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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: _____ / 40

Duration: 45 minutes
Total Marks: 40

Instructions:

Answer all questions.
Write your answers in the spaces provided.
Show all working for calculation questions.
Use $g = 10 \text{ N/kg}$ where needed.
The number of marks is given in brackets [ ] at the end of each question or part question.

Section A: Multiple Choice Questions [10 marks]

Answer all questions. Choose the correct option and write the letter (A, B, C, or D) in the box provided.

1. A transformer has a primary coil of 500 turns and a secondary coil of 2000 turns. The primary voltage is 240 V. What is the secondary voltage? [1]

☐ A. 60 V
☐ B. 240 V
☐ C. 480 V
☐ D. 960 V

2. Which of the following correctly describes the function of a fuse in a household circuit? [1]

☐ A. It prevents the current from exceeding a safe value by melting when the current is too high.
☐ B. It increases the resistance of the circuit to reduce the current.
☐ C. It stores electrical energy for use during power surges.
☐ D. It converts AC to DC for household appliances.

3. A current-carrying wire is placed in a uniform magnetic field as shown. The wire experiences a force directed out of the page. What is the direction of the current in the wire?

<image_placeholder> id: Q3-fig1 type: diagram linked_question: Q3 description: A straight wire placed horizontally in a uniform magnetic field directed from left to right (N to S). The force on the wire is shown as an arrow pointing out of the page (towards the viewer). The wire is perpendicular to the magnetic field. labels: Magnetic field direction (left to right), Force direction (out of page), Wire orientation values: None must_show: Wire perpendicular to magnetic field lines, force arrow out of page, magnetic field lines with N and S labels </image_placeholder>

☐ A. Left to right
☐ B. Right to left
☐ C. Into the page
☐ D. Out of the page

4. The diagram shows a simple AC generator. At the instant shown, the coil is horizontal and rotating clockwise. What is the direction of the induced current in side AB of the coil?

<image_placeholder> id: Q4-fig1 type: diagram linked_question: Q4 description: A rectangular coil ABCD rotating clockwise in a uniform magnetic field directed from left to right (N to S). The coil is shown in a horizontal position with side AB at the top and side CD at the bottom. Slip rings and brushes are shown. labels: N (left), S (right), A (top left), B (top right), C (bottom right), D (bottom left), rotation arrow (clockwise), slip rings, brushes values: None must_show: Rectangular coil in horizontal position, magnetic field left to right, clockwise rotation arrow, slip rings and brushes labelled </image_placeholder>

☐ A. A to B
☐ B. B to A
☐ C. No current induced at this instant
☐ D. Current alternates too quickly to determine

5. A household circuit has a 13 A fuse. The mains voltage is 240 V. What is the maximum power of an appliance that can be safely connected to this circuit? [1]

☐ A. 18.5 W
☐ B. 312 W
☐ C. 3120 W
☐ D. 31200 W

6. The diagram shows a plotting compass placed near a current-carrying wire. The wire is vertical and the current flows upwards. Which diagram shows the correct direction of the compass needle?

<image_placeholder> id: Q6-fig1 type: diagram linked_question: Q6 description: A vertical wire with current flowing upwards. A plotting compass is placed to the east of the wire. Four options show compass needle orientations: (A) pointing north, (B) pointing south, (C) pointing towards the wire, (D) pointing away from the wire. labels: Wire (vertical), Current direction (upwards), Compass position (east of wire), North direction values: None must_show: Vertical wire with upward current arrow, compass to the east, four needle orientations as options </image_placeholder>

☐ A. Option A
☐ B. Option B
☐ C. Option C
☐ D. Option D

7. A step-down transformer has 800 turns on the primary coil and 100 turns on the secondary coil. The primary current is 0.5 A. Assuming 100% efficiency, what is the secondary current? [1]

☐ A. 0.0625 A
☐ B. 0.5 A
☐ C. 4 A
☐ D. 8 A

8. Which of the following statements about electromagnetic induction is correct? [1]

☐ A. An induced current flows only when a magnet moves towards a coil.
☐ B. The magnitude of the induced e.m.f. is proportional to the rate of change of magnetic flux linkage.
☐ C. A stationary magnet inside a coil produces a steady induced current.
☐ D. The direction of induced current is always the same as the direction of the changing magnetic field.

9. The resistance of a 60 W filament lamp at its operating temperature is 960 Ω. What is the current through the lamp when connected to a 240 V supply? [1]

☐ A. 0.0625 A
☐ B. 0.25 A
☐ C. 4 A
☐ D. 16 A

10. A wire of length 0.5 m carries a current of 4 A at right angles to a magnetic field of flux density 0.2 T. What is the magnitude of the force on the wire? [1]

☐ A. 0.1 N
☐ B. 0.4 N
☐ C. 1.6 N
☐ D. 4 N

Section B: Structured Questions [30 marks]

Answer all questions in the spaces provided.

11. The diagram shows a simple DC motor.

<image_placeholder> id: Q11-fig1 type: diagram linked_question: Q11 description: A simple DC motor with a rectangular coil ABCD in a uniform magnetic field between two poles N and S. The coil is connected to a split-ring commutator and brushes. Current flows from the positive terminal through brush X, into the coil, and out through brush Y to the negative terminal. The coil is shown in a vertical position. labels: N (left), S (right), A (top left), B (top right), C (bottom right), D (bottom left), split-ring commutator, brushes X and Y, battery, current direction arrows values: Magnetic flux density = 0.5 T, Current = 2 A, Length of side AB = 0.1 m, Length of side BC = 0.08 m must_show: Rectangular coil in vertical position, magnetic field left to right, split-ring commutator with brushes, current direction, dimensions labelled </image_placeholder>

(a) On the diagram, draw an arrow on side AB to show the direction of the force acting on it. [1]

(b) Explain why the coil experiences a turning effect (torque). [2]

(d) State and explain what happens to the torque when the coil reaches the horizontal position. [2]

12. A student sets up an experiment to investigate electromagnetic induction. A bar magnet is dropped through a long copper tube. The magnet takes 2.5 s to fall through the tube. When a non-magnetic metal block of the same mass and size is dropped through the same tube, it takes 0.8 s.

(a) Explain why the magnet takes longer to fall through the copper tube. [3]

(b) State the direction of the induced current in the copper tube as the magnet approaches a section of the tube, as viewed from above. [1]

(d) If the copper tube is replaced with a plastic tube of the same dimensions, state and explain how the time taken for the magnet to fall would change. [2]

13. The diagram shows a transformer used to charge a mobile phone. The primary coil is connected to a 240 V AC mains supply. The secondary coil outputs 5.0 V to charge the phone. The charging current is 2.0 A. The transformer is 80% efficient.

<image_placeholder> id: Q13-fig1 type: diagram linked_question: Q13 description: A step-down transformer with primary coil connected to 240 V AC mains, secondary coil connected to a mobile phone charging circuit. Primary turns = 4800, secondary turns = unknown. labels: Primary voltage = 240 V, Secondary voltage = 5.0 V, Charging current = 2.0 A, Efficiency = 80%, Primary turns = 4800 values: Vp = 240 V, Vs = 5.0 V, Is = 2.0 A, η = 80%, Np = 4800 must_show: Transformer symbol with primary and secondary coils, input/output voltages labelled, efficiency noted </image_placeholder>

(a) Calculate the number of turns on the secondary coil. [2]

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

(c) The phone battery has a capacity of 3000 mAh. Calculate the time taken to fully charge the battery from empty, assuming the charging current remains constant at 2.0 A. [2]

(d) Explain why the transformer core is laminated. [1]

14. A household circuit has three appliances connected in parallel: a 2000 W kettle, a 1500 W iron, and a 100 W lamp. The mains voltage is 240 V. The circuit is protected by a 20 A circuit breaker.

(a) Calculate the current drawn by each appliance. [3]

(b) Calculate the total current drawn from the mains. [1]

(d) The kettle is used for 15 minutes each day. Calculate the energy consumed by the kettle in one week (7 days), giving your answer in kWh. [2]

15. The diagram shows a cathode-ray oscilloscope (CRO) trace of an AC voltage signal. The time-base is set to 5 ms/div and the Y-gain is set to 2 V/div.

<image_placeholder> id: Q15-fig1 type: graph linked_question: Q15 description: A sinusoidal wave on a CRO screen. The wave spans 4 horizontal divisions for one complete cycle. The peak-to-peak amplitude spans 3 vertical divisions. The grid has 10 horizontal divisions and 8 vertical divisions. labels: Time-base = 5 ms/div, Y-gain = 2 V/div, Horizontal divisions = 10, Vertical divisions = 8 values: Time-base = 5 ms/div, Y-gain = 2 V/div must_show: Sinusoidal wave with clear peaks and troughs, grid lines, 4 divisions per period, 3 divisions peak-to-peak </image_placeholder>

(a) Determine the period of the AC signal. [1]

(b) Determine the frequency of the AC signal. [1]

(d) Determine the root-mean-square (r.m.s.) voltage of the AC signal. [1]

(e) Sketch on the same axes what the trace would look like if the time-base is changed to 10 ms/div and the Y-gain is changed to 4 V/div. [2]

16. A straight wire of length 0.3 m carries a current of 5 A. It is placed at an angle of 30° to a uniform magnetic field of flux density 0.4 T.

(a) Calculate the magnitude of the force on the wire. [2]

(b) State the direction of the force relative to the wire and the magnetic field. [1]

(c) The wire is now bent into a semicircle of radius 0.1 m, with the same current flowing. The magnetic field remains uniform and in the same direction. Explain whether the magnitude of the total force on the wire changes. [2]

17. The diagram shows a solenoid connected to a battery and a switch. A small plotting compass is placed at point P, near the end of the solenoid.

<image_placeholder> id: Q17-fig1 type: diagram linked_question: Q17 description: A solenoid with multiple turns, connected to a battery and an open switch. A plotting compass is placed at point P near the right end of the solenoid. The solenoid axis is horizontal. labels: Battery, Switch (open), Solenoid, Point P (near right end), Compass at P values: Number of turns = 200, Length of solenoid = 0.2 m, Current when switch closed = 1.5 A must_show: Solenoid with labelled turns, battery, open switch, compass at point P near right end </image_placeholder>

(a) The switch is closed. State the polarity of the right end of the solenoid. [1]

(b) Draw the direction of the compass needle at point P when the switch is closed. [1]

(d) The switch is opened. Describe and explain what happens to the compass needle. [2]

18. A student investigates the magnetic field pattern around a current-carrying wire using iron filings.

(a) Sketch the magnetic field pattern around a long straight wire carrying a current into the page. Show at least three field lines with direction arrows. [2]

(b) The current in the wire is doubled. State the effect on the magnetic field strength at a fixed distance from the wire. [1]

(c) The student places a second wire parallel to the first, carrying the same current in the same direction. Describe the force between the two wires. [2]

19. The diagram shows a simple AC generator with a rectangular coil of 50 turns rotating in a uniform magnetic field of flux density 0.3 T. The coil has an area of $2.0 \times 10^{-3} \text{ m}^2$ and rotates at 60 revolutions per second.

<image_placeholder> id: Q19-fig1 type: diagram linked_question: Q19 description: An AC generator with a rectangular coil rotating in a uniform magnetic field. Slip rings and brushes are shown. The coil is at an angle to the magnetic field. labels: N (left), S (right), Coil (50 turns), Area = 2.0 × 10⁻³ m², B = 0.3 T, Rotation = 60 rev/s, Slip rings, Brushes values: N = 50, A = 2.0 × 10⁻³ m², B = 0.3 T, f = 60 Hz must_show: Rectangular coil in magnetic field, slip rings and brushes, rotation direction, magnetic field direction </image_placeholder>

(a) Calculate the maximum e.m.f. induced in the coil. [2]

(b) Calculate the r.m.s. value of the induced e.m.f. [1]

20. A power station generates electricity at 25 kV. The electricity is transmitted through cables to a town 50 km away. The total resistance of the transmission cables is 10 Ω. The power delivered to the town is 10 MW.

(a) Calculate the current in the transmission cables if the transmission voltage is 25 kV. [1]

(b) Calculate the power loss in the cables at this voltage. [2]

(c) A step-up transformer is used to increase the transmission voltage to 400 kV. Calculate the new power loss in the cables. [2]

(d) Explain why high voltage transmission reduces power loss. [2]

End of Quiz

Answers

Secondary 4 Pure Physics Quiz - Electricity Magnetism (Answer Key)

Total Marks: 40

Section A: Multiple Choice Questions [10 marks]

1. Answer: D [1]
Working:
For a transformer: $\frac{V_s}{V_p} = \frac{N_s}{N_p}$
$V_s = V_p \times \frac{N_s}{N_p} = 240 \times \frac{2000}{500} = 240 \times 4 = 960 \text{ V}$

2. Answer: A [1]
Explanation: A fuse contains a thin wire that melts when the current exceeds its rating, breaking the circuit and preventing overheating or fire. It does not increase resistance, store energy, or convert AC to DC.

3. Answer: B [1]
Working:
Use Fleming's Left-Hand Rule:

First finger (Field): Left to right (N to S)
Thumb (Motion/Force): Out of the page
Second finger (Current): Right to left
The current flows from right to left.

4. Answer: C [1]
Explanation: When the coil is horizontal, the sides AB and CD are moving parallel to the magnetic field lines. The rate of change of magnetic flux linkage is zero at this instant, so no e.m.f. is induced. Maximum e.m.f. occurs when the coil is vertical.

5. Answer: C [1]
Working:
$P_{\text{max}} = V \times I_{\text{fuse}} = 240 \times 13 = 3120 \text{ W}$

6. Answer: C [1]
Working:
Use the right-hand grip rule: Thumb points up (current direction), fingers curl around the wire. At a point east of the wire, the magnetic field points towards the wire (into the page). The compass needle aligns with the magnetic field, pointing towards the wire.

7. Answer: C [1]
Working:
For an ideal transformer: $V_p I_p = V_s I_s$ and $\frac{V_s}{V_p} = \frac{N_s}{N_p}$
$\frac{I_s}{I_p} = \frac{N_p}{N_s} = \frac{800}{100} = 8$
$I_s = 8 \times I_p = 8 \times 0.5 = 4 \text{ A}$

8. Answer: B [1]
Explanation: Faraday's law states that the magnitude of induced e.m.f. is proportional to the rate of change of magnetic flux linkage. Lenz's law gives the direction. A stationary magnet produces no induced current. The induced current direction opposes the change producing it.

9. Answer: B [1]
Working:
$P = VI \Rightarrow I = \frac{P}{V} = \frac{60}{240} = 0.25 \text{ A}$
(Alternatively: $P = I^2R \Rightarrow I = \sqrt{\frac{P}{R}} = \sqrt{\frac{60}{960}} = \sqrt{0.0625} = 0.25 \text{ A}$ )

10. Answer: B [1]
Working:
$F = BIL \sin\theta = 0.2 \times 4 \times 0.5 \times \sin 90^\circ = 0.4 \text{ N}$

Section B: Structured Questions [30 marks]

11. (a) Answer: Arrow on side AB pointing into the page (or downwards into the page) [1]
Marking: Correct direction = 1 mark. Use Fleming's Left-Hand Rule: Field left to right, Current A to B (top to bottom on front side), Force into page.

(b) Answer: [2]

The current in sides AB and CD flows in opposite directions. [1]
The forces on AB and CD are equal in magnitude but opposite in direction, creating a couple (turning effect). [1]
Explanation: Side AB has current flowing towards B (into page on front face), side CD has current flowing towards C (out of page on front face). Forces form a couple.

(c) Answer: [2]
Working:
Maximum torque $\tau = B I A N$ (for N turns, here N=1)
Area $A = 0.1 \times 0.08 = 0.008 \text{ m}^2$
$\tau = 0.5 \times 2 \times 0.008 \times 1 = 0.008 \text{ N·m}$
Alternative: Force on each side $F = BIL = 0.5 \times 2 \times 0.1 = 0.1 \text{ N}$ (on AB and CD)
Perpendicular distance between forces = 0.08 m
$\tau = F \times d = 0.1 \times 0.08 = 0.008 \text{ N·m}$ [1 for method, 1 for answer with unit]

(d) Answer: [2]

When the coil is horizontal, the forces on AB and CD act along the same line (through the axis of rotation). [1]
The perpendicular distance between the forces is zero, so the torque becomes zero. [1]
Explanation: At horizontal position, the moment arm is zero. The coil has maximum kinetic energy here and continues rotating due to inertia. The split-ring commutator reverses current to maintain rotation direction.

12. (a) Answer: [3]

As the magnet falls, the magnetic flux through the copper tube changes. [1]
This induces eddy currents in the copper tube (Faraday's law). [1]
The induced currents create a magnetic field that opposes the motion of the magnet (Lenz's law), producing an upward magnetic force that slows the fall. [1]
Key concept: Electromagnetic induction → eddy currents → opposing magnetic force (Lenz's law).

(b) Answer: Anticlockwise (as viewed from above) [1]
Explanation: As the magnet approaches (say, North pole down), the magnetic flux downwards increases. The induced current creates a magnetic field opposing this increase (upwards). Using right-hand grip rule, an upward field requires anticlockwise current as viewed from above.

(c) Answer: Lenz's law states that the direction of the induced current is such that it opposes the change in magnetic flux that produces it. [1]
Acceptable alternatives: The induced e.m.f. acts in a direction to oppose the change producing it.

(d) Answer: [2]

The magnet would fall faster / take less time (approximately 0.8 s, same as the non-magnetic block). [1]
Plastic is an insulator, so no eddy currents can be induced. Without induced currents, there is no opposing magnetic force. The magnet falls freely under gravity (with only air resistance). [1]

13. (a) Answer: [2]
Working:
$\frac{V_s}{V_p} = \frac{N_s}{N_p}$
$N_s = N_p \times \frac{V_s}{V_p} = 4800 \times \frac{5.0}{240} = 4800 \times \frac{1}{48} = 100 \text{ turns}$ [1 for formula/substitution, 1 for answer]

(b) Answer: [2]
Working:
Efficiency $\eta = \frac{P_{\text{out}}}{P_{\text{in}}} = \frac{V_s I_s}{V_p I_p}$
$0.80 = \frac{5.0 \times 2.0}{240 \times I_p}$
$I_p = \frac{10}{240 \times 0.80} = \frac{10}{192} = 0.0521 \text{ A}$ (or 52.1 mA) [1 for formula/substitution, 1 for answer with unit]

(c) Answer: [2]
Working:
Battery capacity = 3000 mAh = 3.0 Ah
Charging current = 2.0 A
Time = $\frac{\text{Capacity}}{\text{Current}} = \frac{3.0}{2.0} = 1.5 \text{ hours}$ [1 for unit conversion, 1 for answer]

(d) Answer: To reduce eddy currents in the core. [1]
Explanation: Laminations (thin insulated sheets) increase the resistance to eddy current paths, reducing energy loss as heat. This improves transformer efficiency.

14. (a) Answer: [3]
Working:
$I = \frac{P}{V}$
Kettle: $I = \frac{2000}{240} = 8.33 \text{ A}$
Iron: $I = \frac{1500}{240} = 6.25 \text{ A}$
Lamp: $I = \frac{100}{240} = 0.417 \text{ A}$
[1 mark each, correct formula and answers]

(b) Answer: [1]
Total current = $8.33 + 6.25 + 0.417 = 15.0 \text{ A}$ (or 14.997 A ≈ 15.0 A)

No, the circuit breaker will not trip. [1]
Total current (15.0 A) is less than the circuit breaker rating (20 A). [1]
Note: Circuit breaker trips when current exceeds its rating.

(d) Answer: [2]
Working:
Power = 2000 W = 2.0 kW
Time per day = 15 min = 0.25 h
Energy per day = $2.0 \times 0.25 = 0.5 \text{ kWh}$
Energy per week = $0.5 \times 7 = 3.5 \text{ kWh}$ [1 for daily energy, 1 for weekly energy with unit]

15. (a) Answer: [1]
Period = 4 divisions × 5 ms/div = 20 ms = 0.020 s

(b) Answer: [1]
Frequency $f = \frac{1}{T} = \frac{1}{0.020} = 50 \text{ Hz}$

(c) Answer: [1]
Peak voltage = $\frac{\text{peak-to-peak}}{2} = \frac{3 \text{ div} \times 2 \text{ V/div}}{2} = \frac{6}{2} = 3.0 \text{ V}$
Or: Amplitude = 1.5 div × 2 V/div = 3.0 V

(d) Answer: [1]
$V_{\text{rms}} = \frac{V_{\text{peak}}}{\sqrt{2}} = \frac{3.0}{\sqrt{2}} = 2.12 \text{ V}$ (3 s.f.)

(e) Answer: [2]

New time-base: 10 ms/div → period spans 2 divisions (half the horizontal width) [1]
New Y-gain: 4 V/div → peak-to-peak spans 1.5 divisions (half the vertical height) [1]
Sketch description: Sinusoidal wave with same shape but horizontally compressed by factor of 2 and vertically compressed by factor of 2. Period now 2 divisions, peak-to-peak 1.5 divisions.

16. (a) Answer: [2]
Working:
$F = BIL \sin\theta = 0.4 \times 5 \times 0.3 \times \sin 30^\circ$
$F = 0.4 \times 5 \times 0.3 \times 0.5 = 0.3 \text{ N}$ [1 for formula/substitution, 1 for answer with unit]

(b) Answer: [1]
The force is perpendicular to both the wire and the magnetic field (direction given by Fleming's Left-Hand Rule or right-hand cross product rule).

The magnitude of the total force remains the same. [1]
The force on a current-carrying conductor in a uniform magnetic field depends only on the straight-line distance between the ends of the conductor (the vector displacement), not the shape of the wire. For a semicircle of radius 0.1 m, the straight-line distance between ends is 0.2 m (diameter). The effective length for force calculation is this displacement. Original wire length 0.3 m at 30° gives vertical component $0.3 \sin 30^\circ = 0.15 \text{ m}$ . Wait — let me recalculate.
Correction: The original wire is straight at 30° to field. The force depends on the component of length perpendicular to field: $L_\perp = L \sin\theta = 0.3 \times \sin 30^\circ = 0.15 \text{ m}$ .
For the semicircle: radius 0.1 m, diameter = 0.2 m. The ends are separated by 0.2 m. If the diameter is perpendicular to the field, $L_\perp = 0.2 \text{ m}$ . The force would be different.
Better answer: The force on a curved wire in a uniform field equals the force on a straight wire joining its endpoints. The semicircle has endpoints separated by the diameter (0.2 m). If the diameter is oriented at the same angle to the field, the perpendicular component is $0.2 \sin 30^\circ = 0.1 \text{ m}$ , giving a different force. The question says "the magnetic field remains uniform and in the same direction" but doesn't specify the orientation of the semicircle.
Expected answer: The total force is unchanged if the straight-line distance between the ends and its orientation relative to the field are unchanged. The force depends only on the vector displacement between the ends, not the path. [1 for concept, 1 for explanation]

17. (a) Answer: North pole [1]
Working: Use right-hand grip rule: fingers curl in direction of current (conventional current from + to -), thumb points to North pole. Current flows from battery + through solenoid. At right end, current appears anticlockwise → North pole.

(b) Answer: [1]
Compass needle points away from the solenoid (towards the right) — the North pole of the compass points towards the South pole of the solenoid? Wait: The right end is North. The compass North pole is repelled by the solenoid's North pole, so it points away (to the right).
Marking: Arrow pointing away from solenoid (to the right) = 1 mark.

(c) Answer: [2]
Working:
$B = \mu_0 n I = \mu_0 \frac{N}{L} I$
$B = (4\pi \times 10^{-7}) \times \frac{200}{0.2} \times 1.5$
$B = (4\pi \times 10^{-7}) \times 1000 \times 1.5$
$B = 6\pi \times 10^{-4} \approx 1.88 \times 10^{-3} \text{ T}$ [1 for formula/substitution, 1 for answer with unit]

(d) Answer: [2]

The compass needle returns to pointing North (aligning with Earth's magnetic field). [1]
When the switch is opened, the current stops, the magnetic field of the solenoid disappears, and the only magnetic field at point P is Earth's magnetic field. [1]

18. (a) Answer: [2]
Sketch requirements:

Concentric circles around the wire [1]
Direction arrows clockwise (current into page → right-hand grip rule: thumb into page, fingers curl clockwise) [1]
At least three field lines with increasing radius
Field lines closer near wire, farther apart further away

(b) Answer: The magnetic field strength doubles. [1]
Explanation: $B \propto I$ for a long straight wire ( $B = \frac{\mu_0 I}{2\pi r}$ ). Doubling current doubles B at a fixed distance.

The wires attract each other. [1]
Each wire produces a magnetic field that exerts a force on the other wire. With currents in the same direction, the magnetic fields between the wires oppose each other, resulting in an attractive force (or use Fleming's Left-Hand Rule on each wire). [1]

19. (a) Answer: [2]
Working:
Maximum e.m.f. $\mathcal{E}_0 = N B A \omega$
$\omega = 2\pi f = 2\pi \times 60 = 120\pi \text{ rad/s}$
$\mathcal{E}_0 = 50 \times 0.3 \times (2.0 \times 10^{-3}) \times 120\pi$
$\mathcal{E}_0 = 50 \times 0.3 \times 2.0 \times 10^{-3} \times 120\pi$
$\mathcal{E}_0 = 3.6\pi \approx 11.3 \text{ V}$ [1 for formula/substitution, 1 for answer with unit]

(b) Answer: [1]
$V_{\text{rms}} = \frac{V_0}{\sqrt{2}} = \frac{11.3}{\sqrt{2}} \approx 8.0 \text{ V}$

Sinusoidal wave (sine or cosine) [1]
Two complete cycles shown [1]
Axes labelled: y-axis "e.m.f. / V" with peak marked ±11.3 V, x-axis "time / s" with period marked $T = \frac{1}{60} \approx 0.0167 \text{ s}$ [1]
Note: Period $T = \frac{1}{f} = \frac{1}{60} \text{ s}$ . Two cycles = $\frac{2}{60} = \frac{1}{30} \text{ s}$ .

20. (a) Answer: [1]
Working:
$P = VI \Rightarrow I = \frac{P}{V} = \frac{10 \times 10^6}{25 \times 10^3} = 400 \text{ A}$

(b) Answer: [2]
Working:
Power loss $P_{\text{loss}} = I^2 R = (400)^2 \times 10 = 160000 \times 10 = 1.6 \times 10^6 \text{ W} = 1.6 \text{ MW}$ [1 for formula/substitution, 1 for answer with unit]

(c) Answer: [2]
Working:
New voltage = 400 kV. Power delivered same = 10 MW.
New current $I' = \frac{P}{V'} = \frac{10 \times 10^6}{400 \times 10^3} = 25 \text{ A}$
New power loss $P'_{\text{loss}} = (I')^2 R = (25)^2 \times 10 = 6250 \text{ W} = 6.25 \text{ kW}$ [1 for new current, 1 for power loss with unit]

(d) Answer: [2]

For a fixed power delivered, increasing the transmission voltage reduces the current ( $P = VI$ ). [1]
Power loss in cables is $I^2 R$ , so reducing current significantly reduces power loss (proportional to $I^2$ ). [1]
Key concept: High voltage → low current → low $I^2R$ losses.

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