Fleming's left-hand rule, F = BIl, motors and loudspeakers (Higher) – GCSE Physics exam questions with mark scheme

1Grade 6–7Higher3 marks

The diagram shows a wire placed between the poles of a magnet. The current in the wire is into the page.

(a) The magnetic flux density is 0.15 T. A 0.050 m length of the wire is in the field and the current is 2.4 A.

Calculate the force on the wire. Use the Physics Equations Sheet. [2]

(b) Use Fleming's left-hand rule to determine the direction of the force on the wire. [1]

(a) F = 0.15 × 2.4 × 0.050(1)
(a) 0.018 N(1)
(b) downwards (towards the bottom of the page)(1)
Answer: (a) 0.018 N (b) downwards
Examiner tip: In Fleming's left-hand rule the First finger is Field (N to S), seCond finger is Current, thuMb is Motion.

2Grade 6–7Higher6 marks

(a) A wire 25 cm long carries a current of 3.0 A at right angles to a magnetic field. The force on the wire is 0.36 N.

Calculate the magnetic flux density. Give the unit. [3]

(b) A simple electric motor has a rectangular coil in a magnetic field. Explain why the coil rotates when there is a current in it. [3]

(a) 0.36 = B × 3.0 × 0.25 (allow B = 0.36 ÷ (3.0 × 0.25); length must be converted to 0.25 m)(1)
(a) B = 0.48(1)
(a) tesla / T (allow N/(A m))(1)
(b) the current-carrying sides of the coil are in the magnetic field (at right angles to it) so each experiences a force (allow: motor effect / Fleming's left-hand rule gives a force on each side)(1)
(b) the current flows in opposite directions in the two sides, so the forces are in opposite directions (one up, one down)(1)
(b) this produces a turning effect (moment) on the coil (allow: the coil rotates about its axis)(1)
Answer: (a) 0.48 T (b) Opposite sides carry current in opposite directions, so the forces act in opposite directions and produce a turning effect on the coil.
Examiner tip: Lengths in F = BIL must be in metres — 25 cm is 0.25 m.

3Grade 4–5Higher4 marks

A wire carrying a current is placed between the poles of a magnet. The wire moves.

(a) What is the name of this effect? [1]

(b) Fleming's left-hand rule can be used to find the direction in which the wire moves.

Complete the table to show what each finger of the left hand represents. [2]

Part of left hand	Represents the direction of
thumb
first finger
second finger

What is the size of the force on the wire now? [1]

(a) (the) motor effect do not accept generator effect / electromagnetic induction / magnetism(1)
(b) all three correct: thumb – force / motion (of the wire); first finger – (magnetic) field; second finger – current allow thrust for force; allow N to S for field; allow + to − (conventional) current(1)
(b) any one correct (1 mark max if only one or two correct)(1)
(c) zero allow there is no force / 0 N; ignore 'it does not move' on its own(1)
Answer: (a) motor effect (b) thumb – force/motion; first finger – field; second finger – current (c) zero
Examiner tip: Many students swap the first and second fingers – remember First finger = Field, seCond finger = Current.

4Grade 4–5Higher4 marks

The diagram shows a wire between the poles of a magnet. The magnetic field is shown by the arrows. The current in the wire is out of the page (towards you).

(a) Use Fleming's left-hand rule to determine the direction of the force on the wire. [1]

(b) Give two different changes that would make the force on the wire act in the opposite direction. [2]

State the direction of the force on the wire now. [1]

(a) to the right (of the page) allow towards the right-hand side; do not accept east unless clearly the right of the diagram(1)
(b) reverse (the direction of) the current allow swap the connections to the power supply / current into the page(1)
(b) reverse the (direction of the) magnetic field allow swap / turn round the poles (so S is at the top); do not accept 'use a stronger magnet'(1)
(c) to the right allow the same direction as in (a) / unchanged; ecf from (a)(1)
Answer: (a) to the right (b) reverse the current; reverse the magnetic field (swap the poles) (c) to the right (unchanged)
Examiner tip: Point your first finger along the field arrows (N to S) – here that is DOWN the page, not across it.

5Grade 8–9Higher7 marks

A student uses the apparatus shown to investigate the force on a current-carrying wire.

A U-shaped magnet sits on a top-pan balance. A straight wire is clamped so that it passes between the poles at right angles to the magnetic field. The wire does not touch the magnet or the balance. The length of wire in the field is 7.0 cm.

The table shows the increase in the balance reading for different currents in the wire.

Current in A	Increase in balance reading in g
0.50	0.25
1.00	0.50
1.50	0.84
2.00	1.00
2.50	1.25

(a) Explain why the balance reading changes when there is a current in the wire. [2]

(b) Identify the anomalous result in the table. [1]

gravitational field strength = 9.8 N/kg [2]

(d) Calculate the magnetic flux density between the poles of the magnet. Use the Physics Equations Sheet. [2]

(a) there is a force on the wire because it carries a current in a magnetic field allow the motor effect acts on the wire(1)
(a) the wire exerts an equal and opposite force on the magnet (so the magnet pushes harder on the balance) allow the magnet and the wire exert forces on each other; allow reference to Newton's third law(1)
(b) 1.50 A / 0.84 g allow the third result(1)
(c) $F = 0.00100 \times 9.8$ mass must be converted to kg(1)
(c) $F = 0.0098$ (N) allow $9.8 \times 10^{-3}$ N; an answer of 9.8 (mass not converted) scores 1 mark(1)
(d) $0.0098 = B \times 2.00 \times 0.070$ allow ecf from (c); length must be in m(1)
(d) $B = 0.070$ (T) allow 0.07 T; allow ecf from (c), e.g. 9.8 ÷ (2 × 0.070) = 70 scores 2 marks(1)
Answer: (a) The wire has a force on it (motor effect); it exerts an equal and opposite force on the magnet. (b) 1.50 A (0.84 g) (c) 0.0098 N (d) 0.070 T
Examiner tip: The balance measures a mass change in grams – convert to kg and multiply by 9.8 before using F = BIl.

6Grade 6–7HigherTriple only6 marks

The diagram shows a cross-section through a moving-coil loudspeaker.

The coil is attached to the paper cone and sits in the magnetic field of the permanent magnet. The coil is connected to an amplifier that supplies an alternating current.

Explain how the loudspeaker produces a sound wave, and how the sound changes if the amplifier increases the size or the frequency of the current. [6]

the coil carries a current in a magnetic field, so there is a force on the coil (motor effect)(1)
the force is at right angles to the field / along the axis of the cone (Fleming's left-hand rule)(1)
the current is alternating, so it keeps reversing direction(1)
so the direction of the force keeps reversing(1)
the coil and cone move backwards and forwards / vibrate(1)
the cone pushes and pulls on the air, producing compressions and rarefactions(1)
this makes pressure variations in the air, which travel as a (longitudinal) sound wave(1)
the current variations are converted into pressure variations (in the air)(1)
the cone vibrates at the same frequency as the current(1)
larger current → larger force → larger amplitude of vibration → louder sound(1)
higher frequency current → cone vibrates more times per second → higher pitch(1)

Level	Marks	What it looks like
Level 3	5–6	A clear, logically linked explanation of how the alternating current makes the cone vibrate (force on coil in field, force reversing with the current) and produce pressure variations / compressions and rarefactions in the air, with correct effects of BOTH a larger current (louder) and a higher frequency (higher pitch).
Level 2	3–4	Explains that there is a force on the coil and that the cone vibrates because the current alternates, with some reference to the air or sound, OR a partial chain of reasoning plus at least one correct effect of changing the current. Some steps may be missing or not linked.
Level 1	1–2	Isolated relevant statements, e.g. there is a force on the coil, the cone vibrates, or a bigger current makes it louder. Little or no linking.
No relevant content	0	Nothing worthy of credit.

Answer: A current in the coil in the magnetic field gives a force (motor effect). The a.c. reverses, so the force reverses and the coil and cone vibrate, pushing the air to make compressions and rarefactions (pressure variations) – a sound wave at the frequency of the current. Larger current → larger force → larger amplitude → louder; higher frequency → higher pitch.
Examiner tip: Top-band answers link every step: alternating current → force reverses → cone vibrates → pressure variations in the air – and say what BOTH changes do.

7Grade 8–9Higher6 marks

The diagram shows a view from above of a metal rod resting on two horizontal metal rails. The rails are 8.0 cm apart and are connected to a cell. A uniform magnetic field of flux density 0.30 T acts vertically downwards (into the page).

The total resistance of the circuit is 0.60 Ω and the potential difference of the cell is 1.5 V. The mass of the rod is 12 g.

(a) Use Fleming's left-hand rule to determine which way the rod moves when the circuit is complete. [1]

(b) Calculate the acceleration of the rod at the instant the circuit is completed. Ignore friction. [5]

(a) to the right / away from the cell(1)
(b) $I = \dfrac{1.5}{0.60} = 2.5$ (A)(1)
(b) $F = 0.30 \times 2.5 \times 0.080$ allow ecf for I; length must be in m (0.080)(1)
(b) $F = 0.060$ (N) allow ecf(1)
(b) $a = \dfrac{0.060}{0.012}$ allow ecf for F; mass must be in kg(1)
(b) $a = 5.0$ (m/s²) allow 5 m/s²; an answer of 0.005 (mass in g) scores 4 marks; an answer of 500 (length in cm) scores 4 marks(1)
Answer: (a) to the right (away from the cell) (b) 5.0 m/s²
Examiner tip: This needs three equations in a chain – find I from V = IR first, then F = BIl, then a = F/m, converting cm and g as you go.

8Grade 6–7HigherTriple only5 marks

A headphone contains a small coil attached to a thin plastic diaphragm. The coil sits in the magnetic field of a permanent magnet. The wire of the coil is at right angles to the field.

(a) The total length of wire in the field is 2.5 m and the magnetic flux density is 1.2 T.

Calculate the force on the coil when the current is 25 mA. Use the Physics Equations Sheet. [2]

(b) The current in the coil alternates with a frequency of 500 Hz. State the frequency of the sound produced. [1]

Explain why no continuous sound is produced. [2]

(a) $F = 1.2 \times 0.025 \times 2.5$ current must be converted to A(1)
(a) $F = 0.075$ (N) allow $7.5 \times 10^{-2}$ N; an answer of 75 (current not converted) scores 1 mark(1)
(b) 500 Hz unit required(1)
(c) the force on the coil is (constant and) always in the same direction allow the current / force does not reverse(1)
(c) (so) the diaphragm moves once and then stays still / does not vibrate, so there are no (continuous) pressure variations in the air allow a single click only; do not accept 'there is no force'(1)
Answer: (a) 0.075 N (b) 500 Hz (c) The direct current gives a force in one direction only, so the diaphragm moves once and does not vibrate – no pressure variations.
Examiner tip: Sound needs vibration: a d.c. current gives a force that never reverses, so there is nothing to keep the diaphragm moving to and fro.

9Grade 6–7Higher4 marks

A straight wire carries a current of 850 mA. Part of the wire is at right angles to a uniform magnetic field of magnetic flux density 60 mT.

The force on the wire is 10.2 mN.

(a) Calculate the length of wire in the magnetic field. Give your answer in metres. Use the Physics Equations Sheet. [3]

(b) The length of wire in the field is doubled and the current is halved. The magnetic flux density is unchanged.

State the effect on the force on the wire. [1]

(a) $0.0102 = 0.060 \times 0.85 \times l$ all three values converted to N, T and A(1)
(a) $l = \dfrac{0.0102}{0.060 \times 0.85}$ correct rearrangement; allow this mark with unconverted values(1)
(a) $l = 0.20$ (m) allow 0.2 (m); an answer of 0.20 m scores 3 marks; an answer with one power-of-ten (milli-unit) error, e.g. $2.0 \times 10^{-4}$ or 200, scores 2 marks(1)
(b) (force is) unchanged allow it stays the same / 10.2 mN; ignore reasons(1)
Answer: (a) 0.20 m (b) the force stays the same
Examiner tip: Convert every milli- unit before substituting – 850 mA is 0.85 A and 60 mT is 0.060 T.

10Grade 6–7Higher4 marks

The diagram shows a rectangular coil ABCD between the poles of a magnet in a simple electric motor. The coil is free to turn about the dashed axis. The arrows show the direction of the current in the coil.

(a) Determine the direction of the force on side AB. [1]

(b) Explain why there is no force on side BC. [1]

(a) into the page allow away from you / away from the reader(1)
(b) (side BC / the current in BC is) parallel to the magnetic field allow BC is not at right angles to the field; do not accept 'BC is not in the field'(1)
(c) any one from: increase the current (allow increase the pd / voltage of the supply); use a stronger magnet (allow increase the magnetic flux density); more turns on the coil; longer sides AB and CD (so a longer length of wire in the field); a wider coil / longer side BC (so the forces act further from the axis) ignore 'bigger magnet'; ignore 'iron core'(1)
(c) a second, different change from the list above(1)
Answer: (a) into the page (b) BC is parallel to the magnetic field (c) any two of: larger current; stronger magnet (greater flux density); more turns on the coil; longer sides AB and CD; wider coil
Examiner tip: A force only acts on the parts of the coil that cut across the field – sides parallel to the field feel no force.

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