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

This guide is aligned to the OCR GCSE (9-1) Gateway Science Physics A (J249) specification, Version 5.0 (July 2026), topic P4.2 Uses of magnetism, statements P4.2a to P4.2d. All four statements are assessed in both Foundation and Higher Tier papers. Induction, generators and transformers (the later statements in P4.2) have their own pages. Start with the OCR GCSE Physics hub, tick off what you know on the checklist, and test yourself with the free diagnostics. You need the fields from P4.1 first, so read Magnets, magnetic fields and electromagnets if poles and field lines are shaky.

Companion pages: revision notes and practice questions.

What this unit covers

Statement	What you must be able to do
P4.2a	Describe how a magnet and a current-carrying conductor exert a force on one another
P4.2b	Show that Fleming’s left-hand rule represents the relative orientations of the force, the current and the magnetic field
P4.2c	Apply the equation linking force on a conductor to magnetic flux density, current and length, to calculate the forces involved
P4.2d	Explain how the force from a magnet on a current-carrying conductor causes rotation in electric motors

For P4.2d the specification says you need to understand how electric motors work, but you are not expected to know the structure of a motor. The suggested practical work is the jumping wire demonstration and building simple motors.

P4.2a: the force between a magnet and a current-carrying conductor

A wire carrying a current has its own magnetic field (see P4.1). Put that wire in the field of a magnet and the two fields interact. The result is a force on the wire. This is called the motor effect.

Key points:

• The force exists only while there is a current. Switch the current off and the force vanishes.
• The magnet exerts a force on the wire, and the wire exerts an equal and opposite force on the magnet. These are a Newton’s third law pair. They act on different objects.
• The force is largest when the wire is at right angles to the field lines. The equation in P4.2c is written for this case.
• A wire lying along the field lines (parallel to them) experiences no force from the field.

The jumping wire demonstration

A typical set-up has a short length of wire resting across two supports (for example two metal rails) between the poles of a horseshoe magnet, with a low-voltage supply and a switch.

1. Close the switch. A current flows along the wire.
2. The wire jumps out of the gap between the poles. It moves at right angles to both the wire and the field.
3. Swap the supply terminals. The current reverses and the wire jumps the opposite way.
4. Turn the magnet over. The field reverses and the wire again jumps the opposite way.
5. Use a stronger magnet or increase the current. The wire jumps more strongly.

Describe this in an exam as a cause and effect chain: current in the wire gives a field around the wire, this interacts with the field of the magnet, and a force acts on the wire.

P4.2b: Fleming’s left-hand rule

The rule links three directions that are all at right angles to one another. Use your left hand and spread the thumb and the first two fingers so that each is perpendicular to the other two.

Part of the hand	Represents	Direction convention
First finger	Field	From the north pole to the south pole of the magnet
Second finger	Current	From the positive terminal to the negative terminal (conventional current)
Thumb	Force (motion)	The direction the wire is pushed

A memory aid: First finger = Field, seCond finger = Current, thuMb = Motion.

The word “show that” in P4.2b means you should be able to use the experiment to justify the rule. Each reversal in the jumping wire demonstration agrees with the rule:

• Reverse the current only: the second finger flips, so the thumb flips.
• Reverse the field only: the first finger flips, so the thumb flips.
• Reverse both: the thumb points the same way as before. The force does not change.

Worked example: finding a direction

A horizontal wire lies on a table, between a north pole on the west side and a south pole on the east side. The current flows from south to north along the wire. In which direction is the force?

1. The field runs from north pole to south pole, so it points east. First finger east.
2. Current flows south to north, so it points north. Second finger north.
3. Turn your left hand so the first finger points east and the second finger points north. The thumb then points down, towards the table.

The force is downwards, into the table. Sketch the three arrows and check them against your own left hand. Practise this until you can do it without hesitation.

P4.2c: the size of the force

For a conductor at right angles to the magnetic field:

force = magnetic flux density × current × length

F = B I l

Symbol	Quantity	Unit
F	Force on the conductor	newton, N
B	Magnetic flux density (field strength)	tesla, T
I	Current	ampere, A
l	Length of conductor in the field	metre, m

Three things to notice. The force is directly proportional to each of B, I and l. The length is only the part of the wire that is inside the field. And the tesla is the unit that makes the equation work: 1 T is 1 N per ampere per metre.

Rearranged forms:

• I = F / (B l)
• B = F / (I l)
• l = F / (B I)

Worked example 1: find the force

A straight wire of length 0.080 m lies at right angles to a uniform field of flux density 0.25 T. The current is 3.0 A. Calculate the force.

F = B I l
F = 0.25 × 3.0 × 0.080
F = 0.060 N

F = 0.060 N

Worked example 2: find the current

A wire of length 0.12 m in a field of 0.15 T experiences a force of 0.45 N. The wire is at right angles to the field. Calculate the current.

I = F / (B l)
I = 0.45 / (0.15 × 0.12)
I = 0.45 / 0.018
I = 25 A

I = 25 A

Worked example 3: find the flux density

A wire carries 1.5 A and has 0.060 m in a uniform field. The force on it is 0.036 N. Find the flux density.

B = F / (I l)
B = 0.036 / (1.5 × 0.060)
B = 0.036 / 0.090
B = 0.40 T

B = 0.40 T

Worked example 4: convert units first

A conductor 4.0 cm long carries 2.5 A in a field of 0.50 T. Find the force.

l = 4.0 cm = 0.040 m
F = 0.50 × 2.5 × 0.040
F = 0.050 N

F = 0.050 N

Worked example 5: find the length

A force of 0.072 N acts on a wire carrying 1.8 A in a field of 0.40 T. How much of the wire is in the field?

l = F / (B I)
l = 0.072 / (0.40 × 1.8)
l = 0.072 / 0.72
l = 0.10 m

l = 0.10 m (10 cm)

Method in steps

1. Write the equation with symbols.
2. Convert all quantities to N, T, A and m. Centimetres to metres is divide by 100. Milliamps to amps is divide by 1000.
3. Rearrange before substituting.
4. Substitute, calculate, and give the unit.
5. Check that the answer is sensible: doubling any one of B, I or l doubles F.

P4.2d: how the motor effect makes a motor turn

An electric motor turns because the motor effect pushes in opposite directions on opposite sides of a coil.

Picture a rectangular coil of wire between the poles of a magnet, with a d.c. supply connected to it.

• Current flows along one side of the coil in one direction, and back along the opposite side in the opposite direction.
• The field of the magnet points the same way across both sides.
• By Fleming’s left-hand rule, the forces on the two sides are in opposite directions: one side is pushed up, the other is pushed down.
• These two forces act on opposite sides of the axle. Together they turn the coil.

The problem of a half turn

When the coil has turned through half a revolution, the side that was on the left is now on the right. The current in that side still flows the same way relative to the supply, but its position has swapped. The force on it would now push the coil back the way it came, and the coil would stop and rock.

To keep turning in one direction, the current in the coil must reverse every half turn. Then the side now going up always has a current that pushes it up. In a d.c. motor this reversal is done by a component that switches the connections to the coil as it turns, normally called a commutator. The specification says you do not need to know the structure of the motor, so what matters is the idea: the current in each side of the coil is reversed at the right moment so each force keeps pushing its side round in one sense of rotation.

Explaining what changes a motor’s behaviour

Change	Effect	Reason
Increase the current	Larger turning force, so faster rotation	F = B I l, F is proportional to I
Stronger magnet	Larger turning force	F is proportional to B
Reverse the supply connections	Motor turns the opposite way	Current reverses, so force reverses
Reverse the magnet	Motor turns the opposite way	Field reverses, so force reverses
Reverse both	Same direction as before	Two reversals cancel

Worked example 6: forces on a coil

A coil has two sides, each with 0.050 m of wire at right angles to a uniform field of 0.30 T. The current is 2.0 A.

1. Force on one side: F = B I l = 0.30 × 2.0 × 0.050 = 0.030 N.
2. The force on the other side is also 0.030 N, in the opposite direction.
3. If the current is doubled to 4.0 A, each force becomes 0.30 × 4.0 × 0.050 = 0.060 N. The forces double.

Each side feels 0.030 N, and doubling the current doubles it to 0.060 N.

Common errors

• Using the right hand. The motor effect needs the left hand.
• Taking the current direction from negative to positive. Use conventional current, positive to negative.
• Mixing up which finger is which. Use the memory aid every time.
• Putting centimetres into F = B I l without converting.
• Writing “the wire is attracted to the magnet”. The force is at right angles to both the field and the current.
• Saying the force reverses when both the current and the field reverse. It does not.
• Explaining a motor without saying the current reverses every half turn.
• Describing the force on a wire as one-sided. The magnet also feels an equal and opposite force.

Next steps

Work through the practice questions and use the revision notes to consolidate. The free diagnostics show which topics to prioritise.

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.