OCR GCSE Physics A: Electromagnetic induction, alternators and dynamos (J249) – 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 P4.2e and P4.2f from the OCR GCSE (9-1) Gateway Science Physics A (J249) specification, Version 5.0 (July 2026). Both statements are Higher Tier only, so every question here is Higher Tier only. Read the study guide and revision notes first if you need to. Course pages: OCR GCSE Physics hub, the checklist and the free diagnostics.

Questions

1. (Higher Tier only) A coil is connected to a sensitive voltmeter. State two changes, to the magnet or the coil, that would increase the induced p.d. when the magnet is pushed in. [2]

2. (Higher Tier only) A bar magnet is held still inside a coil that is part of a complete circuit.

(a) State the reading on a sensitive ammeter in the circuit. [1] (b) Explain your answer to (a). [2]

3. (Higher Tier only) The south pole of a bar magnet is pushed towards end X of a coil that is part of a complete circuit. State the polarity that end X takes while the magnet approaches, and explain, in terms of energy, why it takes that polarity. [3]

4. (Higher Tier only) The same magnet is pushed into a coil that is not connected to anything else. Describe what is induced, and what is not. [2]

5. (Higher Tier only) A simple alternator has a coil that rotates between the poles of a magnet.

(a) Name the parts that connect the rotating coil to the external circuit. [1] (b) Explain why the output is alternating current. [3]

6. (Higher Tier only) An oscilloscope shows the output of an alternator. The time-base is 2.0 ms per division. The trace covers 10 divisions and shows exactly 4 complete cycles. The vertical setting is 4.0 V per division and the peak is 1.5 divisions above the centre line.

(a) Calculate the period of the output. [2] (b) Calculate the frequency. [1] (c) Calculate the peak p.d. [1]

7. (Higher Tier only) A different alternator turns at 1800 revolutions per minute. One revolution gives one cycle.

(a) Calculate the frequency of the output. [2] (b) Calculate the period, to 2 significant figures. [1] (c) Calculate the number of complete cycles in 2.0 s. [1]

8. (Higher Tier only) A dynamo produces direct current.

(a) Explain how the split-ring commutator makes the external current flow in one direction only. [3] (b) The dynamo turns at 9 revolutions per second. Calculate the number of pulses of p.d. each second. [1] (c) Describe the shape of the output p.d. over one rotation. [1]

9. (Higher Tier only) An alternator turns at 1200 revolutions per minute and gives a peak p.d. of 8.0 V. In this simple model the peak p.d. is proportional to the rotation speed.

(a) Calculate its frequency. [2] (b) The speed is reduced to 900 revolutions per minute. Calculate the new frequency and the new peak p.d. [2]

10. (Higher Tier only) A wind-up torch contains a dynamo.

(a) State one similarity between the construction of a dynamo and an alternator. [1] (b) State two differences between the output of a dynamo and the output of an alternator. [2] (c) The handle is harder to turn when the bulb is lit than when the bulb is disconnected. Explain why. [3]

11. (Higher Tier only) A student drops a small magnet through a coil at different speeds and records the largest p.d. induced. The coil is connected to a sensitive voltmeter.

Speed of magnet through coil (m/s)	Peak p.d. (mV)
0.5	0.8
1.0	1.6
1.5	2.4
2.0	3.2

(a) Describe the pattern in the results. [2] (b) Predict the peak p.d. at 2.5 m/s. Show your working. [2] (c) The magnet leaves the bottom of the coil. Explain why the p.d. is reversed compared with when it entered. [2] (d) The coil is now joined into a complete circuit and the magnet falls through it more slowly than before. Explain why. [2]

Answers

1. Move the magnet faster [1]. Use a stronger magnet, or more turns on the coil [1]. [2] Examiner insight: Name changes that make the field change faster or act on more wire. Vague answers such as “make it stronger” do not say what is made stronger.

2. (a) Zero [1]. (b) The magnetic field through the coil is not changing [1]. A p.d. is induced only when the field changes, so no current is driven [1]. [3] Examiner insight: Saying the field is “not strong enough” misses the point. State that the field is steady, not changing.

3. End X becomes a south pole [1]. The field through the coil is increasing, and the induced current gives a field that opposes this, repelling the approaching south pole [1]. Work has to be done to push the magnet against this force, and that work is the energy transferred to the circuit; if the force helped, energy would appear from nothing [1]. [3] Examiner insight: A pole alone gains little. Link it to opposing the change and to the work done on the magnet.

4. A p.d. is induced across the ends of the coil [1]. There is no current, because the circuit is not complete [1]. [2] Examiner insight: Keep p.d. and current separate. Writing “a current is induced” for an open circuit is a typical slip.

5. (a) Slip rings (and brushes) [1]. (b) The coil’s sides cut through the magnetic field lines as it turns [1]. In each half turn a given side moves through the field in the opposite direction, so the induced p.d. reverses [1]. Slip rings keep each brush connected to the same end of the coil, so the external current also reverses every half turn [1]. [4] Examiner insight: Mention the direction of the sides’ movement through the field. “It spins” on its own does not explain the reversal.

6. (a) Total time = 10 × 2.0 = 20 ms [1]. Period = 20 ÷ 4 = 5.0 ms (0.0050 s) [1]. (b) Frequency = 1 ÷ 0.0050 = 200 Hz [1]. (c) Peak = 1.5 × 4.0 = 6.0 V [1]. [4] Examiner insight: Convert milliseconds to seconds before using frequency = 1 ÷ period, and show the unit with every answer.

7. (a) 1800 ÷ 60 = 30 revolutions per second [1], so the frequency is 30 Hz [1]. (b) Period = 1 ÷ 30 = 0.033 s [1]. (c) 30 × 2.0 = 60 cycles [1]. [4] Examiner insight: A common slip is to give 1800 Hz because the revolutions per minute value is copied across. Convert to per second first.

8. (a) The p.d. induced in the coil reverses every half turn [1]. The commutator halves swap which brush they touch at that moment [1]. This cancels the reversal, so the external current always flows the same way [1]. (b) 2 × 9 = 18 pulses per second [1]. (c) The p.d. rises from zero to a peak and falls back to zero, twice per rotation, always in the same direction [1]. [5] Examiner insight: Say that the connections are swapped. “The commutator changes the current” is too vague to explain the mechanism.

9. (a) 1200 ÷ 60 = 20 revolutions per second [1], so the frequency is 20 Hz [1]. (b) 900 ÷ 60 = 15 Hz [1]. Peak p.d. = 8.0 × (900 ÷ 1200) = 6.0 V [1]. [4] Examiner insight: When the question says to assume proportionality, show the ratio you used. A bare answer gives the marker nothing to follow.

10. (a) Both have a coil rotating in a magnetic field (or both transfer kinetic energy to electrical energy) [1]. (b) The dynamo gives d.c. and the alternator gives a.c. [1]. The dynamo output pulses in one direction; the alternator output reverses direction each half turn [1]. (c) When the bulb is lit a current flows in the coil [1]. The current’s magnetic field opposes the change that induces it, so there is a force opposing the rotation [1]. More work must be done on the handle, and that work is transferred to light and heat [1]. [6] Examiner insight: Link the extra effort to the opposing effect of the induced current. “Because the bulb needs energy” does not explain why the handle itself resists.

11. (a) The peak p.d. increases as the speed increases [1]. It is directly proportional, with 1.6 mV for each m/s [1]. (b) 3.2 ÷ 2.0 = 1.6 mV per m/s [1]; 2.5 × 1.6 = 4.0 mV [1]. (c) The magnet is now leaving, so the field through the coil is decreasing rather than increasing [1]. The induced p.d. reverses direction [1]. (d) A current now flows [1]. Its magnetic field opposes the magnet’s motion, giving an upward force that slows the fall [1]. [8] Examiner insight: Use numbers from the table when you describe a pattern. Saying “it goes up” is weaker than showing a constant ratio.

Where marks are usually lost

• Writing “current” where “potential difference” is needed, especially for open circuits.
• Saying a strong magnet at rest induces a p.d., instead of stating that the field must change.
• Naming the induced pole without saying that it opposes the change.
• Explaining the alternator output with “the coil spins” and no mention of field lines being cut in opposite directions.
• Writing that the commutator “converts” a.c. to d.c. without saying it swaps connections every half turn.
• Forgetting to divide by 60 for revolutions per minute.
• Leaving milliseconds in the calculation when finding frequency.
• Missing the units, or giving Hz for a period.

Next steps

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Official syllabus

OCR GCSE (9-1) Gateway Science Physics A (J249), specification Version 5.0 (July 2026), first assessment 2018, published by OCR. This page covers statements P4.2e and P4.2f.