Fleming's left-hand rule (HT only) - AQA GCSE Physics Exam Questions

Practise 30 exam-style questions on Fleming's left-hand rule (HT only) for AQA GCSE Physics. Write your answer as you would in the exam and get it marked in seconds against the mark scheme, point by point, with an AI tutor on hand if you get stuck.

30 questions
1–6 marks each
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20 questions, marked as you go

Sample questions with model answers

Here are 4 of the questions in this topic, each with a model answer, an explanation of why it earns the marks and the mark scheme an examiner would use.

Question 1

2 marks

What is meant by the motor effect?

Model answer

When a conductor carrying a current is placed in a magnetic field, the magnet producing the field and the conductor exert a force on each other. This is the motor effect.

Explanation

The current produces its own magnetic field, which interacts with the field of the magnet, so there is a force. The force acts on both the wire and the magnet, in opposite directions.

Mark scheme

1 mark
a conductor (wire) carrying a current is placed in a magnetic field
1 mark
the conductor and the magnet exert a force on each other
allow the conductor experiences a force

Question 2

2 marks

A horizontal overhead power cable runs from east to west and carries a current of 500 A. The Earth's magnetic field at the cable has a horizontal part of flux density 2.0 × 10⁻⁵ T, at right angles to the cable.

Use the equation:
force = magnetic flux density × current × length

Calculate the force on a 100 m length of the cable due to this part of the Earth's field.

Model answer

F = B I l
F = 2.0 × 10⁻⁵ × 500 × 100
F = 1.0 N

Explanation

The Earth's magnetic field is very weak, but a large current over a long length still produces a measurable force. Take care entering standard form into your calculator.

Mark scheme

1 mark
F = 2.0 × 10⁻⁵ × 500 × 100
an answer of 1.0 (N) scores 2 marks
1 mark
F = 1.0 N

Question 3

3 marks

A light aluminium rod rests across two horizontal rails in a magnetic field. When there is a current in the rod, it rolls along the rails.

Suggest two changes that would make the rod accelerate faster. Explain your answer.

Model answer

Increase the current in the rod, or use a stronger magnet with a greater magnetic flux density. Both increase the force on the rod (F = B I l), and a larger resultant force on the same mass gives a larger acceleration (F = m a).

Explanation

This links the motor effect to Newton's second law. Anything that increases B, I or the length of rod in the field increases the force, and so the acceleration.

Mark scheme

1 mark
increase the current (in the rod)
max 3
1 mark
use a stronger magnet / greater magnetic flux density
allow use a wider magnet so a longer length of rod is in the field
1 mark
these increase the force on the rod, and a bigger resultant force gives a bigger acceleration (F = m a)
1 mark
use a rod with a smaller mass (for the same force)

Question 4

4 marks

A horizontal copper wire of mass 3.0 g has 10 cm of its length in a horizontal magnetic field of flux density 0.15 T. The wire is at right angles to the field. A current in the wire produces an upward force on it.

The gravitational field strength is 9.8 N/kg.

Use the equation:
force = magnetic flux density × current × length

Calculate the current needed for the magnetic force to just support the weight of the wire.

Model answer

m = 3.0 g = 0.0030 kg
W = m g = 0.0030 × 9.8 = 0.0294 N
The magnetic force must equal the weight: F = B I l
0.0294 = 0.15 × I × 0.10
I = 0.0294 ÷ 0.015
I = 1.96 A ≈ 2.0 A

Explanation

This combines two equations: W = m g to find the force needed, then F = B I l to find the current. The forces are balanced when the upward magnetic force equals the weight. Convert grams to kilograms and centimetres to metres.

Mark scheme

1 mark
weight = 0.0030 × 9.8 = 0.0294 N
an answer of 2.0 (A) scores 4 marks; allow 1.96 A
1 mark
0.0294 = 0.15 × I × 0.10
allow ecf from an incorrect weight; length must be 0.10 m
1 mark
I = 0.0294 ÷ (0.15 × 0.10)
allow I = 0.0294 ÷ 0.015
1 mark
I = 2.0 A
allow 1.96 A; allow ecf

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