OCR GCSE Physics A: The motor effect and electric motors (J249) – Revision Notes

Condensed notes for the final weeks, aligned to the OCR GCSE (9-1) Gateway Science Physics A (J249) specification, Version 5.0 (July 2026), topic P4.2, statements P4.2a to P4.2d. All four are assessed in both Foundation and Higher Tier papers. For full explanations, use the study guide, then try the practice questions. See the OCR GCSE Physics hub, the checklist and the free diagnostics.

Key definitions

Term	Meaning
Motor effect	The force on a wire that carries a current while it sits in a magnet’s field
Magnetic flux density, B	Measure of the strength of a magnetic field; unit tesla (T)
Conventional current	Taken to flow from the positive terminal to the negative terminal
Uniform field	Parallel, equally spaced field lines, so B is the same everywhere in the gap
Fleming’s left-hand rule	Hand model for the directions of field, current and force
d.c. motor	Uses the motor effect on a coil to turn a shaft, with the coil current reversed each half turn

The motor effect (P4.2a)

• The current gives the wire its own magnetic field. This interacts with the field of the magnet. The result is a force on the wire.
• No current, no force.
• The force on the wire and the force on the magnet are equal and opposite (Newton’s third law). Each acts on a different object.
• The force acts at right angles to both the wire and the field.
• A wire parallel to the field has no force on it.

Jumping wire demonstration, key results:

Change	Result
Close the switch	Wire jumps out of the gap
Reverse current	Jumps the other way
Reverse field (flip magnet)	Jumps the other way
Reverse both	Same direction as the original
Stronger magnet or larger current	Bigger movement

Fleming’s left-hand rule (P4.2b)

Digit	Stands for	Direction
First finger	Field	North pole to south pole
Second finger	Current	Positive to negative
Thumb	Force (motion)	Way the conductor is pushed

Memory aid: Field, Current, Motion. Left hand only. All three digits at right angles.

Method in steps:

1. Mark the direction of the field with an arrow, north to south.
2. Mark the direction of the current, positive to negative.
3. Point your first finger along the field and your second finger along the current.
4. Read off the thumb. That is the force.

Showing that the rule fits the experiment: reverse one input and the thumb reverses. Reverse both and the thumb stays put.

The equation (P4.2c)

F = B I l

Symbol	Quantity	Unit
F	Force on the conductor	N
B	Flux density of the field	T
I	Current in the conductor	A
l	Length of the conductor inside the field	m

Rearrangements:

To find	Use
Current	I = F / (B l)
Flux density	B = F / (I l)
Length	l = F / (B I)

• Works for a conductor at right angles to the field.
• Proportional to each of B, I and l.
• Convert first: cm to m is divide by 100. mA to A is divide by 1000.

Worked reminder

A wire has 0.15 m in a uniform field with B = 0.40 T and carries 5.0 A at right angles.

F = B I l = 0.40 × 5.0 × 0.15 = 0.30 N

Reversing the idea: the same 0.30 N in a 0.50 T field over the same 0.15 m needs I = 0.30 / (0.50 × 0.15) = 4.0 A.

Electric motors (P4.2d)

• A coil of wire sits between two magnet poles, with the current in opposite directions along opposite sides.
• Fleming’s rule gives opposite forces on the two sides: one up, one down. They turn the coil.
• After half a turn, the sides have swapped positions. Without a change, the forces would pull the coil back.
• So the current in the coil is reversed every half turn. Each side then keeps being pushed the same way round.
• You need the idea, not the structure. A part called a commutator does the reversing in a typical d.c. motor, but the specification states that knowledge of motor structure is not expected.

Changing the motor:

Change	Effect
Larger current	Larger forces, faster turning
Stronger magnets	Larger forces
Reverse supply	Turns the other way
Reverse magnet	Turns the other way
Reverse both	No change in direction

Interpreting data

If you measure the force on a fixed length of wire in a fixed field at different currents, the points on a force-current graph lie on a straight line through the origin. That shows F is directly proportional to I. The gradient equals B × l, so dividing the gradient by l gives B.

Worked reminder: a line has gradient 0.060 N/A for a wire with 0.20 m in the field.

gradient = B × l
B = 0.060 / 0.20 = 0.30 T

Other checks:

• Fixed current, varying length in the field: F against l is also a straight line through the origin, with gradient B × I.
• A line that does not pass through the origin suggests a systematic error, such as a zero error on the balance or newton meter.
• Control variables: keep the other two quantities of B, I and l constant while you vary one.

Explaining a motor in four marks

Plan your answer in this order:

1. There is a current in the coil, so the coil is a current-carrying conductor in a magnetic field.
2. A force acts on each side of the coil, in opposite directions (Fleming’s left-hand rule).
3. The two forces turn the coil about its axis.
4. The current is reversed every half turn, so each force keeps pushing its side round in one sense of rotation.

Must-know distinctions

• Field direction (north to south) versus current direction (positive to negative): both are conventions, and both go into the rule.
• Left hand (motor effect, forces on conductors) is for the force. Do not use it for field direction around a wire, which is covered in P4.1.
• The force on the wire versus the force on the magnet: equal, opposite, different objects.
• Length of the wire versus length in the field: use only the part in the field.
• Flux density B versus force F: B is a property of the field, F is what the wire feels.

Quick self-test

1. A 0.10 m wire carrying 6.0 A lies perpendicular to a 0.20 T field. Calculate the force.
2. A force of 0.18 N acts on a wire of length 0.15 m in a 0.30 T field, at right angles. Calculate the current.
3. A wire carries 2.0 A in a field of 0.45 T. The force is 0.054 N. Calculate the length of wire in the field.
4. A wire of length 6.0 cm carries 0.50 A in a field of 0.80 T. Calculate the force.
5. The force on a 0.10 m wire carrying 1.2 A is 0.024 N. Calculate the flux density.
6. A force of 0.040 N acts on a wire. The current doubles. State the new force.
7. A 2.5 cm wire carries 3.0 A in a 0.40 T field. Calculate the force.
8. Which digit of the left hand represents the current?
9. Name the unit of magnetic flux density.
10. The current in a jumping wire and the magnet are both reversed. What happens to the direction of the force?
11. Why must the current in a motor coil reverse every half turn?
12. What happens to the force on a wire if it is lying parallel to the field?

Answers

1. F = 0.20 × 6.0 × 0.10 = 0.12 N.
2. I = 0.18 / (0.30 × 0.15) = 4.0 A.
3. l = 0.054 / (0.45 × 2.0) = 0.060 m.
4. l = 0.060 m, F = 0.80 × 0.50 × 0.060 = 0.024 N.
5. B = 0.024 / (1.2 × 0.10) = 0.20 T.
6. Force is proportional to current, so 0.080 N.
7. l = 0.025 m, F = 0.40 × 3.0 × 0.025 = 0.030 N.
8. Second finger.
9. Tesla (T).
10. It stays the same. Two reversals cancel.
11. Otherwise the force on each side would reverse its turning effect after half a turn and the coil would rock back. Reversing the current keeps the forces turning it in one direction.
12. There is no force.

Where marks are usually lost

• Using the right hand instead of the left.
• Using electron flow instead of conventional current in the rule.
• Not converting centimetres to metres or milliamps to amps.
• Leaving off units, or writing T where N is needed.
• Forgetting the current must reverse in a motor coil, and so not explaining continuous rotation.
• Saying the wire “is attracted” or “repelled” by the magnet without giving the direction at right angles to field and current.
• Writing that reversing both the field and the current reverses the force.
• Rearranging incorrectly: write the rearranged equation first, then substitute.
• Using the whole length of the wire, not the length in the field.

Official syllabus

OCR GCSE (9-1) Gateway Science Physics A (J249) specification, Version 5.0 (July 2026), first assessment 2018, published by OCR, topic P4.2 Uses of magnetism, statements P4.2a to P4.2d.