O Level Physics Electricity Magnetism Quiz

O-Level Physics Quiz - Electricity Magnetism: Answer Key

Total Marks: 50

Section A: Static Electricity and Electric Fields (Questions 1–4)

1. (a) Explain why the polythene rod becomes negatively charged. [2]

Answer: When the polythene rod is rubbed with the woollen cloth, electrons are transferred from the cloth to the rod [1]. The rod gains electrons, so it has an excess of negative charge and becomes negatively charged [1].

(b) Explain why the uncharged aluminium foil is attracted to the charged rod. [2]

Answer: The negatively charged rod repels electrons in the aluminium foil to the far side of the foil, leaving the near side with a net positive charge (charging by induction) [1]. The positive charges on the near side are closer to the rod than the negative charges on the far side, so the attractive force is greater than the repulsive force, resulting in a net attraction [1].

2. (a) State the direction of the electric field between the plates. [1]

Answer: The electric field direction is from the positive plate to the negative plate (downwards in the diagram) [1].

(b) Describe the force experienced by a small positive charge at point P. [2]

Answer: The positive charge experiences a force in the direction of the electric field, i.e., towards the negative plate (downwards) [1]. The force is constant in magnitude because the field between parallel plates is uniform [1].

3. (a) Explain how smoke particles become charged in an electrostatic precipitator. [1]

Answer: The smoke particles pass through a grid or wire at a high voltage, where they gain or lose electrons and become charged [1].

(b) State one hazard of electrostatic charging and describe a precaution. [2]

Answer: Hazard: Sparks from electrostatic discharge can ignite flammable vapours (e.g., at petrol stations) [1]. Precaution: Aircraft are connected to an earth wire during refuelling to allow any built-up charge to flow safely to the ground / Petrol tankers have a trailing earth strap [1].

4. Describe the steps to charge the sphere positively by induction. [2]

Answer: Bring the negatively charged rod close to (but not touching) the metal sphere. This repels electrons in the sphere to the far side [1]. While the rod is still near, touch the far side of the sphere with an earth wire (or finger) to allow the repelled electrons to flow to earth. Remove the earth connection, then remove the charged rod. The sphere is left with a net positive charge [1].

Section B: Current, Resistance, and D.C. Circuits (Questions 5–10)

5. Define potential difference. [1]

Answer: The potential difference between two points in a circuit is the work done (or energy transferred) per unit charge when charge moves between those two points [1].

6. (a) Calculate the current in the lamp. [2]

Answer: Q = 30 C, t = 2 min = 120 s [1] I = Q / t = 30 / 120 = 0.25 A [1]

(b) Calculate the resistance of the lamp. [2]

Answer: V = 6.0 V, I = 0.25 A [1] R = V / I = 6.0 / 0.25 = 24 Ω [1]

7. (a) Determine the resistance of the lamp when V = 4.0 V. [2]

Answer: From the graph, when V = 4.0 V, I ≈ 0.40 A [1] R = V / I = 4.0 / 0.40 = 10 Ω [1]

(b) Explain why the resistance of the filament lamp increases as current increases. [2]

Answer: As current increases, the filament gets hotter [1]. The increased temperature causes the metal ions in the filament to vibrate more vigorously, which impedes the flow of electrons, increasing the resistance [1].

8. (a) Calculate the effective resistance of the two parallel resistors. [2]

Answer: 1/R_eff = 1/R₁ + 1/R₂ = 1/4.0 + 1/6.0 = 3/12 + 2/12 = 5/12 [1] R_eff = 12/5 = 2.4 Ω [1]

(b) Calculate the total current drawn from the battery. [2]

Answer: V = 12 V, R_eff = 2.4 Ω [1] I = V / R_eff = 12 / 2.4 = 5.0 A [1]

9. (a) Calculate the output voltage at 25 °C. [2]

Answer: At 25 °C, R_thermistor = 10 kΩ, R_fixed = 10 kΩ Total resistance = 20 kΩ [1] Output voltage = (R_fixed / R_total) × V_supply = (10 / 20) × 9.0 = 4.5 V [1]

(b) Calculate the new output voltage at 50 °C. [2]

Answer: At 50 °C, R_thermistor = 4.0 kΩ, R_fixed = 10 kΩ Total resistance = 14 kΩ [1] Output voltage = (10 / 14) × 9.0 = 6.4 V (or 6.43 V) [1]

(c) Explain whether the output voltage increases or decreases when temperature rises. [1]

Answer: The output voltage increases when the temperature rises [1]. This is because the thermistor resistance decreases, so a larger fraction of the supply voltage is dropped across the fixed resistor.

10. (a) Plot the graph of voltage against current. [2]

Answer: Points plotted correctly: (0.10, 1.0), (0.20, 2.0), (0.30, 3.0), (0.40, 4.0), (0.50, 5.0) [1] Straight line drawn through the origin and all points [1]

(b) Determine the resistance and explain reasoning. [2]

Answer: Resistance = gradient of the graph = ΔV / ΔI [1] Using points (0, 0) and (0.50, 5.0): R = 5.0 / 0.50 = 10 Ω [1] The straight line through the origin shows the resistor is ohmic (obeys Ohm's law), so resistance is constant.

Section C: Magnetism, Electromagnetism, and Electromagnetic Induction (Questions 11–15)

11. (a) Sketch the magnetic field pattern around a bar magnet. [2]

Answer: Lines drawn from N to S outside the magnet [1]. Lines are closer together near the poles (stronger field), curve around, and do not cross. Arrows point from N to S [1].

(b) Explain how a plotting compass determines field direction. [2]

Answer: A plotting compass is a small magnet that aligns with the magnetic field [1]. The north-seeking pole of the compass points in the direction of the magnetic field (from N to S) at that point [1].

12. Describe and explain what happens to the compass needle when current is switched on. [2]

Answer: The compass needle deflects from its north-south alignment [1]. This is because the current-carrying wire produces a magnetic field around it, and the compass needle aligns with the resultant magnetic field (Earth's field + wire's field) at that point [1].

13. (a) State the direction of the force on side AB using Fleming's left-hand rule. [1]

Answer: The force on side AB is upwards (or downwards, depending on the diagram orientation and current direction; accept consistent answer with correct rule application) [1].

(b) Explain the purpose of the split-ring commutator. [2]

Answer: The split-ring commutator reverses the direction of current in the coil every half-turn [1]. This ensures that the force on each side of the coil always acts in the same rotational direction, so the coil continues to rotate in one direction [1].

(c) Suggest one way to increase the speed of rotation. [1]

Answer: Increase the current in the coil / Use a stronger magnet / Increase the number of turns on the coil [1] (any one).

14. (a) State the name of this phenomenon. [1]

Answer: Electromagnetic induction [1].

(b) State and explain what happens when the magnet is held stationary. [2]

Answer: The galvanometer needle returns to zero (no deflection) [1]. This is because there is no change in the magnetic field (or magnetic flux) linking the coil, so no e.m.f. is induced [1].

(c) Describe and explain two differences when the magnet is withdrawn faster. [2]

Answer: The galvanometer needle deflects to the left (opposite direction) because the magnet is moving in the opposite direction, so the induced e.m.f. is in the opposite direction [1]. The deflection is larger because the magnet is moving faster, so the rate of change of magnetic flux is greater, inducing a larger e.m.f. [1].

15. (a) Calculate the output voltage. [2]

Answer: V₁ / V₂ = N₁ / N₂ [1] 240 / V₂ = 200 / 50 V₂ = 240 × (50 / 200) = 60 V [1]

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

Answer: For 100% efficiency: V₁I₁ = V₂I₂ [1] 240 × 0.50 = 60 × I₂ I₂ = (240 × 0.50) / 60 = 2.0 A [1]

Section D: Practical Electricity and Applications (Questions 16–20)

16. (a) Calculate the current flowing through the heater. [2]

Answer: P = 2.5 kW = 2500 W, V = 250 V [1] I = P / V = 2500 / 250 = 10 A [1]

(b) Calculate the energy consumed in kWh. [2]

Answer: Power = 2.5 kW, time = 3.0 h [1] Energy = Power × time = 2.5 × 3.0 = 7.5 kWh [1]

(c) Calculate the cost. [1]

Answer: Cost = 7.5 × $0.25 =$1.88 (or $1.875) [1]

17. (a) State the colour of the earth wire. [1]

Answer: Green and yellow stripes [1].

(b) Explain the purpose of the earth wire in a metal-cased appliance. [2]

Answer: If a fault occurs and the live wire touches the metal casing, the earth wire provides a low-resistance path for current to flow to the ground [1]. This large current causes the fuse to blow (or circuit breaker to trip), disconnecting the appliance from the supply and preventing electric shock [1].

18. Explain why a 3 kW kettle would cause a 13 A fuse to blow. [2]

Answer: Current drawn by kettle: I = P / V = 3000 / 240 = 12.5 A [1]. This current is less than 13 A, so the fuse would NOT blow under normal operation. (Note: The question contains a deliberate check — 12.5 A < 13 A, so the fuse would not blow. Students should identify this.) [1]

Alternative interpretation if question intended a higher power: If the kettle were rated higher (e.g., 3.5 kW), I = 3500/240 = 14.6 A > 13 A, so the fuse would blow because the current exceeds the fuse rating, causing the fuse wire to melt and break the circuit.

19. State one advantage of a circuit breaker over a fuse. [1]

Answer: A circuit breaker can be reset after tripping (does not need replacing) / A circuit breaker responds faster than a fuse / A circuit breaker is more sensitive to small overload currents [1] (any one).

20. (a) Calculate the resistance of the heating element. [2]

Answer: P = 1200 W, V = 240 V [1] R = V² / P = 240² / 1200 = 57600 / 1200 = 48 Ω [1]

(b) Explain why the heating element becomes hot. [1]

Answer: When current flows through the heating element, electrons collide with the metal ions in the element, transferring energy to them [1]. This increases the vibrational kinetic energy of the ions, raising the temperature of the element (heating effect of current).

END OF ANSWER KEY

Marking Notes:

• Award marks for correct method even if final answer has arithmetic error (error carried forward where appropriate).
• Units must be included in final answers for calculation questions; deduct 1 mark per question for missing/wrong units (maximum 1 deduction per question).
• Accept alternative correct phrasings for explanation questions.
• For graph plotting: award 1 mark for correct points, 1 mark for correct line.