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O Level Physics Practice Paper 2

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TuitionGoWhere Practice Paper - Physics O-Level

TuitionGoWhere Practice Paper (AI)

Subject: Physics Level: O-Level (6091) Paper: Practice Paper – Electricity & Magnetism Version: 2 of 5 Duration: 1 hour 45 minutes Total Marks: 80

Name: _________________________ Class: _________________________ Date: _________________________

Instructions to Candidates

This paper consists of three sections: Section A, Section B, and Section C.
Answer all questions in Section A and Section B.
In Section C, answer any two of the three questions.
Write your answers in the spaces provided.
Show all working clearly; marks are awarded for correct method.
State units in all final numerical answers.
The number of marks is given in brackets [ ] at the end of each question or part question.
You may use a scientific calculator.
Take g = 10 N/kg and g = 10 m/s² where required.

Section A: Structured Questions (30 marks)

Answer ALL questions in this section. Write your answers in the spaces provided.

1. A student rubs a polythene rod with a woollen cloth. The rod becomes negatively charged.

(a) Explain, in terms of electron movement, why the polythene rod becomes negatively charged. [2]

(b) The negatively charged rod is brought near a small piece of uncharged aluminium foil. The foil is attracted to the rod. Explain why this attraction occurs. [2]

(c) State one hazard of electrostatic charging and describe a situation where this hazard may occur. [2]

2. A circuit contains a 12 V battery connected in series with a fixed resistor of resistance 6.0 Ω and a variable resistor. The variable resistor is set to 4.0 Ω.

(a) Calculate the total resistance of the circuit. [1]

(b) Calculate the current flowing in the circuit. [2]

(c) Calculate the potential difference across the 6.0 Ω resistor. [2]

(d) The variable resistor is now adjusted so that its resistance increases. State and explain what happens to the potential difference across the fixed 6.0 Ω resistor. [2]

3. A student investigates the I-V characteristic of a filament lamp. The results are shown in the table below.

Potential difference / V	0.0	1.0	2.0	3.0	4.0	5.0	6.0
Current / A	0.00	0.12	0.18	0.22	0.25	0.27	0.29

(a) On the grid below, plot a graph of current (y-axis) against potential difference (x-axis). Draw a smooth curve through the points. [3]

[Grid space for graph – 6 cm × 6 cm]

(b) Use your graph to determine the resistance of the filament lamp when the potential difference across it is 3.0 V. [2]

(c) Explain why the resistance of the filament lamp changes as the potential difference increases. [2]

4. A bar magnet is pushed into a coil of wire connected to a sensitive centre-zero galvanometer. The galvanometer needle deflects to the right.

(a) Explain why the galvanometer needle deflects when the magnet moves into the coil. [2]

(b) State two ways in which the size of the deflection can be increased. [2]

(c) The magnet is now pulled out of the coil at the same speed. State the direction of the galvanometer deflection and explain your answer. [2]

Section B: Data-Based and Application Questions (20 marks)

Answer ALL questions in this section. Write your answers in the spaces provided.

5. A student designs a potential divider circuit using a thermistor and a fixed resistor to control a cooling fan. The circuit is shown below.

[Circuit diagram: 9 V battery connected across a thermistor (top) and a 2.0 kΩ fixed resistor (bottom) in series. A voltmeter is connected across the fixed resistor. The output voltage across the fixed resistor is connected to a switch that turns on a fan when the voltage exceeds 4.5 V.]

The thermistor has the following resistance-temperature characteristic:

Temperature / °C	10	20	30	40	50	60
Resistance / kΩ	8.0	5.0	3.0	2.0	1.4	1.0

(a) Calculate the output voltage across the fixed resistor when the temperature is 20 °C. [3]

(b) Determine the temperature at which the fan will first switch on. [3]

(c) The student wants the fan to switch on at a lower temperature. Suggest and explain one modification that could be made to the circuit to achieve this. [2]

6. An electric kettle is rated at 240 V, 2000 W.

(a) Calculate the current drawn by the kettle when operating at its rated voltage. [2]

(b) Calculate the resistance of the heating element in the kettle. [2]

(c) The kettle is used to heat 1.5 kg of water from 25 °C to 100 °C. The specific heat capacity of water is 4200 J/(kg °C).

(i) Calculate the energy required to heat the water. [2]

(ii) The kettle has an efficiency of 85%. Calculate the time taken to heat the water. [3]

(d) The kettle is fitted with a 13 A fuse in its plug. Explain why a 3 A fuse would not be suitable for this kettle. [3]

Section C: Free Response Questions (30 marks)

Answer any TWO questions from this section. Write your answers on separate paper.

Each question carries 15 marks.

7. (a) State Faraday's law of electromagnetic induction. [2]

(b) A simple a.c. generator consists of a rectangular coil rotating in a uniform magnetic field. The coil is connected to an external circuit via slip rings and carbon brushes.

(i) Explain why an e.m.f. is induced in the coil as it rotates. [3]

(ii) Sketch a graph showing how the induced e.m.f. varies with time for one complete rotation of the coil. Label the axes clearly. [3]

(iii) Explain why the induced e.m.f. is zero at two positions during each rotation. [2]

(c) A 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. mains supply.

(i) Calculate the output voltage of the transformer. [2]

(ii) Explain why the transformer will not work if the primary coil is connected to a 240 V d.c. supply. [3]

8. (a) A straight wire of length 0.30 m carries a current of 4.0 A. The wire is placed perpendicular to a uniform magnetic field of flux density 0.50 T.

(i) Calculate the force acting on the wire. [2]

(ii) State the rule used to determine the direction of the force on the wire. [1]

(b) A d.c. motor is used to lift a load. The motor consists of a rectangular coil placed in a magnetic field, with current supplied through a split-ring commutator and carbon brushes.

(i) Explain the purpose of the split-ring commutator in a d.c. motor. [3]

(ii) Explain why the coil experiences a turning effect when current flows through it. [3]

(iii) Suggest two ways in which the speed of rotation of the motor can be increased. [2]

(c) An electromagnet is made by winding insulated copper wire around a soft iron core. When a current flows through the coil, the electromagnet can pick up iron objects.

(i) Explain why a soft iron core is used rather than a steel core. [2]

(ii) State one practical application of an electromagnet and explain how it works. [2]

9. (a) A student investigates the magnetic field pattern around a long straight wire carrying a current. Describe how the student can determine the direction of the magnetic field around the wire. [3]

(b) Two long straight parallel wires are placed 0.050 m apart. Wire P carries a current of 3.0 A and wire Q carries a current of 5.0 A in the same direction.

(i) State whether the force between the wires is attractive or repulsive. [1]

(ii) Explain your answer to (b)(i) in terms of the magnetic fields around the wires. [3]

(c) A solenoid of length 0.40 m has 800 turns of wire and carries a current of 2.5 A.

(i) Describe the magnetic field pattern inside and around the solenoid. [3]

(ii) State two ways in which the strength of the magnetic field inside the solenoid can be increased. [2]

(iii) The solenoid is used to magnetise a steel bar placed inside it. Explain why the steel bar becomes a permanent magnet while a soft iron bar placed in the same solenoid would only be a temporary magnet. [3]

END OF PAPER

Copyright Notice: This practice paper is generated by TuitionGoWhere AI for educational purposes. It is not derived from any specific past examination paper.

Answers

TuitionGoWhere Practice Paper - Physics O-Level

Answer Key and Marking Scheme

Subject: Physics Level: O-Level (6091) Paper: Practice Paper – Electricity & Magnetism Version: 2 of 5 Total Marks: 80

Section A: Structured Questions (30 marks)

Question 1: Static Electricity [6 marks]

(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, giving it an overall negative charge [1].

Marking notes:

Award [1] for stating electrons are transferred from cloth to rod.
Award [1] for linking gain of electrons to negative charge.
Accept: "electrons move from wool to polythene" or equivalent.

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

Answer: The negative charges on the rod repel the free electrons in the aluminium foil to the far side of the foil [1]. The side of the foil nearest the rod becomes positively charged (induced charge). Since unlike charges attract, the foil is attracted to the rod [1].

Marking notes:

Award [1] for describing charge separation/induction in the foil.
Award [1] for stating attraction between unlike charges.
Accept: reference to "electrostatic induction" with correct explanation.

(c) State one hazard of electrostatic charging and describe a situation where this hazard may occur. [2]

Answer: Hazard: Sparks produced by electrostatic discharge can ignite flammable vapours or gases, causing fires or explosions [1]. Situation: This can occur when refuelling an aircraft or a car at a petrol station, where fuel vapour may be present and a spark from static buildup on the vehicle or fuel nozzle could cause ignition [1].

Marking notes:

Award [1] for identifying a valid hazard (sparks causing fire/explosion, electric shock, damage to electronic components).
Award [1] for describing a plausible situation.
Accept any valid hazard and situation combination.

Question 2: D.C. Circuits [7 marks]

(a) Calculate the total resistance of the circuit. [1]

Answer: R_total = 6.0 + 4.0 = 10.0 Ω [1]

Marking notes:

Award [1] for correct answer with unit.
Accept 10 Ω.

(b) Calculate the current flowing in the circuit. [2]

Answer: I = V / R_total = 12 / 10.0 = 1.2 A [1] for formula/substitution, [1] for correct answer with unit.

Marking notes:

Award [1] for correct formula or substitution.
Award [1] for correct numerical answer with unit (1.2 A).

(c) Calculate the potential difference across the 6.0 Ω resistor. [2]

Answer: V = I × R = 1.2 × 6.0 = 7.2 V [1] for formula/substitution, [1] for correct answer with unit.

Marking notes:

Award [1] for correct formula or substitution.
Award [1] for correct numerical answer with unit (7.2 V).

(d) State and explain what happens to the potential difference across the fixed 6.0 Ω resistor when the variable resistor's resistance increases. [2]

Answer: The potential difference across the 6.0 Ω resistor decreases [1]. When the variable resistor's resistance increases, the total resistance increases, so the current in the circuit decreases. Since V = IR, and I decreases while R (6.0 Ω) is constant, the p.d. across the fixed resistor decreases [1].

Marking notes:

Award [1] for stating p.d. decreases.
Award [1] for correct explanation linking increased total resistance → decreased current → decreased p.d. across fixed resistor.
Accept alternative explanation using potential divider principle.

Question 3: I-V Characteristics [7 marks]

(a) Plot a graph of current against potential difference. [3]

Answer: Graph should show:

Correctly labelled axes: Current / A (y-axis), Potential difference / V (x-axis) [1]
All 7 points plotted accurately (± half small square) [1]
Smooth curve drawn through points showing increasing gradient (resistance decreasing) or curve that is not a straight line through origin [1]

Marking notes:

Award [1] for correct axes labels with units.
Award [1] for accurate plotting of all points.
Award [1] for smooth best-fit curve (not dot-to-dot straight lines).
The curve should show that current increases at a decreasing rate as p.d. increases (non-ohmic behaviour).

(b) Determine the resistance of the filament lamp when the p.d. is 3.0 V. [2]

Answer: From graph, at V = 3.0 V, I = 0.22 A [1] R = V / I = 3.0 / 0.22 = 13.6 Ω (accept 13 to 14 Ω depending on graph reading) [1]

Marking notes:

Award [1] for reading correct current value from graph (± 0.01 A).
Award [1] for correct calculation of resistance with unit.
Allow error carried forward from graph reading.

(c) Explain why the resistance of the filament lamp changes as the p.d. increases. [2]

Answer: As the p.d. increases, the current increases, causing the filament temperature to increase [1]. The increased temperature causes increased vibration of the metal atoms in the filament, which impedes the flow of electrons, resulting in higher resistance [1].

Marking notes:

Award [1] for linking increased p.d./current to increased temperature.
Award [1] for explaining that higher temperature causes greater resistance (more atomic vibrations impede electron flow).
Accept: reference to positive temperature coefficient of resistance for metals.

Question 4: Electromagnetic Induction [6 marks]

(a) Explain why the galvanometer needle deflects when the magnet moves into the coil. [2]

Answer: When the magnet moves into the coil, the magnetic field lines passing through the coil change (the magnetic flux through the coil changes) [1]. According to Faraday's law, a changing magnetic field/flux induces an e.m.f. in the coil, which drives a current through the galvanometer, causing the needle to deflect [1].

Marking notes:

Award [1] for stating that the magnetic field/flux through the coil changes.
Award [1] for linking changing flux to induced e.m.f. and current.

(b) State two ways in which the size of the deflection can be increased. [2]

Answer: Any two from:

Move the magnet faster [1]
Use a stronger magnet [1]
Use a coil with more turns [1]
Use a coil with a soft iron core [1]

Marking notes:

Award [1] for each valid method (maximum 2 marks).
Accept any valid method that increases the rate of change of magnetic flux.

(c) State the direction of the galvanometer deflection when the magnet is pulled out and explain. [2]

Answer: The galvanometer needle deflects to the left (opposite direction) [1]. When the magnet is pulled out, the magnetic flux through the coil decreases. According to Lenz's law, the induced current flows in a direction that opposes the change causing it. Since the flux is decreasing (magnet moving away), the induced current creates a magnetic field that attracts the magnet back, so the current direction is opposite to that in part (a), causing deflection in the opposite direction [1].

Marking notes:

Award [1] for stating deflection is in the opposite direction.
Award [1] for correct explanation using Lenz's law or describing that decreasing flux induces current in opposite direction.
Accept: "deflects to the left" or "deflects in the opposite direction".

Section B: Data-Based and Application Questions (20 marks)

Question 5: Potential Divider with Thermistor [8 marks]

(a) Calculate the output voltage across the fixed resistor when the temperature is 20 °C. [3]

Answer: At 20 °C, thermistor resistance R_T = 5.0 kΩ [1] Total resistance = 5.0 + 2.0 = 7.0 kΩ [1] Output voltage V_out = (R_fixed / R_total) × V_supply = (2.0 / 7.0) × 9.0 = 2.57 V (accept 2.6 V) [1]

Marking notes:

Award [1] for reading correct thermistor resistance from table.
Award [1] for correct potential divider formula or method.
Award [1] for correct numerical answer with unit.

(b) Determine the temperature at which the fan will first switch on. [3]

Answer: Fan switches on when V_out = 4.5 V [1] 4.5 = (2.0 / (R_T + 2.0)) × 9.0 0.5 = 2.0 / (R_T + 2.0) R_T + 2.0 = 4.0 R_T = 2.0 kΩ [1] From table, R_T = 2.0 kΩ corresponds to 40 °C [1]

Marking notes:

Award [1] for setting up equation with V_out = 4.5 V.
Award [1] for solving to find R_T = 2.0 kΩ.
Award [1] for reading correct temperature from table (40 °C).

(c) Suggest and explain one modification to make the fan switch on at a lower temperature. [2]

Answer: Replace the 2.0 kΩ fixed resistor with a resistor of lower resistance [1]. At a lower temperature, the thermistor resistance is higher. With a smaller fixed resistor, the fraction (R_fixed / R_total) is smaller, so a lower thermistor resistance (corresponding to a higher temperature) would be needed to achieve V_out = 4.5 V. Wait — to switch on at a lower temperature, we need V_out to reach 4.5 V when R_T is larger (at lower temperature). Using a larger fixed resistor would increase the fraction (R_fixed / R_total) for a given R_T, so V_out = 4.5 V would be reached at a higher R_T (lower temperature). Therefore, use a larger fixed resistor [1 for correct suggestion, 1 for explanation].

Alternative answer: Increase the supply voltage [1]. With a higher supply voltage, the output voltage across the fixed resistor will be higher for the same thermistor resistance, so V_out = 4.5 V will be reached at a higher thermistor resistance (lower temperature) [1].

Marking notes:

Award [1] for a valid modification (increase fixed resistor value, increase supply voltage, use a different thermistor).
Award [1] for correct explanation linking modification to switching at lower temperature.
Accept any valid modification with correct reasoning.

Question 6: Practical Electricity [12 marks]

(a) Calculate the current drawn by the kettle. [2]

Answer: P = V × I I = P / V = 2000 / 240 = 8.33 A (accept 8.3 A) [1] for formula/substitution, [1] for correct answer with unit.

Marking notes:

Award [1] for correct formula or substitution.
Award [1] for correct numerical answer with unit.

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

Answer: R = V / I = 240 / 8.33 = 28.8 Ω [1] for formula/substitution, [1] for correct answer with unit. OR: R = V² / P = 240² / 2000 = 57600 / 2000 = 28.8 Ω

Marking notes:

Award [1] for correct formula or substitution.
Award [1] for correct numerical answer with unit.
Accept 28.8 Ω or 29 Ω.

(c)(i) Calculate the energy required to heat the water. [2]

Answer: E = m × c × Δθ = 1.5 × 4200 × (100 - 25) [1] E = 1.5 × 4200 × 75 = 472,500 J (or 472.5 kJ) [1]

Marking notes:

Award [1] for correct formula and substitution.
Award [1] for correct numerical answer with unit.

(c)(ii) Calculate the time taken to heat the water. [3]

Answer: Efficiency = Useful energy output / Total energy input 0.85 = 472,500 / E_input E_input = 472,500 / 0.85 = 555,882 J (or ≈ 556,000 J) [1] Power = Energy / time time = E_input / P = 555,882 / 2000 [1] time = 278 s (or 4 min 38 s) [1]

Marking notes:

Award [1] for calculating total energy input using efficiency.
Award [1] for correct formula time = energy/power.
Award [1] for correct numerical answer with unit (accept 278 s, 4.6 min, or 4 min 38 s).

(d) Explain why a 3 A fuse would not be suitable for this kettle. [3]

Answer: The kettle draws a current of 8.33 A during normal operation [1]. A 3 A fuse is rated to blow when current exceeds 3 A [1]. Since the normal operating current (8.33 A) is much greater than the fuse rating (3 A), the fuse would blow immediately when the kettle is switched on, preventing the kettle from operating [1].

Marking notes:

Award [1] for stating the operating current (8.33 A or ~8 A).
Award [1] for explaining that a 3 A fuse blows at currents above 3 A.
Award [1] for concluding that the fuse would blow during normal operation.
Accept: "The fuse rating must be slightly higher than the normal operating current. A 13 A fuse is appropriate; a 3 A fuse is too low."

Section C: Free Response Questions (30 marks)

Answer any TWO questions. Each question carries 15 marks.

Question 7: Electromagnetic Induction and Transformers [15 marks]

(a) State Faraday's law of electromagnetic induction. [2]

Answer: The magnitude of the induced e.m.f. in a circuit is directly proportional to the rate of change of magnetic flux linking (or passing through) the circuit [2].

Marking notes:

Award [2] for complete statement including "rate of change of magnetic flux".
Award [1] if statement mentions changing magnetic field but omits "rate of change" or "flux".
Accept: "An e.m.f. is induced when there is a change in the magnetic flux through a circuit."

(b)(i) Explain why an e.m.f. is induced in the coil as it rotates. [3]

Answer: As the coil rotates, the angle between the plane of the coil and the magnetic field lines changes continuously [1]. This means the magnetic flux passing through the coil changes with time [1]. According to Faraday's law, a changing magnetic flux induces an e.m.f. in the coil [1].

Marking notes:

Award [1] for stating that the coil's orientation relative to the field changes.
Award [1] for linking this to changing magnetic flux through the coil.
Award [1] for stating that changing flux induces e.m.f. (Faraday's law).

(b)(ii) Sketch a graph of induced e.m.f. against time for one complete rotation. [3]

Answer: Graph should show:

x-axis labelled "Time" or "t", y-axis labelled "Induced e.m.f." or "E" [1]
A sinusoidal curve (sine wave) with one complete cycle [1]
Curve crossing zero at start, middle, and end; positive and negative peaks of equal magnitude [1]

Marking notes:

Award [1] for correctly labelled axes.
Award [1] for sinusoidal shape (not triangular or square).
Award [1] for correct zero-crossing positions and equal positive/negative peaks.
The graph should show that e.m.f. is zero at 0°, 180°, and 360° positions, with maximum e.m.f. at 90° and 270°.

(b)(iii) Explain why the induced e.m.f. is zero at two positions during each rotation. [2]

Answer: At two positions during each rotation, the plane of the coil is perpendicular to the magnetic field (the coil is in the vertical position if the field is horizontal) [1]. At these positions, the sides of the coil are moving parallel to the magnetic field lines, so they do not cut across magnetic field lines. The rate of change of magnetic flux through the coil is momentarily zero, so the induced e.m.f. is zero [1].

Marking notes:

Award [1] for identifying the positions (coil perpendicular to field, or coil sides moving parallel to field).
Award [1] for explaining that no flux cutting occurs or rate of change of flux is zero.

(c)(i) Calculate the output voltage of the transformer. [2]

Answer: V_s / V_p = N_s / N_p V_s / 240 = 50 / 500 [1] V_s = 240 × (50/500) = 240 × 0.1 = 24 V [1]

Marking notes:

Award [1] for correct formula or substitution.
Award [1] for correct numerical answer with unit (24 V).

(c)(ii) Explain why the transformer will not work if connected to a 240 V d.c. supply. [3]

Answer: A transformer works on the principle of electromagnetic induction, which requires a changing magnetic flux [1]. A d.c. supply produces a constant current in the primary coil, which creates a constant (unchanging) magnetic field [1]. Since there is no change in magnetic flux through the secondary coil, no e.m.f. is induced in the secondary coil [1].

Marking notes:

Award [1] for stating that electromagnetic induction requires changing magnetic flux.
Award [1] for explaining that d.c. produces a constant magnetic field.
Award [1] for concluding that no e.m.f. is induced in the secondary coil.

Question 8: Force on Current-Carrying Conductor and D.C. Motor [15 marks]

(a)(i) Calculate the force acting on the wire. [2]

Answer: F = B × I × L F = 0.50 × 4.0 × 0.30 [1] F = 0.60 N [1]

Marking notes:

Award [1] for correct formula and substitution.
Award [1] for correct numerical answer with unit (0.60 N).

(a)(ii) State the rule used to determine the direction of the force. [1]

Answer: Fleming's left-hand rule [1].

Marking notes:

Award [1] for "Fleming's left-hand rule".
Accept: "left-hand motor rule".

(b)(i) Explain the purpose of the split-ring commutator in a d.c. motor. [3]

Answer: The split-ring commutator reverses the direction of current in the coil every half rotation [1]. This ensures that the direction of the force on each side of the coil is reversed each time the coil passes the vertical position [1]. As a result, the coil continues to rotate in the same direction, producing continuous rotation rather than oscillating back and forth [1].

Marking notes:

Award [1] for stating that the commutator reverses current direction every half turn.
Award [1] for linking this to reversing the force direction.
Award [1] for explaining that this produces continuous rotation in one direction.

(b)(ii) Explain why the coil experiences a turning effect. [3]

Answer: When current flows through the coil, each side of the coil that is perpendicular to the magnetic field experiences a force, as given by Fleming's left-hand rule [1]. The current flows in opposite directions in the two sides of the coil, so the forces on the two sides are in opposite directions [1]. These two equal and opposite forces, acting at different points on the coil, form a couple, which produces a turning effect (torque) that rotates the coil [1].

Marking notes:

Award [1] for stating that forces act on current-carrying sides in the magnetic field.
Award [1] for explaining that forces on opposite sides are in opposite directions.
Award [1] for explaining that this pair of forces forms a couple producing rotation.

(b)(iii) Suggest two ways to increase the speed of rotation of the motor. [2]

Answer: Any two from:

Increase the current flowing through the coil [1]
Use a stronger magnetic field (stronger magnets) [1]
Use a coil with more turns [1]
Increase the area of the coil [1]

Marking notes:

Award [1] for each valid method (maximum 2 marks).
Accept any method that increases the force or torque on the coil.

(c)(i) Explain why a soft iron core is used rather than a steel core. [2]

Answer: Soft iron is a magnetically soft material, meaning it is easily magnetised and easily demagnetised [1]. When the current is switched off, the soft iron core loses most of its magnetism quickly, so the electromagnet can be turned on and off. Steel is magnetically hard; it retains magnetism after the current is switched off, so the electromagnet would remain magnetised even when not wanted [1].

Marking notes:

Award [1] for stating that soft iron is easily magnetised and demagnetised.
Award [1] for explaining that steel would retain magnetism (not suitable for an on/off electromagnet).

(c)(ii) State one practical application of an electromagnet and explain how it works. [2]

Answer: Application: Electric bell [1]. How it works: When the bell push is pressed, current flows through the electromagnet, which attracts a soft iron armature. The armature moves, striking the bell and simultaneously breaking the circuit. The electromagnet demagnetises, the armature springs back, remaking the circuit, and the process repeats, producing a continuous ringing sound [1].

Marking notes:

Award [1] for naming a valid application (electric bell, relay, circuit breaker, magnetic crane, loudspeaker).
Award [1] for a brief but correct explanation of how the electromagnet functions in that application.
Accept any valid application with correct explanation.

Question 9: Magnetism and Electromagnetism [15 marks]

(a) Describe how to determine the direction of the magnetic field around a long straight current-carrying wire. [3]

Answer: Place a small plotting compass near the wire [1]. The compass needle will align with the magnetic field at that point. Mark the position of the compass and the direction of the needle. Move the compass so that its south pole is at the previous position of the north pole, and repeat to trace a field line [1]. The direction of the magnetic field is from the north-seeking pole to the south-seeking pole of the compass. Alternatively, use the right-hand grip rule: if the thumb points in the direction of conventional current, the curled fingers show the direction of the magnetic field (concentric circles around the wire) [1].

Marking notes:

Award [1] for describing use of a plotting compass.
Award [1] for describing how to trace the field pattern.
Award [1] for stating how to determine direction (compass N pole points in field direction, or right-hand grip rule).

(b)(i) State whether the force between the wires is attractive or repulsive. [1]

Answer: Attractive [1].

Marking notes:

Award [1] for "attractive".
Parallel currents in the same direction attract.

(b)(ii) Explain your answer in terms of the magnetic fields around the wires. [3]

Answer: Each current-carrying wire produces a magnetic field around it [1]. Wire P produces a magnetic field at the location of wire Q, and wire Q (carrying current) experiences a force in this magnetic field. Similarly, wire Q produces a field at wire P [1]. Using the right-hand grip rule and Fleming's left-hand rule, the direction of the force on each wire is towards the other wire. The magnetic field lines between the wires are in opposite directions and cancel, while the field lines on the outer sides reinforce, pushing the wires together [1].

Marking notes:

Award [1] for stating that each wire produces a magnetic field.
Award [1] for explaining that each wire experiences a force due to the other's field.
Award [1] for explaining why the force is attractive (field cancellation between wires, or correct application of rules).
Accept: clear explanation using right-hand grip rule and Fleming's left-hand rule to show attraction.

(c)(i) Describe the magnetic field pattern inside and around the solenoid. [3]

Answer: Inside the solenoid: The magnetic field lines are parallel, evenly spaced, and straight, indicating a strong, uniform magnetic field [1]. The field direction inside is from the south pole to the north pole of the solenoid [1]. Outside the solenoid: The magnetic field pattern is similar to that of a bar magnet, with field lines emerging from the north pole, curving around, and entering the south pole. The field is weaker and non-uniform outside [1].

Marking notes:

Award [1] for describing the uniform field inside (parallel, evenly spaced lines).
Award [1] for stating field direction inside (S to N).
Award [1] for describing the external field pattern (similar to bar magnet).

(c)(ii) State two ways to increase the strength of the magnetic field inside the solenoid. [2]

Answer: Any two from:

Increase the current flowing through the solenoid [1]
Increase the number of turns of wire (per unit length) [1]
Insert a soft iron core inside the solenoid [1]

Marking notes:

Award [1] for each valid method (maximum 2 marks).

(c)(iii) Explain why the steel bar becomes a permanent magnet while the soft iron bar would only be a temporary magnet. [3]

Answer: Steel is a magnetically hard material, meaning it is difficult to magnetise but retains its magnetism well once magnetised [1]. When placed in the solenoid's magnetic field, the magnetic domains in the steel align with the field, and they remain aligned even after the current is switched off, making it a permanent magnet [1]. Soft iron is a magnetically soft material; its magnetic domains align easily in the field but return to random orientations when the field is removed, so it loses its magnetism quickly and is only a temporary magnet [1].

Marking notes:

Award [1] for stating that steel is magnetically hard (retains magnetism).
Award [1] for explaining domain alignment in steel remains after field removal.
Award [1] for explaining that soft iron domains randomise when field is removed (magnetically soft).
Accept: reference to "retentivity" or "coercivity" at O-Level standard.

END OF ANSWER KEY

Marking Scheme Summary:

Section	Questions	Marks
A	1–4	30
B	5–6	20
C	7–9 (answer 2 of 3)	30
Total		80

Grade Boundaries (indicative):

A1: 70–80
A2: 60–69
B3: 50–59
B4: 45–49
C5: 40–44
C6: 35–39
D7: 30–34
E8: 25–29
F9: Below 25

Copyright Notice: This answer key is generated by TuitionGoWhere AI for educational purposes. It is not derived from any specific past examination paper.