Physics questions, chapter 4

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Electromagnetism

45 marks. Use Practice to check each answer as you go, or Exam to answer everything against the clock first.

Section A: objective questions

11 marks

Choose the best answer for each question.

Question 1

What produces the force on a current-carrying conductor in a magnetic field?

The heating effect of the current
The interaction between the magnetic field of the current and the magnetic field of the magnet
The electric field of the power supply
The gravitational pull of the magnet

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The two magnetic fields combine into a catapult field, which exerts the force (p. 137).

For the teacher: Remembering, low difficulty, standard 4.1.1

Textbook p. 137

Question 2

A straight wire lies between the N pole (on the left) and the S pole (on the right) of a magnet. The current flows into the plane of the paper. What is the direction of the force on the wire?

Upwards
To the left
Downwards
To the right

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Field to the right (N to S), current into the page. Fleming's left-hand rule gives the force downwards. (With current out of the page the force is upwards, as in Figure 4.3.)

For the teacher: Applying, medium difficulty, standard 4.1.2

Textbook p. 139

Question 3

Which change will NOT increase the magnitude of the force on a current-carrying conductor in a magnetic field?

Reversing the direction of the current
Increasing the current
Using stronger magnets
Bringing the magnetic poles closer together

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Reversing the current only reverses the direction of the force. The force grows with a larger current and a stronger field.

For the teacher: Understanding, low difficulty, standard 4.1.3

Textbook p. 141

Question 4

What is the function of the commutator in a direct current motor?

To increase the voltage of the supply
To change the magnetic field of the magnets
To reverse the current in the coil every half rotation
To store charge for the coil

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The split ring commutator reverses the current in the coil every half rotation, so the coil keeps rotating in the same direction.

For the teacher: Understanding, low difficulty, standard 4.1.5

Textbook p. 144

Question 5

Which is an advantage of a brushless motor compared with a brushed motor?

No friction or sparking from carbon brushes
The coil rotates instead of the magnet
It produces louder operation
It uses a commutator

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A brushless motor has no carbon brushes or commutator, so there is no wear and no sparking, and it is quieter (Table 4.2).

For the teacher: Remembering, low difficulty, standard 4.1.6

Textbook p. 147

Question 6

What is electromagnetic induction?

Production of a magnetic field by a current in a coil
Production of heat when a current passes through a wire
Production of a force on a current in a magnetic field
Production of an induced e.m.f. when there is relative motion between a conductor and a magnetic field

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Electromagnetic induction is the production of an induced e.m.f. in a conductor when there is relative motion between the conductor and a magnetic field, or the conductor is in a changing field.

For the teacher: Remembering, low difficulty, standard 4.2.1

Textbook p. 153

Question 7

A bar magnet and a solenoid are connected to a galvanometer. In which situation does the galvanometer show NO deflection?

The magnet is pushed into the solenoid
The magnet is pulled out of the solenoid
The magnet is held stationary inside the solenoid
The solenoid is moved towards a stationary magnet

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There must be relative motion. When both are stationary no magnetic flux is cut, so no e.m.f. is induced.

For the teacher: Understanding, low difficulty, standard 4.2.1

Textbook p. 151

Question 8

The north pole of a bar magnet is pushed towards one end of a solenoid. What is the polarity induced at that end of the solenoid?

North pole
South pole
No pole is induced
Both poles

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By Lenz's law the induced current opposes the motion, so the end becomes a north pole and repels the approaching north pole.

For the teacher: Applying, medium difficulty, standard 4.2.3

Textbook p. 159

Question 9

Which part makes an a.c. generator give alternating current in the external circuit?

Split-ring commutator
Carbon brushes
Soft iron core
Two slip rings

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The ends of the coil connect to two slip rings, so the external current alternates. A d.c. generator uses a split-ring commutator.

For the teacher: Understanding, low difficulty, standard 4.2.4

Textbook p. 160

Question 10

A transformer has 200 turns in the primary coil and 1 000 turns in the secondary coil. The primary voltage is 12 V. What is the secondary voltage?

2.4 V
12 V
60 V
6 000 V

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V_S = 12 × 1000 / 200 = 60 V (a step-up transformer).

For the teacher: Applying, medium difficulty, standard 4.3.1

Textbook p. 163

Question 11

Why is electrical energy transmitted at a very high voltage?

To make the cables lighter
To reduce the current so less energy is lost as heat in the cables
To increase the power produced by the power station
To make the frequency of the supply higher

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For the same power, a higher voltage gives a smaller current, so I²R heating loss in the cables is reduced.

For the teacher: Understanding, low difficulty, standard 4.3.4

Textbook p. 167

Section B: structured questions

34 marks

Answer all questions. Write your answers in the spaces provided.

Question 12

A direct current motor has a rectangular coil, a split-ring commutator and carbon brushes.

[2 marks] (a)Explain why the coil rotates when current flows through it.

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The current in the coil and the magnetic field of the magnet form a catapult field [1 mark]1
Forces of equal size and opposite direction act on the two sides of the coil, producing a turning effect [1 mark]1

Tick each point you got.

[2 marks] (b)Why is the commutator split into two halves?

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So the current in the coil reverses every half rotation [1 mark]1
So the force keeps turning the coil in the same direction [1 mark]1

Tick each point you got.

[3 marks] (c)State three ways to make the motor rotate faster.

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Increase the current in the coil [1 mark]1
Use stronger magnets [1 mark]1
Increase the number of turns of the coil [1 mark]1

Tick each point you got.

For the teacher: Understanding, low difficulty, standard 4.1.5

Textbook p. 144

Question 13

A bar magnet is pushed quickly, north pole first, into a solenoid connected to a sensitive galvanometer.

[2 marks] (a)Explain why the galvanometer pointer deflects.

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The solenoid cuts magnetic flux (relative motion between magnet and solenoid) [1 mark]1
An e.m.f. is induced and an induced current flows in the complete circuit [1 mark]1

Tick each point you got.

[1 mark] (b)State the polarity at the end of the solenoid nearest the magnet.

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North pole1

Tick each point you got.

[2 marks] (c)Name the law that explains your answer in (b) and state it.

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Lenz's law [1 mark]1
The induced current flows in a direction that opposes the change of magnetic flux that causes it [1 mark]1

Tick each point you got.

For the teacher: Analysing, low difficulty, standard 4.2.3

Textbook p. 159

Question 14

A transformer, assumed ideal, changes 230 V to 11.5 V. The secondary coil has 100 turns and delivers 23 W to a lamp.

A simple transformer with primary and secondary coils on an iron core — Diagram 1

[1 mark] (a)Is this a step-up or a step-down transformer?

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Step-down (output voltage is less than input voltage)1

Tick each point you got.

[2 marks] (b)Calculate the number of turns of the primary coil.

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N_P = (V_P / V_S) × N_S = 230 / 11.5 × 100 [1 mark]1
= 2 000 turns [1 mark]1

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[2 marks] (c)Calculate the current in the secondary coil.

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I_S = P / V_S = 23 / 11.5 [1 mark]1
= 2.0 A [1 mark]1

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[2 marks] (d)Calculate the current in the primary coil.

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V_PI_P = V_SI_S (ideal): 230 × I_P = 23 [1 mark]1
I_P = 0.10 A [1 mark]1

Tick each point you got.

For the teacher: Applying, medium difficulty, standard 4.3.2

Textbook p. 164

Question 15

A pupil tests a transformer. The primary voltage is 240 V and the primary current is 0.50 A. The secondary voltage is 12 V and the secondary current is 8.5 A.

[3 marks] (a)Calculate the efficiency of the transformer.

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Input power = 240 × 0.50 = 120 W [1 mark]1
Output power = 12 × 8.5 = 102 W [1 mark]1
Efficiency = 102 / 120 × 100% = 85% [1 mark]1

Tick each point you got.

[2 marks] (b)Explain why the transformer is not ideal. State two causes.

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Resistance of the coils causes heating [1 mark]1
Eddy currents / hysteresis / leakage of magnetic flux in the core [1 mark]1

Tick each point you got.

[2 marks] (c)Suggest two improvements to the design to increase the efficiency.

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Use thicker copper wire for the coils [1 mark]1
Use a laminated soft iron core / wind the secondary coil on the primary coil [1 mark]1

Tick each point you got.

For the teacher: Evaluating, high difficulty, standard 4.3.3

Textbook p. 164

Question 16

A power station delivers 50 kW through cables of total resistance 0.50 Ω.

[1 mark] (a)Calculate the current in the cables if the power is transmitted at 250 V.

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I = P / V = 50 000 / 250 = 200 A1

Tick each point you got.

[1 mark] (b)Calculate the current if a step-up transformer raises the voltage to 25 kV.

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I = 50 000 / 25 000 = 2 A1

Tick each point you got.

[2 marks] (c)Calculate the power lost in the cables (P = I²R) in each case.

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At 250 V: 200² × 0.50 = 20 000 W [1 mark]1
At 25 kV: 2² × 0.50 = 2 W [1 mark]1

Tick each point you got.

For the teacher: Applying, low difficulty, standard 4.3.4

Textbook p. 167

Question 17

Two identical metal balls look the same. One is a neodymium magnet and the other is a steel ball. A long copper tube is available.

[2 marks] (a)Describe an activity to identify the magnet.

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Drop each ball through the copper tube and compare the time taken to fall [1 mark]1
The ball that takes longer to fall is the neodymium magnet [1 mark]1

Tick each point you got.

[2 marks] (b)Explain the physics principle used.

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The falling magnet induces an e.m.f. and eddy currents in the copper tube (electromagnetic induction) [1 mark]1
By Lenz's law the induced current opposes the motion, so the magnet falls slowly; the steel ball falls freely [1 mark]1

Tick each point you got.

For the teacher: Creating, medium difficulty, standard 4.2.3

Textbook p. 170

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