OCR GCSE Physics A: Electromagnetic induction, alternators and dynamos (J249) – Revision Notes

These notes condense the full study guide for the OCR GCSE (9-1) Gateway Science Physics A (J249) specification, Version 5.0 (July 2026), for first assessment in 2018. They cover statements P4.2e and P4.2f, both of which are Higher Tier only. Test yourself with the practice questions, tick off statements on the checklist, see the course hub, or take a free diagnostic to find weak spots.

Key ideas at a glance

Statement	Core idea	Tier
P4.2e	A changing magnetic field around a conductor induces a p.d. across its ends. A current, if the circuit is complete, makes a field that opposes the change.	Higher Tier only
P4.2f	An alternator uses induction to give a.c. (slip rings). A dynamo gives d.c. (split-ring commutator).	Higher Tier only

Induced p.d. (Higher Tier only)

• Needs a change. Moving a magnet, moving a conductor across field lines, or switching or reversing the field. A steady field gives nothing.
• P.d. first, current second. The p.d. is always induced. A current flows only if the circuit is complete.
• Bigger induced p.d. from: faster movement, a stronger magnet, more turns on the coil.
• Reversed induced p.d. from: reversing the direction of motion, or turning the magnet round.
• Opposing effect. The induced current’s own field opposes the change that caused it. This is Lenz’s law, and the energy you do work with is the energy that appears in the circuit.

Method: polarity of the induced field

1. Decide whether the field through the coil is increasing or decreasing.
2. Increasing: the near end of the coil takes the same pole as the approaching magnet (repels it).
3. Decreasing: the near end takes the opposite pole to the departing magnet (attracts it).
4. Check your answer: the coil always resists what you are doing.

Magnet movement	Near end of coil
North pole approaches	North
North pole leaves	South
South pole approaches	South
South pole leaves	North

Galvanometer demonstration

Action	Reading
Magnet still inside coil	Zero
Push in, then pull out	Opposite deflections
Faster movement	Larger deflection
Magnet reversed	Deflection reversed

Alternator (Higher Tier only)

• Coil rotates in a magnetic field. Output taken through slip rings and brushes.
• Each ring is complete, so each brush stays joined to the same end of the coil.
• The p.d. reverses every half turn, because each side of the coil moves through the field in the opposite direction.
• Output is a.c. One rotation of the coil gives one cycle in the simple model.
• Maximum p.d. when the plane of the coil is along the field lines (sides cutting fastest).
• Zero p.d. when the plane of the coil is at right angles to the field lines (sides moving along the lines).

Dynamo (Higher Tier only)

• Same coil and field, but the output goes through a split-ring commutator and brushes.
• The commutator swaps the connections each half turn, just as the p.d. in the coil reverses.
• External current always flows the same way, which is d.c.
• The d.c. is pulsing: two pulses per rotation, each rising from zero to a peak and back to zero.
• Used in wind-up torches and radios, as suggested in the specification.

Alternator and dynamo side by side

	Alternator	Dynamo
Contacts	Slip rings	Split-ring commutator
Output	a.c.	d.c.
Per rotation	One cycle	Two pulses
Faster rotation	Higher frequency, higher peak	More pulses per second, higher peak

A motor transfers electrical energy to kinetic energy. A dynamo or alternator transfers kinetic energy to electrical energy. The layouts look alike, but the energy transfer is reversed.

Model explanations to learn

Longer answers earn credit for a chain of linked points. Practise writing these in your own words.

Why does a p.d. appear when a magnet is pushed into a coil? The magnet’s field through the coil is changing. The wires of the coil are cutting through field lines. A potential difference is therefore induced across the ends of the coil.

Why is it harder to push the magnet in when the coil is in a complete circuit? A current flows. The current makes a magnetic field that opposes the approaching pole, so there is a repelling force. Work must be done against it, and that work supplies the electrical energy.

Why does an alternator give a.c.? The coil turns in a magnetic field, so its sides cut field lines. In each half turn a given side moves in the opposite direction through the field. The induced p.d. reverses every half turn. Slip rings keep each brush on the same end of the coil, so the external current reverses too.

Why does a dynamo give d.c.? The p.d. in the coil still reverses every half turn. The split-ring commutator swaps the connections to the brushes at that moment, so the external current always flows in the same direction.

Effect of changing the generator

Change	Alternator output
Rotate faster	Higher frequency and higher peak p.d.
Rotate slower	Lower frequency and lower peak p.d.
Stronger magnet	Higher peak p.d., same frequency
More turns on the coil	Higher peak p.d., same frequency
Reverse the direction of rotation	Output starts by going the opposite way

Worked reminder 4

An alternator gives a peak p.d. of 6.0 V at 1500 revolutions per minute. At 1000 revolutions per minute, in the simple model where peak p.d. is proportional to speed, the peak p.d. is 6.0 × (1000 ÷ 1500) = 4.0 V, and the frequency is 1000 ÷ 60 = 16.7 Hz (3 significant figures).

Calculation reminders

Quantity	Relationship
Revolutions per second	revolutions per minute ÷ 60
Frequency (Hz)	1 ÷ period (s)
Period (s)	1 ÷ frequency (Hz)
Dynamo pulses per second	2 × revolutions per second
Time shown on a trace	divisions × time-base setting

Worked reminder 1

A trace shows 4 cycles in 80 ms. Period = 80 ÷ 4 = 20 ms = 0.020 s. Frequency = 1 ÷ 0.020 = 50 Hz.

Worked reminder 2

An alternator turns at 750 revolutions per minute. Revolutions per second = 750 ÷ 60 = 12.5, so the frequency is 12.5 Hz and the period is 1 ÷ 12.5 = 0.080 s.

Worked reminder 3

A dynamo turns at 8 revolutions per second. Pulses per second = 2 × 8 = 16. Time between pulses = 1 ÷ 16 = 0.0625 s.

Must-know distinctions

• Induced p.d. versus induced current: the first always, the second only in a complete circuit.
• Slip rings versus split-ring commutator: complete rings keep the a.c.; a split ring turns it into d.c.
• Alternator versus motor: opposite energy transfers.
• Change in field versus size of field: only the change matters.

Quick self-test

1. State what must happen to the magnetic field around a conductor for a p.d. to be induced.
2. A magnet is held still inside a coil. What is the induced p.d.?
3. The north pole of a magnet is pulled out of a coil. Which pole forms at the near end of the coil?
4. Why must the induced field oppose the change, in terms of energy?
5. Name the parts that connect an alternator coil to the outside circuit.
6. Why does the output of an alternator reverse direction?
7. What does the split-ring commutator in a dynamo do every half turn?
8. A dynamo turns at 15 revolutions per second. How many pulses per second does it give?
9. An alternator turns at 600 revolutions per minute. Give its frequency.
10. An a.c. trace has a period of 0.025 s. Give its frequency.

Answers

1. It must change, for example through movement of the magnet or conductor.
2. Zero, because the field is not changing.
3. A south pole, which attracts the departing north pole.
4. If it helped the change, energy would be created from nothing. Opposing it means work must be done to keep going.
5. Slip rings and brushes.
6. Each side of the coil moves through the field in the opposite direction every half turn, so the induced p.d. reverses.
7. It swaps the connections between the coil and the brushes, so the external current stays in one direction.
8. 2 × 15 = 30 pulses per second.
9. 600 ÷ 60 = 10 rev/s, so 10 Hz.
10. 1 ÷ 0.025 = 40 Hz.

Where marks are usually lost

• Writing that a stationary magnet in a coil induces a p.d. because it is strong.
• Saying “current is induced” for an open circuit instead of “p.d. is induced”.
• Giving the induced pole in the wrong direction, or not saying that it opposes the change.
• Forgetting to say that the commutator swaps connections every half turn.
• Saying a dynamo produces a “smooth” d.c. instead of pulsing d.c.
• Dividing instead of multiplying (or the reverse) when converting between period and frequency.
• Forgetting to divide by 60 when the speed is given in revolutions per minute.
• Giving no unit, or the wrong unit, for frequency and period.

Official syllabus

These notes are based on the OCR GCSE (9-1) Gateway Science Physics A (J249) specification, Version 5.0 (July 2026), for first assessment in 2018, published by OCR, topic P4.2 Uses of magnetism, statements P4.2e and P4.2f.