – As the magnet falls, the flux in the tube wall changes and eddy currents are induced. By Lenz’s law these currents oppose the motion, exerting an upward force that slows the fall. No currents are induced in the insulating plastic tube.
Induced EMF Calculations
Q21. The flux linkage of a coil changes from 0.12 Wb turns to 0.04 Wb turns in a time of 0.20 s. The average induced EMF is
- (A) 0.40 V
- (B) 0.80 V
- (C) 0.16 V
- (D) 0.024 V
Show Answer
Answer: A – Δ(NΦ) = 0.12 − 0.04 = 0.08 Wb turns. E = Δ(NΦ)/Δt = 0.08 / 0.20 = 0.40 V.
Q22. A coil has 400 turns. The flux through each turn changes by 5.0 × 10−4 Wb in 0.10 s. The average induced EMF in the coil is
- (A) 5.0 × 10−3 V
- (B) 2.0 V
- (C) 20 V
- (D) 0.20 V
Show Answer
Answer: B – Change of flux linkage = 400 × 5.0 × 10−4 = 0.20 Wb turns. E = 0.20 / 0.10 = 2.0 V. Forgetting the 400 turns gives the distractor of 5.0 × 10−3 V.
Q23. A metal rod of length 0.50 m moves at 4.0 m s−1 at right angles to a uniform magnetic field of flux density 0.30 T, cutting the field lines. The EMF induced between the ends of the rod is
- (A) 6.0 V
- (B) 1.2 V
- (C) 0.60 V
- (D) 0.060 V
Show Answer
Answer: C – Motional EMF E = B L v = 0.30 × 0.50 × 4.0 = 0.60 V.
Q24. The EMF induced in a moving metal rod arises because
- (A) the rod becomes magnetised as it moves
- (B) the field heats the rod and the temperature rise produces a voltage
- (C) friction between the rod and the air separates the charges
- (D) the free electrons in the rod experience a magnetic force F = B Q v that drives them towards one end, and this separation of charge produces the EMF
Show Answer
Answer: D – The charges in the rod move with it through the field, so each free electron experiences a force BQv along the rod. Electrons collect at one end until the resulting electric field balances the magnetic force; the charge separation is the induced EMF, E = B L v.
Q25. A metal rod of length 0.25 m slides at 2.0 m s−1 along parallel rails in a perpendicular field of 0.80 T. The rails are joined by a resistor of resistance 2.0 Ω. The induced current is
- (A) 0.20 A
- (B) 0.40 A
- (C) 0.80 A
- (D) 2.0 A
Show Answer
Answer: A – Induced EMF E = B L v = 0.80 × 0.25 × 2.0 = 0.40 V. Current I = E / R = 0.40 / 2.0 = 0.20 A.
Q26. A simple AC generator has a coil of 150 turns and area 0.020 m2 rotating in a field of flux density 0.50 T at a frequency of 50 Hz. The peak EMF is (peak EMF = B A N ω, with ω = 2πf)
- (A) 47 V
- (B) 471 V
- (C) 150 V
- (D) 3140 V
Show Answer
Answer: B – ω = 2π × 50 = 314 rad s−1. E0 = B A N ω = 0.50 × 0.020 × 150 × 314 = 471 V.
Q27. In a simple AC generator, the induced EMF is at its maximum value at the instant when
- (A) the flux linkage of the coil is at its maximum
- (B) the plane of the coil is perpendicular to the field
- (C) the plane of the coil is parallel to the field, so the flux linkage is zero but is changing at its fastest rate
- (D) the coil momentarily stops turning
Show Answer
Answer: C – Flux linkage varies as cos ωt, so it is zero when the coil plane is parallel to the field, but at that instant its rate of change is greatest. EMF follows the rate of change, so EMF is maximum there and zero when the flux linkage is maximum.
Q28. Eddy currents are unwanted in a transformer core mainly because they waste energy as heat and exert opposing forces. They are reduced by
- (A) making the core from one solid block of iron
- (B) making the core from a material of lower resistivity
- (C) removing the core completely
- (D) laminating the core, building it from thin insulated layers that increase the resistance of the eddy current paths
Show Answer
Answer: D – The changing flux induces circulating eddy currents in any bulk conductor. Thin insulated laminations break up the paths and greatly increase their resistance, so the eddy currents, and the heating and braking they cause, are much reduced.
Q29. The flux linkage of a coil decreases uniformly from 0.90 Wb turns to 0.30 Wb turns in 0.15 s. The average induced EMF is
- (A) 0.60 V
- (B) 2.0 V
- (C) 4.0 V
- (D) 90 V
Show Answer
Answer: C – Δ(NΦ) = 0.90 − 0.30 = 0.60 Wb turns. E = Δ(NΦ)/Δt = 0.60 / 0.15 = 4.0 V. The EMF depends on the change of linkage, not on either value alone.
Q30. A rod moves through a magnetic field and an EMF is induced between its ends. Which single change would double this EMF?
- (A) Moving the rod at twice the speed
- (B) Halving the speed and doubling the flux density
- (C) Doubling the flux density and halving the speed
- (D) Halving both the speed and the flux density
Show Answer
Answer: A – E = B L v is proportional to v, so doubling the speed doubles the EMF. In B and C the two changes cancel, leaving E unchanged; in D the EMF falls to one quarter.
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These are original practice questions written for revision; they are not Cambridge past paper questions, and this page is not affiliated with Cambridge Assessment International Education.