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Secondary 4 Pure Physics Electricity Magnetism Quiz
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Questions
Secondary 4 Pure Physics Quiz - Electricity Magnetism
Name: _________________________ Class: _________________________ Date: _________________________ Score: ______ / 50
Duration: 45 minutes Total Marks: 50
Instructions:
- This quiz contains 20 questions on Electricity and Magnetism.
- Answer ALL questions in the spaces provided.
- Show all working for calculation questions.
- Use g = 10 m/s² where applicable.
- The number of marks is shown in brackets [ ].
Section A: Static Electricity and Current of Electricity (Questions 1–5)
10 marks
1. State the SI unit of electric charge and explain what is meant by an electric field. [2 marks]
2. A positively charged rod is brought near an uncharged metal sphere mounted on an insulating stand. The sphere is then earthed briefly, and the earth connection is removed before the rod is taken away.
(a) State the final charge on the metal sphere. [1 mark]
(b) Explain the process that causes the sphere to acquire this charge. [2 marks]
3. A current of 0.50 A flows through a lamp for 3.0 minutes. Calculate the total charge that passes through the lamp. [2 marks]
4. The current in a wire is 2.0 A. Calculate the number of electrons flowing past a point in the wire in 1.0 second. (Charge on one electron = 1.6 × 10⁻¹⁹ C) [2 marks]
5. State Ohm's Law and identify one circuit component that does NOT obey Ohm's Law. [1 mark]
Section B: D.C. Circuits and Practical Electricity (Questions 6–10)
12 marks
6. Three resistors of 2.0 Ω, 3.0 Ω, and 6.0 Ω are connected in parallel across a 12 V battery.
(a) Calculate the effective resistance of the parallel combination. [2 marks]
(b) Calculate the total current drawn from the battery. [1 mark]
7. A potential divider consists of two resistors, R₁ = 4.0 kΩ and R₂ = 6.0 kΩ, connected in series across a 10 V supply. Calculate the output voltage across R₂. [2 marks]
8. An electric kettle is rated at 240 V, 2000 W.
(a) Calculate the current drawn by the kettle when operating at its rated voltage. [1 mark]
(b) Calculate the resistance of the heating element. [1 mark]
(c) The kettle is used for 5.0 minutes to heat water. Calculate the electrical energy consumed in joules. [2 marks]
9. State the colour and function of each of the following wires in a three-pin plug:
(a) Live wire [1 mark]
(b) Neutral wire [1 mark]
(c) Earth wire [1 mark]
Section C: Magnetism and Electromagnetism (Questions 11–15)
13 marks
10. Sketch the magnetic field pattern around a bar magnet. Label the poles and indicate the direction of the field lines. [3 marks]
11. A straight wire carries a current vertically upwards. Sketch the magnetic field pattern around the wire when viewed from above. Indicate the direction of the field lines. [2 marks]
12. State two ways to increase the strength of an electromagnet. [2 marks]
13. A current-carrying conductor of length 0.50 m is placed perpendicular to a uniform magnetic field of flux density 0.40 T. The current in the conductor is 3.0 A.
(a) Calculate the force acting on the conductor. [2 marks]
(b) State the direction of this force relative to the current and the magnetic field. [1 mark]
14. Explain, with the aid of a labelled diagram, how a simple D.C. motor works. Your explanation should include the role of the split-ring commutator. [3 marks]
Section D: Electromagnetic Induction and Transformers (Questions 16–20)
15 marks
15. State Faraday's Law of Electromagnetic Induction. [2 marks]
16. A bar magnet is pushed into a coil connected to a galvanometer. The galvanometer needle deflects.
(a) Explain why the needle deflects. [2 marks]
(b) State what happens to the needle when the magnet is held stationary inside the coil. [1 mark]
(c) State two ways to increase the magnitude of the induced EMF. [2 marks]
17. An ideal transformer has 500 turns on its primary coil and 50 turns on its secondary coil. The primary coil is connected to a 240 V a.c. supply.
(a) Calculate the output voltage across the secondary coil. [2 marks]
(b) State whether this is a step-up or step-down transformer. [1 mark]
18. A transformer has an efficiency of 80%. The primary voltage is 240 V and the primary current is 2.0 A. The secondary voltage is 12 V. Calculate the secondary current. [3 marks]
19. State one advantage of transmitting electrical power at high voltages over long distances. [1 mark]
20. Explain why the core of a transformer is laminated. [1 mark]
END OF QUIZ
Answers
Secondary 4 Pure Physics Quiz - Electricity Magnetism
Answer Key and Marking Scheme
Total Marks: 50
Section A: Static Electricity and Current of Electricity (Questions 1–5)
1. State the SI unit of electric charge and explain what is meant by an electric field. [2 marks]
Answer:
- SI unit of electric charge: coulomb (C) [1 mark]
- An electric field is a region of space where an electric charge experiences an electric force. [1 mark]
2. (a) State the final charge on the metal sphere. [1 mark]
Answer: Negative / Negatively charged. [1 mark]
(b) Explain the process that causes the sphere to acquire this charge. [2 marks]
Answer:
- The positively charged rod attracts electrons from the earth to the sphere when earthed. [1 mark]
- When the earth connection is removed, the excess electrons are trapped on the sphere. When the rod is removed, the sphere is left with a net negative charge. [1 mark]
3. A current of 0.50 A flows through a lamp for 3.0 minutes. Calculate the total charge that passes through the lamp. [2 marks]
Answer:
- t = 3.0 × 60 = 180 s [1 mark for conversion]
- Q = I × t = 0.50 × 180 = 90 C [1 mark for correct answer with unit]
4. The current in a wire is 2.0 A. Calculate the number of electrons flowing past a point in the wire in 1.0 second. [2 marks]
Answer:
- Q = I × t = 2.0 × 1.0 = 2.0 C [1 mark]
- Number of electrons = Q / e = 2.0 / (1.6 × 10⁻¹⁹) = 1.25 × 10¹⁹ [1 mark for correct answer]
- Accept: 1.3 × 10¹⁹
5. State Ohm's Law and identify one circuit component that does NOT obey Ohm's Law. [1 mark]
Answer:
- Ohm's Law: The current through a conductor is directly proportional to the potential difference across it, provided temperature and other physical conditions remain constant. [0.5 marks]
- Component that does not obey Ohm's Law: filament lamp / diode / thermistor / LDR (any one) [0.5 marks]
Section B: D.C. Circuits and Practical Electricity (Questions 6–10)
6. (a) Calculate the effective resistance of the parallel combination. [2 marks]
Answer:
- 1/R = 1/2.0 + 1/3.0 + 1/6.0 = 3/6.0 + 2/6.0 + 1/6.0 = 6/6.0 = 1.0 [1 mark for correct substitution]
- R = 1.0 Ω [1 mark for correct answer with unit]
(b) Calculate the total current drawn from the battery. [1 mark]
Answer:
- I = V / R = 12 / 1.0 = 12 A [1 mark for correct answer with unit]
7. A potential divider consists of two resistors, R₁ = 4.0 kΩ and R₂ = 6.0 kΩ, connected in series across a 10 V supply. Calculate the output voltage across R₂. [2 marks]
Answer:
- V_out = V_supply × [R₂ / (R₁ + R₂)] [1 mark for correct formula]
- V_out = 10 × [6.0 / (4.0 + 6.0)] = 10 × 0.60 = 6.0 V [1 mark for correct answer with unit]
8. (a) Calculate the current drawn by the kettle when operating at its rated voltage. [1 mark]
Answer:
- I = P / V = 2000 / 240 = 8.33 A (accept 8.3 A) [1 mark]
(b) Calculate the resistance of the heating element. [1 mark]
Answer:
- R = V / I = 240 / 8.33 = 28.8 Ω [1 mark]
- OR: R = V² / P = 240² / 2000 = 57,600 / 2000 = 28.8 Ω
(c) Calculate the electrical energy consumed in joules. [2 marks]
Answer:
- t = 5.0 × 60 = 300 s [1 mark for conversion]
- E = P × t = 2000 × 300 = 600,000 J (or 600 kJ) [1 mark for correct answer with unit]
9. State the colour and function of each wire:
(a) Live wire: Brown; carries the alternating current from the supply to the appliance. [1 mark]
(b) Neutral wire: Blue; provides the return path for current, completing the circuit at approximately zero potential. [1 mark]
(c) Earth wire: Green and yellow stripes; provides a low-resistance path to ground for fault currents, protecting users from electric shock. [1 mark]
Section C: Magnetism and Electromagnetism (Questions 11–15)
10. Sketch the magnetic field pattern around a bar magnet. [3 marks]
Answer:
- Field lines emerge from the North pole and enter the South pole. [1 mark]
- Lines are closer together near the poles (stronger field). [1 mark]
- Arrows indicate direction from N to S outside the magnet. [1 mark]
- Diagram should show curved lines looping from N to S, with correct labelling of poles.
11. Sketch the magnetic field pattern around a straight current-carrying wire (viewed from above). [2 marks]
Answer:
- Concentric circles around the wire. [1 mark]
- Direction indicated as anticlockwise (using right-hand grip rule for current upwards). [1 mark]
12. State two ways to increase the strength of an electromagnet. [2 marks]
Answer (any two, 1 mark each):
- Increase the current in the coil.
- Increase the number of turns in the coil.
- Insert a soft iron core.
- Use a core material with higher permeability.
13. (a) Calculate the force acting on the conductor. [2 marks]
Answer:
- F = B I L [1 mark for correct formula]
- F = 0.40 × 3.0 × 0.50 = 0.60 N [1 mark for correct answer with unit]
(b) State the direction of this force relative to the current and the magnetic field. [1 mark]
Answer:
- The force is perpendicular to both the current direction and the magnetic field direction (given by Fleming's Left-Hand Rule). [1 mark]
14. Explain how a simple D.C. motor works, including the role of the split-ring commutator. [3 marks]
Answer:
- A current-carrying coil is placed in a magnetic field. The interaction between the magnetic field of the permanent magnet and the magnetic field around the coil produces a pair of forces (Fleming's Left-Hand Rule) that create a turning effect (moment) on the coil. [1 mark]
- The split-ring commutator reverses the direction of current in the coil every half-turn. [1 mark]
- This ensures the forces continue to act in the same rotational direction, maintaining continuous rotation. [1 mark]
- (Award marks for a correctly labelled diagram showing magnet, coil, commutator, and brushes.)
Section D: Electromagnetic Induction and Transformers (Questions 16–20)
15. State Faraday's Law of Electromagnetic Induction. [2 marks]
Answer:
- The magnitude of the induced EMF in a circuit is directly proportional to the rate of change of magnetic flux linking (or cutting) the circuit. [2 marks]
- Accept: An EMF is induced when there is a change in the magnetic flux linking a conductor/coil.
16. (a) Explain why the needle deflects. [2 marks]
Answer:
- Moving the magnet changes the magnetic flux linking the coil. [1 mark]
- By Faraday's Law, this changing flux induces an EMF, which drives a current through the galvanometer, causing the needle to deflect. [1 mark]
(b) State what happens to the needle when the magnet is held stationary inside the coil. [1 mark]
Answer:
- The needle returns to zero / shows no deflection. [1 mark]
- Reason: No change in magnetic flux, so no induced EMF.
(c) State two ways to increase the magnitude of the induced EMF. [2 marks]
Answer (any two, 1 mark each):
- Move the magnet faster.
- Use a stronger magnet.
- Increase the number of turns in the coil.
- Use a coil with a larger cross-sectional area.
17. (a) Calculate the output voltage across the secondary coil. [2 marks]
Answer:
- V_s / V_p = N_s / N_p [1 mark for correct formula]
- V_s = V_p × (N_s / N_p) = 240 × (50 / 500) = 240 × 0.10 = 24 V [1 mark for correct answer with unit]
(b) State whether this is a step-up or step-down transformer. [1 mark]
Answer:
- Step-down transformer (since N_s < N_p and V_s < V_p). [1 mark]
18. A transformer has an efficiency of 80%. Calculate the secondary current. [3 marks]
Answer:
- Input power: P_in = V_p × I_p = 240 × 2.0 = 480 W [1 mark]
- Output power: P_out = η × P_in = 0.80 × 480 = 384 W [1 mark for efficiency conversion and calculation]
- I_s = P_out / V_s = 384 / 12 = 32 A [1 mark for correct answer with unit]
19. State one advantage of transmitting electrical power at high voltages over long distances. [1 mark]
Answer:
- For a given power, higher voltage means lower current. Lower current reduces I²R (resistive/heating) losses in the transmission cables. [1 mark]
- Accept: Reduces power loss / increases transmission efficiency.
20. Explain why the core of a transformer is laminated. [1 mark]
Answer:
- Lamination reduces eddy currents induced in the core. This reduces energy losses due to heating of the core, improving transformer efficiency. [1 mark]
END OF ANSWER KEY