OCR A Level Physics A: Electromagnetic induction, generators and transformers (H556) – Practice Questions

These are original questions written for Marlbridge, for revision and practice on this content. They are not reproduced past-paper questions, and they do not replicate the exam’s exact structure, question count or mark tariffs – examination boards hold copyright in their own papers. Use these alongside the official past papers from your board or school.

These questions cover statements 6.3.3(a) to 6.3.3(f) of the OCR Advanced GCE in Physics A specification (H556), Version 3.0 (March 2026), for first assessment in 2017. Start with the study guide or the revision notes.

Questions

1. Define magnetic flux and state what is meant by one weber. [2]

2. A circular loop of radius 3.0 cm lies in a uniform magnetic field of flux density 0.37 T. The field lines make an angle of 25° with the plane of the loop. Calculate the magnetic flux through the loop. [3]

3. A coil has 600 turns and a cross-sectional area of 2.5 × 10⁻⁴ m². Its plane is perpendicular to a uniform field of flux density 0.060 T.

(a) Calculate the magnetic flux linkage of the coil. [2] (b) The coil is pulled out of the field in 0.15 s. Calculate the mean induced e.m.f. [2]

4. The north pole of a bar magnet is dropped, pole first, into the top of a vertical coil that is part of a closed circuit.

(a) State the direction of the induced current as seen from above, and explain how you decided. [2] (b) Explain, using energy, why the magnet falls with a smaller acceleration than it would without the coil. [2]

5. The flux linkage of a coil rises uniformly from zero to 6.0 mWb turns in 3.0 ms, stays constant for 4.0 ms, then falls uniformly to zero in 2.0 ms.

(a) Calculate the magnitude of the induced e.m.f. in the first and the last stage. [2] (b) State the e.m.f. in the middle stage and give the reason. [1] (c) Explain why the e.m.f. in the first and the last stage has opposite signs. [2]

6. A search coil has 250 turns and a cross-sectional area of 3.0 × 10⁻⁴ m². It is placed in the alternating field of a solenoid supplied at 100 Hz, with its normal along the axis. An oscilloscope shows a peak e.m.f. of 46 mV.

(a) Explain why the induced e.m.f. alternates at the same frequency as the field. [2] (b) Calculate the peak flux density at the search coil. [3]

7. A simple a.c. generator has a rectangular coil of 80 turns measuring 5.0 cm by 8.0 cm. It rotates at 40 Hz in a uniform field of flux density 0.12 T.

(a) State the purpose of the slip rings and brushes. [2] (b) Calculate the peak e.m.f. [3] (c) The rotation frequency is increased to 60 Hz. Calculate the new peak e.m.f. and the new period of the output. [2]

8. An ideal transformer reduces a mains supply of 230 V to 6.0 V. The secondary coil has 90 turns.

(a) Calculate the number of turns on the primary coil. [2] (b) The secondary supplies a current of 1.5 A. Calculate the current in the primary coil. [2]

9. Explain why a transformer needs an alternating supply, and why its iron core is laminated. [4]

10. A student investigates a transformer with a fixed primary of 300 turns connected to a 6.0 V a.c. supply. She changes the number of turns on the secondary coil and measures the secondary p.d.

Secondary turns Ns	75	150	300	450
Secondary p.d. Vs / V	1.42	2.85	5.70	8.55

(a) Describe how she should take readings, including one safety precaution. [3] (b) State what graph she should plot and the gradient she expects for an ideal transformer, and comment on the data. [3] (c) With 150 secondary turns, the primary takes 3.5 A and the secondary supplies 6.0 A. Calculate the efficiency of the transformer. [2]

11. The flux through each turn of a 50-turn generator coil varies as Φ = Φ₀ cos(2πft), where Φ₀ = 4.0 mWb and f = 50 Hz.

(a) Calculate the peak e.m.f. of the generator. [3] (b) Explain why the e.m.f. is zero when the flux is at its maximum value. [2] (c) The output is connected to the primary of an ideal transformer with 800 primary turns and 150 secondary turns. Calculate the peak p.d. across the secondary. [2] (d) A resistor of 24 Ω is connected across the secondary. Calculate the peak current in the secondary and in the primary. [3]

Answers

1. Magnetic flux is the product of the flux density and the area perpendicular to the field, Φ = BA cos θ [1]. One weber is the flux through an area of 1 m² that is perpendicular to a uniform field of flux density 1 T [1]. [2] Examiner insight: A definition of the weber is stronger when it names the perpendicular area and the field strength.

2. A = π × (0.030)² = 2.83 × 10⁻³ m² [1]. The angle to the normal is 90° − 25° = 65° [1]. Φ = 0.37 × 2.83 × 10⁻³ × cos 65° = 4.4 × 10⁻⁴ Wb [1]. [3] Examiner insight: The angle given is to the plane, so using 25° straight in the cosine is the usual slip. Show the 65° step.

3. (a) Φ = BA = 0.060 × 2.5 × 10⁻⁴ = 1.5 × 10⁻⁵ Wb [1]. NΦ = 600 × 1.5 × 10⁻⁵ = 9.0 × 10⁻³ Wb turns [1]. (b) |E| = ΔNΦ/Δt = 9.0 × 10⁻³ / 0.15 [1] = 0.060 V [1]. [4] Examiner insight: Taking the final flux linkage as zero is the key reading of “pulled out of the field”. Give the unit Wb turns.

4. (a) The coil must repel the approaching north pole, so the top face of the coil is a north pole [1]. A north face means the current is anticlockwise as seen from above [1]. (b) The induced current opposes the motion, so there is an upward force on the magnet [1]. The magnet does work against it, so some gravitational potential energy becomes electrical energy instead of kinetic energy [1]. [4] Examiner insight: State what is being opposed. “Opposes the motion of the magnet” or “opposes the increase in flux” is stronger than “opposes the magnet”.

5. (a) First stage: |E| = 6.0 mWb turns / 3.0 ms = 2.0 V [1]. Last stage: |E| = 6.0 mWb turns / 2.0 ms = 3.0 V [1]. (b) The e.m.f. is zero because the flux linkage is not changing [1]. (c) The flux linkage is increasing in the first stage and decreasing in the last [1]. The induced e.m.f. opposes the change, so it acts in opposite directions (opposite signs, E = −gradient) [1]. [5] Examiner insight: mWb turns divided by ms gives volts directly. Give a zero e.m.f. a reason that refers to the rate of change.

6. (a) The field varies sinusoidally with time, so the flux linkage of the coil varies in the same way [1]. The e.m.f. depends on how fast the flux linkage changes, which varies sinusoidally at the same frequency [1]. (b) ω = 2π × 100 = 628 rad s⁻¹ [1]. B₀ = E₀/(NAω) = 0.046 / (250 × 3.0 × 10⁻⁴ × 628) [1] = 9.8 × 10⁻⁴ T [1]. [5] Examiner insight: Convert millivolts to volts before substituting.

7. (a) The slip rings rotate with the coil and let the current pass out while the coil turns continuously [1]. The fixed brushes press on the rings and connect them to the external circuit [1]. (b) A = 0.050 × 0.080 = 4.0 × 10⁻³ m² [1]. ω = 2π × 40 = 251 rad s⁻¹ [1]. E₀ = NBAω = 80 × 0.12 × 4.0 × 10⁻³ × 251 = 9.7 V [1]. (c) E₀ is proportional to ω, so the new value is 9.65 × 60/40 = 14.5 V [1] and the period is 1/60 = 0.017 s [1]. [7] Examiner insight: Using the ratio 60/40 in part (c) avoids repeating the full calculation.

8. (a) Np/Ns = Vp/Vs [1], so Np = 90 × 230/6.0 = 3450 turns [1]. (b) Ip/Is = Ns/Np, so Ip = 1.5 × 90/3450 [1] = 0.039 A [1]. [4] Examiner insight: Checking that power in equals power out (9.0 W each) shows the current ratio is the right way up.

9. An alternating p.d. gives an alternating current in the primary and a changing flux in the core [1]. The changing flux linkage in the secondary induces an e.m.f. [1]. A steady supply would give a constant flux and no induced e.m.f. The core is laminated because the changing flux induces eddy currents in the iron [1]. Thin insulated sheets restrict these currents, so less energy is transferred as heat [1]. [4] Examiner insight: Link each feature to its effect: lamination reduces eddy currents, which reduces heating.

10. (a) Connect the primary to a low-voltage a.c. supply and an a.c. voltmeter across the secondary [1]. Keep Np and Vp constant, change Ns in steps and record Vs each time [1]. Safety: use a low voltage and switch off before changing the coils [1]. (b) Plot Vs against Ns [1]. For an ideal transformer this is a straight line through the origin with gradient Vp/Np = 6.0/300 = 0.020 V per turn [1]. The data give 0.019 V per turn, about 5% below the ideal value, which suggests systematic losses such as heating in the windings or flux that does not link both coils [1]. (c) Output power = 2.85 × 6.0 = 17.1 W and input power = 6.0 × 3.5 = 21.0 W [1]. Efficiency = 17.1/21.0 = 81% [1]. [8] Examiner insight: In comments on data, quote a figure from the data (here the gradient or the percentage difference) rather than only saying “close to the ideal value”.

11. (a) ω = 2π × 50 = 314 rad s⁻¹ [1]. E₀ = NΦ₀ω = 50 × 4.0 × 10⁻³ × 314 [1] = 63 V [1]. (b) The e.m.f. depends on how fast the flux linkage changes [1]. At maximum flux the flux linkage is momentarily not changing (the graph is flat at its peak), so the gradient and the e.m.f. are zero [1]. (c) Vs = E₀ × Ns/Np = 62.8 × 150/800 [1] = 12 V [1]. (d) Is = Vs/R = 11.8/24 = 0.49 A [1]. Ip = Is × Ns/Np = 0.491 × 150/800 [1] = 0.092 A [1]. [10] Examiner insight: Carry unrounded values through the later parts.

Where marks are usually lost

• Using the angle to the plane instead of the angle to the normal in Φ = BA cos θ.
• Leaving out the number of turns when finding flux linkage or e.m.f.
• Not converting centimetres, millivolts or milliseconds before substituting.
• Saying Lenz’s law “stops” the change, or not naming the change being opposed.
• Giving a reason for zero e.m.f. that refers to the flux being zero instead of its rate of change.
• Inverting the current ratio in the transformer equation.
• Describing a transformer investigation without stating what is kept constant.

Next steps

• Revise with the revision notes.
• Re-read the study guide for any question that cost you marks.
• Tick off the topic on the printable checklist and see the course hub.
• Try all free 10-minute diagnostics at /diagnostics/.
• Book a free trial class.

Official syllabus

These questions follow the OCR Advanced GCE in Physics A (H556) specification, Version 3.0 (March 2026), for first assessment in 2017, published by OCR, section 6.3.3 (Electromagnetism), statements (a) to (f).