Secondary 4 Pure Physics Practice Paper 4

TuitionGoWhere Practice Paper — Pure Physics Secondary 4

Answer Key — Electricity & Magnetism (Version 4)

Section A: Multiple Choice [20 marks]

1. (b) 11.5 V [2]

Working: V_s / V_p = N_s / N_p → V_s = (100 / 2000) × 230 = 11.5 V

Common mistake: Confusing step-up with step-down; using N_p/N_s instead of N_s/N_p.

2. (b) It provides a low-resistance path for current to flow to the ground in case of a fault. [2]

Note: The earth wire does not carry current during normal operation. It is a safety feature that prevents the metal casing of an appliance from becoming live.

3. (b) It reverses. [2]

Explanation: The force on a current-carrying conductor in a magnetic field depends on the direction of current (Fleming's left-hand rule). Reversing the current reverses the force direction.

4. (b) 26.5 Ω [2]

Working: P = V²/R → R = V²/P = 230² / 2000 = 52900 / 2000 = 26.45 Ω ≈ 26.5 Ω

Common mistake: Using P = IV without rearranging correctly; forgetting to square the voltage.

5. (c) Increasing the number of turns in the coil [2]

Explanation: By Faraday's law, the induced e.m.f. is proportional to the number of turns. More turns means a greater rate of change of flux linkage.

6. (b) 1.0 A [2]

Working: R_total = 2 + 3 + 5 = 10 Ω; I = V/R = 10/10 = 1.0 A

Note: In series, resistances add directly.

7. (c) They are closer together where the field is stronger. [2]

Note: Field lines never cross; they run from north to south outside a magnet; they show field direction, not particle paths.

8. (c) 108 J [2]

Working: I = Q/t = 12/4 = 3 A; E = VIt = 9 × 3 × 4 = 108 J (or E = VQ = 9 × 12 = 108 J)

Common mistake: Forgetting to calculate current first; using E = V/t without Q.

9. (b) Anticlockwise [2]

Explanation: Using the right-hand grip rule: thumb points in direction of current (upwards), fingers curl anticlockwise when viewed from above.

10. (b) 20 A [2]

Working: η = (V_s × I_s)/(V_p × I_p) → 0.80 = (46 × I_s)/(230 × 5) → 0.80 = 46I_s / 1150 → 46I_s = 920 → I_s = 20 A

Common mistake: Forgetting to convert 80% to 0.80; confusing primary and secondary values.

Section B: Short Answer & Structured Questions [30 marks]

11. [4 marks]

(a) Electromotive force (e.m.f.) is the work done by a cell in driving a unit charge around a complete circuit. [2]

Accept: "Energy transferred per unit charge from the cell to the circuit" or "Total energy supplied per coulomb of charge passing through the cell."

(b) Internal resistance is the resistance within the cell itself, which causes some energy to be lost (as heat) inside the cell when current flows. [2]

Accept: "Resistance of the chemicals/electrolyte inside the cell" or "Resistance that opposes current flow within the source."

12. [5 marks]

(a) Equivalent resistance: [2]

1/R = 1/4 + 1/6 = 3/12 + 2/12 = 5/12

R = 12/5 = 2.4 Ω

Marking: 1 mark for correct formula/substitution; 1 mark for correct answer.

(b) Total current: [2]

I = V/R = 12 / 2.4 = 5.0 A

Marking: 1 mark for correct formula; 1 mark for correct answer.

(c) Current through R₁: [1]

V across R₁ = 12 V (parallel combination has same voltage as battery)

I₁ = V/R₁ = 12/4 = 3.0 A

Accept: Using current divider: I₁ = 5.0 × (6/(4+6)) = 3.0 A

13. [5 marks]

(a) F = BIL (where B is magnetic flux density, I is current, L is length of conductor in the field) [1]

Accept: F = BIL sin θ, with θ = 90° since perpendicular.

(b) Force calculation: [2]

F = BIL = 0.4 × 3.0 × 0.05 = 0.06 N

Marking: 1 mark for substitution; 1 mark for correct answer with unit.

(c) Two ways to increase the force: [2 — 1 each]

1. Increase the current in the conductor.
2. Use a stronger magnet (increase magnetic flux density).

Also accept: Increase the length of the conductor in the field; align the conductor more perpendicular to the field (if not already perpendicular).

14. [6 marks]

(a) Individual currents (using I = P/V): [2]

• Oven: I = 3000/230 = 13.0 A
• Microwave: I = 1200/230 = 5.2 A
• Kettle: I = 2200/230 = 9.6 A
• Refrigerator: I = 400/230 = 1.7 A

Marking: 0.5 mark each for correct values.

(b) Total current: [2]

I_total = 13.0 + 5.2 + 9.6 + 1.7 = 29.5 A

Accept: Using total power: P_total = 6800 W; I = 6800/230 = 29.6 A (rounding difference accepted)

Marking: 1 mark for method; 1 mark for correct answer.

(c) Will the circuit breaker trip? [2]

The total current drawn (29.5 A) is less than the 30 A rating, so the circuit breaker will not trip. However, the margin is very small (only 0.5 A), so if any additional appliance is connected, the breaker would trip.

Marking: 1 mark for correct conclusion; 1 mark for valid explanation/comparison.

15. [5 marks]

(a) Direction of force on AB: Downwards (into the page/screen) [1]

Explanation using Fleming's left-hand rule: First finger (field) points from N to S (left to right), second finger (current) points from A to B, thumb points downwards.

(b) Function of split-ring commutator: [2]

The split-ring commutator reverses the direction of current in the coil every half-rotation. This ensures that the torque on the coil always acts in the same direction, allowing the coil to rotate continuously in one direction.

Marking: 1 mark for "reverses current direction"; 1 mark for linking to continuous rotation.

(c) Two modifications to increase turning effect: [2 — 1 each]

1. Increase the current through the coil.
2. Use a stronger magnet (increase magnetic flux density).

Also accept: Increase the number of turns on the coil; increase the area of the coil.

16. [5 marks]

(a) Number of turns on secondary coil: [2]

V_s / V_p = N_s / N_p

12 / 230 = N_s / 920

N_s = (12 / 230) × 920 = 48 turns

Marking: 1 mark for correct formula/substitution; 1 mark for correct answer.

(b) Primary current: [3]

Using efficiency: η = (V_s × I_s) / (V_p × I_p)

0.90 = (12 × 4.0) / (230 × I_p)

0.90 = 48 / (230 × I_p)

230 × I_p = 48 / 0.90 = 53.33

I_p = 53.33 / 230 = 0.23 A (or 0.232 A)

Marking: 1 mark for correct efficiency equation; 1 mark for correct substitution; 1 mark for correct answer with unit.

Common mistake: Forgetting to use efficiency and simply equating V_p × I_p = V_s × I_s (which would give I_p = 0.21 A — this is the ideal case, not the 90% efficient case).

Section C: Free Response / Application Questions [30 marks]

17. [8 marks]

(a) Explanation using Faraday's law: [3]

As the magnet approaches and passes through the solenoid, the magnetic flux linking the coil changes. According to Faraday's law, a change in magnetic flux through a coil induces an e.m.f. across the coil. The magnitude of the induced e.m.f. is proportional to the rate of change of magnetic flux linkage.

Marking:

• 1 mark: Mentions changing magnetic flux
• 1 mark: References Faraday's law correctly
• 1 mark: Links rate of flux change to induced e.m.f.

(b) Why e.m.f. changes direction: [2]

As the magnet enters the solenoid, the flux through the coil increases in one direction, inducing an e.m.f. in one direction. As the magnet leaves the solenoid, the flux through the coil decreases (or increases in the opposite sense as the opposite pole passes through), so by Lenz's law the induced e.m.f. reverses direction. This produces the two peaks of opposite polarity.

Marking:

• 1 mark: Explains that flux change direction reverses on entry vs. exit
• 1 mark: References Lenz's law or the opposing nature of the flux change

(c) Two differences when dropped from greater height: [3]

1. Greater peak e.m.f. values: The magnet reaches a higher speed as it enters and exits the solenoid (due to gravitational acceleration over a longer distance). A higher speed means a greater rate of change of flux, so by Faraday's law, the induced e.m.f. is larger. Both peaks will be higher.

2. Shorter time between peaks / narrower peaks: Since the magnet is moving faster, it spends less time passing through the solenoid. The time interval between the two peaks will be shorter, and each peak will be narrower (sharper).

Marking: 1.5 marks each — 1 mark for stating the difference, 1 mark for correct explanation (0.5 if explanation is partial).

Also accept: The time between the two peaks decreases because the magnet travels through the solenoid at a higher speed.

18. [10 marks]

(a) Why high-voltage transmission: [4]

When electrical power is transmitted, there is power loss in the cables due to heating, given by P_loss = I²R. For a fixed power P to be delivered, P = VI, so I = P/V. By transmitting at a higher voltage, the current I is reduced. Since power loss is proportional to I², reducing the current dramatically reduces the energy lost as heat in the cables.

Marking:

• 1 mark: States P_loss = I²R
• 1 mark: States P = VI and explains that higher V means lower I for same P
• 1 mark: Explains that lower I reduces I²R losses
• 1 mark: Clear, coherent explanation linking all points

(b) Power loss at 500 A: [2]

P_loss = I²R = 500² × 8 = 250 000 × 8 = 2 000 000 W = 2 MW

Marking: 1 mark for substitution; 1 mark for correct answer with unit.

(c) Current and power loss at 25 000 V: [4]

Assuming the same power is delivered:

At 400 000 V: P = V × I = 400 000 × 500 = 200 000 000 W = 200 MW

At 25 000 V: I = P / V = 200 000 000 / 25 000 = 8000 A

New power loss: P_loss = I²R = 8000² × 8 = 64 000 000 × 8 = 512 000 000 W = 512 MW

Comment: The power loss (512 MW) is far greater than the power being delivered (200 MW), which is completely impractical. This demonstrates why high-voltage transmission is essential — it reduces current and therefore dramatically reduces I²R losses in the cables.

Marking:

• 1 mark: Correct calculation of power delivered
• 1 mark: Correct current at 25 000 V (8000 A)
• 1 mark: Correct power loss (512 MW)
• 1 mark: Valid comment on impracticality / comparison

Note: Students may also calculate the ratio: (8000/500)² = 256 times more power loss, which is an acceptable alternative approach.

19. [6 marks]

(a) Magnetic field pattern: [2]

The field lines around each wire form concentric circles (right-hand grip rule). Between the wires, the fields from the two wires are in opposite directions (since currents are in the same direction), so they partially cancel, creating a weaker field region between them. Outside both wires, the fields reinforce.

    P                          Q
  ↻  ↻  ↻                  ↻  ↻  ↻
  ↻  |  ↻    (weak)       ↻  |  ↻
  ↻  ↻  ↻    region        ↻  ↻  ↻
    ↓                          ↓
  (fields oppose         (fields oppose
   between wires)         between wires)

Marking: 1 mark for correct circular field direction around each wire; 1 mark for showing weaker/field cancellation between the wires.

(b) Force between wires and direction: [2]

Each current-carrying wire produces a magnetic field that exerts a force on the other current-carrying wire (F = BIL). The force on wire Q due to wire P is towards wire P (attractive), because parallel currents in the same direction attract.

Marking: 1 mark for explaining that each wire experiences a force due to the other's magnetic field; 1 mark for correct direction (towards P / attractive).

(c) Effect of reversing both currents: [2]

The force between the wires would still be attractive and in the same direction (towards each other). Reversing both currents reverses the magnetic field direction around each wire, but the force direction depends on the product of both current directions — reversing both leaves the force direction unchanged.

Marking: 1 mark for stating the force remains attractive; 1 mark for correct explanation.

20. [6 marks]

(a) Resistance of the fan: [1]

R = V/I = 6 / 0.5 = 12 Ω

(b) Series resistor value: [3]

The fan needs 6 V across it. The battery supplies 9 V, so the series resistor must drop:

V_resistor = 9 − 6 = 3 V

The current through the series resistor is the same as the fan current (series circuit): I = 0.5 A

R_series = V_resistor / I = 3 / 0.5 = 6 Ω

Marking: 1 mark for correct voltage across resistor (3 V); 1 mark for using same current (0.5 A); 1 mark for correct answer (6 Ω).

(c) Power dissipated by series resistor: [2]

P = V × I = 3 × 0.5 = 1.5 W

Accept: P = I²R = 0.5² × 6 = 1.5 W or P = V²/R = 3²/6 = 1.5 W

Marking: 1 mark for correct formula/substitution; 1 mark for correct answer with unit.

End of Answer Key

Total: 80 marks