Secondary 4 Pure Physics Electricity Magnetism Quiz

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

Total Marks: 40

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Section A: Multiple Choice Questions (10 marks)

1. [1] Answer: C

Working:
For a transformer: $\frac{V_s}{V_p} = \frac{N_s}{N_p}$
$V_s = V_p \times \frac{N_s}{N_p} = 240 \times \frac{2000}{500} = 240 \times 4 = 960 \text{ V}$

Concept: Transformer voltage ratio equals turns ratio.

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2. [1] Answer: A

Working:
Energy = Power × Time
$= 2000 \text{ W} \times \frac{15}{60} \text{ h} = 2 \text{ kW} \times 0.25 \text{ h} = 0.5 \text{ kWh}$

Concept: Convert power to kW and time to hours for kWh.

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3. [1] Answer: A

Working:
Force on current-carrying conductor: $F = BIL \sin\theta$
$\theta = 90^\circ$ (perpendicular), so $\sin\theta = 1$
$F = 0.4 \times 5 \times 0.2 = 0.4 \text{ N}$

Concept: $F = BIL$ when wire is perpendicular to field.

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4. [1] Answer: B

Explanation: The earth wire connects the metal casing of an appliance to ground. If the live wire touches the casing, the earth wire provides a low-resistance path for current to flow to ground, causing the fuse to blow and disconnecting the supply. This prevents electric shock.

Common mistake: Confusing earth wire with neutral wire (which carries normal return current).

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5. [1] Answer: A

Explanation: Induced e.m.f. is maximum when the rate of change of magnetic flux linkage is greatest. This occurs when the plane of the coil is parallel to the field lines (coil cutting field lines at maximum rate). When perpendicular, flux linkage is maximum but rate of change is zero.

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6. [1] Answer: C

Working:
Three 6 Ω resistors in parallel:
$\frac{1}{R_{\text{total}}} = \frac{1}{6} + \frac{1}{6} + \frac{1}{6} = \frac{3}{6} = \frac{1}{2}$
$R_{\text{total}} = 2 \Omega$
$I = \frac{V}{R} = \frac{12}{2} = 6.0 \text{ A}$

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7. [1] Answer: A

Working:
Use Fleming's left-hand rule:

• First finger (Field): Left to right (N to S)
• Second finger (Current): Into page (⊗)
• Thumb (Force): Upwards

Answer: A. Upwards

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8. [1] Answer: C

Working:
For 100% efficient transformer: $V_p I_p = V_s I_s$ and $\frac{V_s}{V_p} = \frac{N_s}{N_p}$
$\frac{I_s}{I_p} = \frac{N_p}{N_s} = \frac{800}{100} = 8$
$I_s = 8 \times 0.5 = 4.0 \text{ A}$

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9. [1] Answer: C

Explanation: Magnetic field lines never cross. They form closed loops from N to S outside the magnet. The density of field lines indicates field strength — closer lines mean stronger field.

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10. [1] Answer: C

Working:
$I = \frac{P}{V} = \frac{3000}{230} = 13.04 \text{ A}$
Standard fuse ratings: 3 A, 5 A, 13 A, 30 A.
Choose the smallest fuse rating greater than the operating current: 13 A.

Note: 30 A would also work but 13 A provides better protection.

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Section B: Structured Questions (30 marks)

11. [4]

(a) [2]
Magnetic field lines are concentric circles around the wire, directed anticlockwise when viewed from above (current upward/out of page).

• 1 mark: Circular field lines centred on wire
• 1 mark: Correct anticlockwise direction (using right-hand grip rule)

(b) [1]
The North pole of each compass points tangent to the circular field lines, in the anticlockwise direction.

• At North position: points West
• At East position: points North
• At South position: points East
• At West position: points South

(c) [1]
The magnetic field pattern remains circular but the direction reverses to clockwise (when viewed from above).

Key concept: Right-hand grip rule — thumb = current direction, fingers = field direction.

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12. [5]

(a) [2]
$\frac{V_s}{V_p} = \frac{N_s}{N_p}$
$N_s = N_p \times \frac{V_s}{V_p} = 1200 \times \frac{12}{240} = 1200 \times 0.05 = 60 \text{ turns}$

(b) [1]
$P = V_s I_s$
$I_s = \frac{P}{V_s} = \frac{24}{12} = 2.0 \text{ A}$

(c) [2]
Efficiency $\eta = 0.90 = \frac{P_{\text{out}}}{P_{\text{in}}} = \frac{V_s I_s}{V_p I_p}$
$I_p = \frac{V_s I_s}{\eta V_p} = \frac{12 \times 2.0}{0.90 \times 240} = \frac{24}{216} = 0.111 \text{ A}$ (or 0.11 A)

Mark breakdown:

• 1 mark: Correct efficiency formula rearrangement
• 1 mark: Correct substitution and answer with unit

Common mistake: Forgetting to convert 90% to 0.90.

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13. [4]

(a) [1]
Force on AB: Upwards
Fleming's left-hand rule:

• First finger (Field): Left to right (N to S)
• Second finger (Current): A to B (left to right along top side)
• Thumb (Force): Upwards (out of page)

(b) [2]
$F = BIL \sin\theta$
Side AB is perpendicular to field ($\theta = 90^\circ$)
$F = 0.5 \times 2 \times 0.1 \times 1 = 0.1 \text{ N}$

Mark breakdown:

• 1 mark: Correct formula and identification of $\theta = 90^\circ$
• 1 mark: Correct calculation with unit

(c) [1]
Forces on AB and CD are equal in magnitude, opposite in direction (AB upward, CD downward), and not along the same line of action. This forms a couple producing a turning effect (torque). Sides BC and DA are parallel to the field, so no force acts on them.

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14. [5]

(a) [2]
Total power = 2000 + 1200 + 300 + 150 = 3650 W
$I_{\text{total}} = \frac{P_{\text{total}}}{V} = \frac{3650}{230} = 15.87 \text{ A} \approx 15.9 \text{ A}$

Mark breakdown:

• 1 mark: Correct total power
• 1 mark: Correct current calculation with unit

(b) [1]
No, the circuit breaker will not trip.
Total current (15.9 A) < Circuit breaker rating (30 A).

(c) [1]
Energy = Power × Time = 0.300 kW × 24 h = 7.2 kWh

(d) [1]
Cost = Energy × Rate = 7.2 kWh/day × 30 days × $0.28/kWh = **$60.48**

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15. [4]

(a) [1]
South pole
Lenz's law: The induced current opposes the change causing it. As N pole approaches, the solenoid end facing it becomes a N pole to repel the approaching magnet. But the galvanometer deflects right — by convention, this means the induced current creates a South pole at that end (attracting the approaching N pole). Wait — let's reconsider.

Correction: If galvanometer deflects right (positive), and using standard convention where current flows from positive terminal of galvanometer... Actually, the question states the needle deflects right. By standard school convention (right-hand grip rule for solenoid), if the induced current creates a North pole at the near end, it would repel the approaching North pole. But the question says the needle deflects right — we need to know the galvanometer connection.

Standard answer expected: The end of the solenoid facing the magnet becomes a North pole (to oppose the approaching North pole). The galvanometer deflection direction depends on winding direction, but the polarity is determined by Lenz's law.

Answer: North pole (to repel the approaching North pole)

(b) [1]
The galvanometer needle deflects to the left (opposite direction), as the induced current reverses to oppose the magnet moving away (now the near end becomes a South pole to attract the receding North pole).

(c) [2]
Any two of:

• Increase the speed of the magnet
• Use a stronger magnet
• Increase the number of turns on the solenoid
• Increase the cross-sectional area of the solenoid
• Use a soft iron core inside the solenoid

Mark breakdown: 1 mark each for any two valid methods.

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16. [4]

(a) [1]
2 cycles occupy 8 horizontal divisions.
1 cycle = 4 divisions.
Period $T = 4 \times 5 \text{ ms} = 20 \text{ ms} = 0.020 \text{ s}$

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

(c) [1]
Peak-to-peak = 3 divisions
Amplitude (peak) = 1.5 divisions
$V_0 = 1.5 \times 2 \text{ V/div} = 3.0 \text{ V}$

(d) [1]
$V_{\text{rms}} = \frac{V_0}{\sqrt{2}} = \frac{3.0}{\sqrt{2}} = 2.12 \text{ V}$

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17. [4]

(a) [2]
$F = BIL \sin\theta$
$F = 0.6 \times 3 \times 0.5 \times \sin 30^\circ$
$F = 0.9 \times 0.5 = 0.45 \text{ N}$

Mark breakdown:

• 1 mark: Correct formula with $\sin 30^\circ$
• 1 mark: Correct calculation with unit

(b) [2]
Maximum force at 90° (perpendicular to field).
$F_{\text{max}} = BIL = 0.6 \times 3 \times 0.5 = 0.9 \text{ N}$

Mark breakdown:

• 1 mark: Correct angle (90°)
• 1 mark: Correct maximum force with unit

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Section C: Data Analysis and Application (10 marks)

18. [5]

(a) [2]
Graph requirements:

• Axes labelled with units: "Speed of magnet (m/s)" and "Induced e.m.f. (mV)"
• Suitable scales (e.g., 1 cm = 0.5 m/s horizontally, 1 cm = 10 mV vertically)
• All 5 points plotted correctly
• Best-fit straight line through origin
• 1 mark: Correct axes, scales, and labels
• 1 mark: All points plotted correctly and best-fit line drawn

(b) [1]
Induced e.m.f. is directly proportional to the speed of the magnet. (Straight line through origin)

(c) [2]
Faraday's law: Induced e.m.f. = rate of change of magnetic flux linkage ($\mathcal{E} = -\frac{d\Phi}{dt}$).
As the magnet moves faster, the rate of change of magnetic flux through the coil increases proportionally. Since flux change per unit time is proportional to speed, induced e.m.f. is proportional to speed.

Mark breakdown:

• 1 mark: Mention Faraday's law / rate of change of flux linkage
• 1 mark: Link speed to rate of change of flux

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19. [5]

(a) [2]
$I = \frac{P}{V}$

• Heater (A): $I_A = \frac{2000}{230} = 8.70 \text{ A}$
• Kettle (B): $I_B = \frac{1000}{230} = 4.35 \text{ A}$
• Oven (C): $I_C = \frac{3000}{230} = 13.04 \text{ A}$

Mark breakdown:

• 1 mark: Correct formula and method
• 1 mark: All three currents calculated correctly with units

(b) [2]
In a ring main, the total current does not simply add up because current splits at the consumer unit and flows both ways around the ring. However, the maximum current in any part of the ring must be considered.

Worst case: All current flows through one path.
Total current = 8.70 + 4.35 + 13.04 = 26.09 A
This is less than 30 A, so the fuse will not blow.

Alternative simpler approach accepted at this level: Total current = 26.1 A < 30 A, so fuse does not blow.

Mark breakdown:

• 1 mark: Correct total current calculation
• 1 mark: Correct conclusion with comparison to 30 A

(c) [1]
Advantage: Thinner/cheaper wire can be used because current divides and flows in two directions around the ring, reducing the maximum current in any single cable.
Or: If a break occurs at one point, appliances still receive power from the other direction.

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20. [5]

(a) [2]
$P = VI$
$I = \frac{P}{V} = \frac{100 \times 10^6}{400 \times 10^3} = \frac{100 \times 10^6}{4 \times 10^5} = 250 \text{ A}$

Mark breakdown:

• 1 mark: Correct formula and unit conversion (MW to W, kV to V)
• 1 mark: Correct answer with unit

(b) [2]
Power loss $P_{\text{loss}} = I^2 R = (250)^2 \times 10 = 62,500 \times 10 = 625,000 \text{ W} = 625 \text{ kW}$

Mark breakdown:

• 1 mark: Correct formula $I^2R$
• 1 mark: Correct calculation with unit

(c) [1]
For a given power $P$, increasing transmission voltage $V$ reduces current $I$ (since $P = VI$). Power loss in cables is $I^2R$, so reducing $I$ significantly reduces power loss (proportional to $I^2$).

Key concept: $P_{\text{loss}} \propto I^2 \propto \frac{1}{V^2}$ — doubling voltage quarters the power loss.

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End of Answer Key