IGCSE Physics Paper 4 - Topic 4: Electricity and Magnetism

(a) [2]

State two properties of magnets.

Model Answer — 1(a)

Like poles repel [1]

Unlike poles attract [1]

Mark 1 — Like poles repel

Mark 2 — Unlike poles attract

(b) [3]

Draw the magnetic field pattern around a bar magnet. Include the direction of the field lines.

Model Answer — 1(b)

Correct shape showing field lines from N to S [1]

Lines not crossing, correct spacing (closer together at poles) [1]

Direction arrows from N to S [1]

Mark 1 — Correct shape showing field lines from N to S

Mark 2 — Lines not crossing, correct spacing

Mark 3 — Direction arrows from N to S

Describe how an electromagnet is made.

Model Answer — 1(c)(i)

Coil of insulated wire [1]

Wound around a soft iron core [1]

Mark 1 — Coil of insulated wire

Mark 2 — Wound around a soft iron core

State two ways to increase the strength of an electromagnet.

Model Answer — 1(c)(ii)

Increase the current [1]

Increase the number of turns on the coil [1]

Mark 1 — Increase the current

Mark 2 — Increase the number of turns on the coil

(d) [3]

Explain why soft iron is used as the core of an electromagnet rather than steel.

Model Answer — 1(d)

Soft iron is easily magnetised [1]

Soft iron is easily demagnetised [1]

So the electromagnet can be switched on and off / steel would remain permanently magnetised [1]

Mark 1 — Soft iron is easily magnetised

Mark 2 — Soft iron is easily demagnetised

Mark 3 — Electromagnet can be switched on and off, steel remains permanently magnetised

(a) [2]

Describe how a polythene rod can be charged by friction.

Model Answer — 2(a)

Rub the rod with a dry cloth or duster [1]

Electrons transfer from cloth to rod giving it a negative charge [1]

Mark 1 — Rub the rod with a dry cloth or duster

Mark 2 — Electrons transfer from cloth to rod giving it a negative charge

(b) [2]

A negatively charged rod is brought close to a small piece of uncharged paper. Explain why the paper is attracted to the rod.

Model Answer — 2(b)

The rod repels electrons in the paper to the far side, inducing positive charge on the near surface [1]

The attraction between the rod and the nearer positive charges is greater than the repulsion with the more distant negative charges [1]

Mark 1 — Rod repels electrons in the paper to the far side, inducing positive charge on near surface

Mark 2 — Attraction between rod and nearer positive charges is greater than repulsion with distant negative charges

Draw the electric field pattern between two parallel plates, one positive and one negative.

Model Answer — 2(c)

Straight parallel lines between the plates [1]

Lines perpendicular to the plates [1]

Direction from positive plate to negative plate, arrows shown correctly [1]

Mark 1 — Straight parallel lines between the plates

Mark 2 — Lines perpendicular to the plates

Mark 3 — Direction from positive plate to negative plate with arrows

(d) [3]

A Van de Graaff generator is used to demonstrate electrostatic effects. A student's hair stands on end when they touch the dome. Explain why.

Model Answer — 2(d)

Charge transfers from dome to student and hair [1]

Each strand of hair receives the same charge (positive) [1]

Like charges repel, so the strands repel each other and stand on end [1]

Mark 1 — Charge transfers from dome to student and hair

Mark 2 — Each strand receives the same charge (positive)

Mark 3 — Like charges repel so strands repel each other and stand on end

(a) (i) [1]

State the equation linking current, charge and time.

Model Answer — 3(a)(i)

I = Q/t or Q = It [1]

I = Q / t

Mark 1 — Correct equation I = Q/t or Q = It stated

(a) (ii) [3]

A current of 2.5 A flows through a lamp for 3 minutes. Calculate the charge that flows through the lamp.

Model Answer — 3(a)(ii)

Convert time: t = 3 × 60 = 180 s [1]

Q = It = 2.5 × 180 [1]

Q = 450 C [1]

Q = 2.5 × 180 = 450 C

Mark 1 — Convert time to seconds: 3 × 60 = 180 s

Mark 2 — Correct substitution Q = 2.5 × 180

Mark 3 — Correct answer Q = 450 C

(b) (i) [1]

State the equation linking voltage, current and resistance.

Model Answer — 3(b)(i)

V = IR or R = V/I [1]

V = I × R

Mark 1 — Correct equation V = IR or R = V/I stated

(b) (ii) [3]

A resistor has a resistance of 47 Ω. The potential difference across it is 9.4 V. Calculate the current flowing through the resistor.

Model Answer — 3(b)(ii)

I = V/R [1]

I = 9.4 / 47 [1]

I = 0.20 A [1]

I = 9.4 / 47 = 0.20 A

Mark 1 — Correct rearrangement I = V/R

Mark 2 — Correct substitution I = 9.4 / 47

Mark 3 — Correct answer I = 0.20 A

The I–V characteristic of a filament lamp is non-linear. Explain why the resistance of a filament lamp increases as the current increases.

Model Answer — 3(c)

As current increases, the filament gets hotter [1]

The metal atoms vibrate more [1]

This causes more frequent collisions with the charge carriers (electrons) [1]

Therefore the resistance increases (V increases more than proportionally to I) [1]

Mark 1 — As current increases the filament gets hotter

Mark 2 — The metal atoms vibrate more

Mark 3 — More frequent collisions with charge carriers or electrons

Mark 4 — Therefore resistance increases, V increases more than proportionally to I

(a) (i) [2]

Calculate the total resistance of the circuit when the resistors are connected in series.

Model Answer — 4(a)(i)

R_total = R1 + R2 [1]

R_total = 4.0 + 8.0 = 12 Ω [1]

R = 4.0 + 8.0 = 12 Ω

Mark 1 — Correct formula R_total = R1 + R2

Mark 2 — Correct answer R_total = 12 ohm

(a) (ii) [2]

Calculate the current flowing through the circuit.

Model Answer — 4(a)(ii)

I = V/R = 12/12 [1]

I = 1.0 A [1]

I = 12 / 12 = 1.0 A

Mark 1 — Correct substitution I = V/R = 12/12

Mark 2 — Correct answer I = 1.0 A

(a) (iii) [2]

Calculate the potential difference across R₂.

Model Answer — 4(a)(iii)

V = IR = 1.0 × 8.0 [1]

V = 8.0 V [1]

V = 1.0 × 8.0 = 8.0 V

Mark 1 — Correct substitution V = IR = 1.0 × 8.0

Mark 2 — Correct answer V = 8.0 V

(b) (i) [3]

Calculate the total resistance of the parallel combination.

Model Answer — 4(b)(i)

1/R = 1/R1 + 1/R2 = 1/4.0 + 1/8.0 [1]

1/R = 0.25 + 0.125 = 0.375 [1]

R = 1/0.375 = 2.67 Ω (accept 2.7 Ω or 8/3 Ω) [1]

1/R = 1/4.0 + 1/8.0 = 0.375 → R = 2.67 Ω

Mark 1 — Correct formula 1/R = 1/R1 + 1/R2 with substitution

Mark 2 — Correct calculation 1/R = 0.25 + 0.125 = 0.375

Mark 3 — Correct answer R = 2.67 ohm (accept 2.7 or 8/3)

(b) (ii) [3]

Calculate the current through each resistor.

Model Answer — 4(b)(ii)

Through R1: I = V/R = 12/4.0 = 3.0 A [1]

Through R2: I = V/R = 12/8.0 = 1.5 A [1]

Total current = 3.0 + 1.5 = 4.5 A [1]

I1 = 12/4.0 = 3.0 A; I2 = 12/8.0 = 1.5 A; I_total = 4.5 A

Mark 1 — Current through R1: I = 12/4.0 = 3.0 A

Mark 2 — Current through R2: I = 12/8.0 = 1.5 A

Mark 3 — Total current = 3.0 + 1.5 = 4.5 A

(a) (i) [1]

State the equation linking power, current and voltage.

Model Answer — 5(a)(i)

P = IV [1]

P = I × V

Mark 1 — Correct equation P = IV stated

(a) (ii) [2]

An electric kettle operates at 230 V and draws a current of 10 A. Calculate the power of the kettle.

Model Answer — 5(a)(ii)

P = IV = 10 × 230 [1]

P = 2300 W (or 2.3 kW) [1]

P = 10 × 230 = 2300 W

Mark 1 — Correct substitution P = 10 × 230

Mark 2 — Correct answer P = 2300 W or 2.3 kW

(a) (iii) [3]

The kettle takes 4 minutes to boil water. Calculate the energy transferred in kilowatt-hours and the cost if electricity costs 15p per kWh.

Model Answer — 5(a)(iii)

Time = 4/60 = 1/15 hours (or 0.0667 h) [1]

Energy = 2.3 × (4/60) = 0.153 kWh [1]

Cost = 0.153 × 15 = 2.3p [1]

E = 2.3 kW × (4/60) h = 0.153 kWh; Cost = 0.153 × 15 = 2.3p

Mark 1 — Convert time: 4/60 = 0.0667 hours

Mark 2 — Energy = 2.3 × (4/60) = 0.153 kWh

Mark 3 — Cost = 0.153 × 15 = 2.3p

(b) (i) [2]

The kettle has a three-pin plug. State the colours of the wires connected to the live, neutral and earth terminals.

Model Answer — 5(b)(i)

Live = brown, Neutral = blue [1]

Earth = green and yellow (striped) [1]

Mark 1 — Live is brown, Neutral is blue

Mark 2 — Earth is green and yellow striped

(b) (ii) [2]

Explain why a fuse is connected in the live wire.

Model Answer — 5(b)(ii)

If the current exceeds the rated value, the fuse melts/blows and breaks the circuit [1]

This must be in the live wire so that the appliance is disconnected from the high voltage supply when the fuse blows / prevents the case becoming live [1]

Mark 1 — Fuse melts and breaks the circuit if current exceeds rated value

Mark 2 — Must be in live wire to disconnect appliance from high voltage supply

The kettle is rated at 2300 W and operates at 230 V. Determine the most appropriate fuse rating. Choose from 3 A, 5 A, or 13 A.

Model Answer — 5(c)

I = P/V = 2300/230 = 10 A [1]

13 A fuse (must be higher than 10 A but closest available rating) [1]

I = 2300 / 230 = 10 A → 13 A fuse

Mark 1 — Correct calculation I = P/V = 2300/230 = 10 A

Mark 2 — 13 A fuse selected (higher than 10 A, closest available)

(a) (i) [2]

State two ways of inducing an e.m.f. in a coil of wire.

Model Answer — 6(a)(i)

Move a magnet into or out of the coil [1]

Change the magnetic field through the coil / vary the current in a nearby coil [1]

Mark 1 — Move a magnet into or out of the coil

Mark 2 — Change the magnetic field through the coil or vary current in a nearby coil

(a) (ii) [2]

State two ways of increasing the magnitude of the induced e.m.f.

Model Answer — 6(a)(ii)

Move the magnet faster / increase the rate of change of flux [1]

Use more turns on the coil / use a stronger magnet [1]

Mark 1 — Move the magnet faster or increase rate of change of flux

Mark 2 — Use more turns on the coil or use a stronger magnet

(b) [1]

State the equation linking the voltages and number of turns in a transformer.

Model Answer — 6(b)

Vp/Vs = Np/Ns [1]

Vp / Vs = Np / Ns

Mark 1 — Correct transformer equation Vp/Vs = Np/Ns

A transformer has 400 turns on its primary coil and 2000 turns on its secondary coil. The input voltage is 25 V. Calculate the output voltage and state whether this is a step-up or step-down transformer.

Model Answer — 6(c)

Vs = Vp × Ns/Np = 25 × 2000/400 [1]

Vs = 125 V [1]

Step-up transformer (because Vs > Vp or Ns > Np) [1]

Vs = 25 × (2000/400) = 25 × 5 = 125 V

Mark 1 — Correct substitution Vs = 25 × 2000/400

Mark 2 — Correct answer Vs = 125 V

Mark 3 — Step-up transformer correctly identified

(d) [4]

Explain why transformers are used in the transmission of electrical power over long distances. Include relevant equations in your answer.

Model Answer — 6(d)

Power loss in cables = I²R (or P = I²R) [1]

Transformers step up the voltage for transmission [1]

For the same power transmitted, P = IV means higher V gives lower I [1]

Lower current means less power wasted as heat in the cables (I²R losses are reduced) [1]

P_loss = I²R; P = IV → higher V, lower I, less loss

Mark 1 — Power loss in cables equals I squared R

Mark 2 — Transformers step up the voltage for transmission

Mark 3 — P = IV so higher voltage gives lower current for same power

Mark 4 — Lower current means less power wasted as heat in cables

(a) [2]

State the rule used to find the direction of force on a current-carrying conductor in a magnetic field. Describe what each finger represents.

Model Answer — 7(a)

Fleming's left-hand rule [1]

First finger = Field, seCond finger = Current, thuMb = Motion/Force [1]

Mark 1 — Fleming's left-hand rule named

Mark 2 — First finger = Field, Second finger = Current, Thumb = Motion or Force

(b) [3]

Describe the structure of a simple DC motor. Include the purpose of the commutator.

Model Answer — 7(b)

A coil of wire (armature) placed between the poles of a magnet [1]

The coil is connected to a power supply via a split-ring commutator and carbon brushes [1]

The commutator reverses the current direction every half turn to maintain continuous rotation in the same direction [1]

Mark 1 — Coil of wire (armature) between poles of a magnet

Mark 2 — Connected via split-ring commutator and carbon brushes

Mark 3 — Commutator reverses current every half turn for continuous rotation

State two ways to increase the speed of rotation of a DC motor.

Model Answer — 7(c)

Increase the current [1]

Use a stronger magnet / increase the number of turns on the coil [1]

Mark 1 — Increase the current

Mark 2 — Use a stronger magnet or increase number of turns on coil

(d) [3]

Describe the differences between a DC motor and an AC generator.

Model Answer — 7(d)

A motor converts electrical energy to kinetic energy; a generator converts kinetic energy to electrical energy [1]

A motor has a split-ring commutator; a generator has slip rings [1]

A motor requires an input current; a generator produces an output e.m.f./current [1]

Mark 1 — Motor converts electrical to kinetic energy; generator converts kinetic to electrical energy

Mark 2 — Motor has split-ring commutator; generator has slip rings

Mark 3 — Motor requires input current; generator produces output e.m.f. or current

IGCSE Physics Paper 4 (Theory / Extended)

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