OCR GCSE Physics A: The motor effect and electric motors (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 are 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 of them are suitable for both Foundation and Higher Tier. Work from the study guide and the revision notes. The OCR GCSE Physics hub and the checklist show where this unit sits, and the free diagnostics help you decide what to do next.

Questions

1. State what the first finger, the second finger and the thumb of the left hand represent in Fleming’s left-hand rule. [3]

2. A wire of length 0.15 m lies at right angles to a uniform magnetic field of flux density 0.35 T. The current in the wire is 4.0 A. Calculate the force on the wire. [3]

3. A straight conductor with 0.30 m inside a uniform 0.20 T field, perpendicular to it, feels a force of 0.90 N. Calculate the current in the conductor. [3]

4. A short wire rests across two metal rails between the poles of a horseshoe magnet. When a switch is closed, the wire jumps out of the gap. Explain why the wire moves. [4]

5. A wire carries a current that flows out of the page towards you. It is in a uniform field that points from left to right across the page. State the direction of the force on the wire and explain how you worked it out. [3]

6. Explain how a d.c. motor keeps a coil turning in the same direction. [5]

7. A wire 2.5 cm long carries a current of 800 mA at right angles to a field of 0.60 T. Calculate the force on the wire. [3]

8. A strong magnet sits on a top-pan balance. A rigid wire is held in a fixed clamp so that 0.050 m of it lies at right angles to the field between the poles. The flux density is 0.12 T. A current of 5.0 A is switched on.

(a) Calculate the force on the wire. [3] (b) Explain why the reading on the balance changes. [1] (c) The force on the magnet is equal to your answer to part (a). Use g = 10 N/kg to calculate the change in the balance reading, in grams. [2]

9. A student measures the force on a wire with 0.10 m in a uniform field for four values of current.

Current (A)	1.0	2.0	3.0	4.0
Force (N)	0.040	0.081	0.119	0.160

(a) State what the data show about how force depends on current, and justify your answer with calculations. [2] (b) Use the data for a current of 3.0 A to calculate the magnetic flux density. [3] (c) Describe the shape of the graph of force against current. [1]

10. A wire in a magnetic field is pushed upwards by the motor effect. State the direction of the force after each change below, taking each change from the starting situation.

(a) The current is reversed. [1] (b) The magnet is turned over so its poles swap. [1] (c) The current is reversed and the magnet is turned over. [1] (d) The wire is turned to lie along the field lines. [1]

11. A rectangular coil in a uniform field of 0.25 T has two opposite sides, each with 0.060 m of wire at right angles to the field. The current in the coil is 1.8 A.

(a) Calculate the force on one of these sides. [3] (b) State the direction of the force on the other side compared with the first, and explain how the two forces make the coil turn. [2] (c) The current is increased to 3.6 A. State the new force on one side. [1] (d) Explain why the current in the coil must be reversed every half turn for the coil to keep turning. [2]

Answers

1. First finger: the magnetic field (north to south) [1]. Second finger: the current (positive to negative) [1]. Thumb: the force or motion [1]. [3] Examiner insight: Listing “field, current, motion” in the wrong order can lose credit, so match each digit to its quantity explicitly.

2. F = B I l [1]. F = 0.35 × 4.0 × 0.15 [1]. F = 0.21 N [1]. [3] Examiner insight: Show the equation and the substitution. A clear substitution lets a reader follow your method and spot a slip.

3. I = F / (B l) [1]. I = 0.90 / (0.20 × 0.30) = 0.90 / 0.060 [1]. I = 15 A [1]. [3] Examiner insight: Rearrange first, then substitute. Substituting into F = B I l and solving afterwards is fine, but keep each line clear.

4. There is a current in the wire, and so the wire has its own magnetic field [1]. This interacts with the field of the magnet [1]. The result is a force on the wire [1]. The force is at right angles to both the field and the current, so the wire is pushed out of the gap [1]. [4] Examiner insight: A four-mark “explain” needs a chain of linked reasons. “It is pushed by the magnet” gives only one point.

5. Field points right: first finger right [1]. Current is towards you: second finger towards you [1]. The thumb points up the page, so the force is upwards [1]. [3] Examiner insight: State both input directions before the result. A correct direction with no reasoning may not gain all the marks on an “explain how” question.

6. The coil carries a current in a magnetic field, so a force acts on each side [1]. The currents in the two sides are in opposite directions, so the forces are in opposite directions [1]. These forces turn the coil [1]. After half a turn, the sides have swapped places, so the forces would turn the coil back [1]. Reversing the current every half turn keeps the forces turning the coil the same way [1]. [5] Examiner insight: The step linking half a turn to the need for a current reversal is the one most often missing. Name the component only if you are sure, because the specification does not require the structure.

7. Convert: l = 0.025 m and I = 0.800 A [1]. F = 0.60 × 0.800 × 0.025 [1]. F = 0.012 N [1]. [3] Examiner insight: Unconverted values give an answer out by a factor of 100,000. Write the converted values on their own line.

8. (a) F = B I l [1]. F = 0.12 × 5.0 × 0.050 [1]. F = 0.030 N [1]. (b) The wire exerts an equal and opposite force on the magnet (Newton’s third law), so the force on the magnet changes the balance reading [1]. (c) Mass change = F / g = 0.030 / 10 = 0.0030 kg [1]. Change = 3.0 g [1]. [6] Examiner insight: Link the balance reading to the third law pair explicitly. Convert kilograms to grams at the end and show the line.

9. (a) Force divided by current is almost constant: 0.040, 0.0405, 0.0397, 0.040 N per A [1]. So force is directly proportional to current [1]. (b) B = F / (I l) [1]. B = 0.119 / (3.0 × 0.10) [1]. B = 0.40 T (to 2 significant figures) [1]. (c) A straight line through the origin [1]. [6] Examiner insight: “Proportional” needs evidence, such as a constant ratio or a line through the origin. Saying “it goes up” is not enough.

10. (a) The force is downwards [1]. (b) The force is downwards [1]. (c) The force is upwards (two reversals cancel) [1]. (d) There is no force [1]. [4] Examiner insight: Parts (a) and (b) each reverse one quantity. Do not carry a change from an earlier part into the next one.

11. (a) F = B I l [1]. F = 0.25 × 1.8 × 0.060 [1]. F = 0.027 N [1]. (b) The force on the other side is the same size but in the opposite direction [1]. Opposite forces on opposite sides of the axle give a turning effect on the coil [1]. (c) 0.054 N (force is proportional to current) [1]. (d) After half a turn the sides have swapped positions, so without a reversal the forces would push the coil back towards where it came from [1]. Reversing the current keeps each force turning the coil in the same direction [1]. [8] Examiner insight: Say that the forces act in opposite directions on opposite sides. “A force turns the coil” alone does not explain why it turns.

Where marks are usually lost

• Using the right hand, or mixing up first finger and second finger.
• Using electron flow instead of conventional current.
• Forgetting to convert cm to m or mA to A.
• Missing the unit on a force, or writing T for a force.
• Giving a direction without the reasoning that earned it.
• Not saying the force is at right angles to both the field and the current.
• Reversing the force when both the current and the field are reversed.
• Describing a motor without saying why the current must be reversed every half turn.
• Writing that doubling the current doubles the flux density. The field is the magnet’s, so B is unchanged.

Next steps

• Revise with the revision notes and re-read the study guide.
• Tick off the topic on the checklist and browse the OCR GCSE Physics hub.
• Take one of the free 10-minute diagnostics.
• Book a free trial class.

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.