Secondary 3 Physics Electricity Magnetism Quiz

Secondary 3 Physics Quiz - Electricity Magnetism (Answer Key)

Total Marks: 40

Section A: Multiple Choice Questions (10 marks)

1. Answer: B
Explanation: Conventional current is defined as the flow of positive charge, which moves from the positive terminal to the negative terminal of a battery (or power supply). This is opposite to the direction of electron flow (which is from negative to positive).
Common mistake: Confusing conventional current direction of electron flow with conventional current.

2. Answer: B
Working:
Using Ohm's Law: $V = IR$
$I = \frac{V}{R} = \frac{12}{10} = 1.2 \ \text{A}$
Mark breakdown: 1 mark for correct substitution and answer with unit.

3. Answer: C
Working:
For resistors in parallel: $\frac{1}{R_{\text{eff}}} = \frac{1}{R} + \frac{1}{R} + \frac{1}{R} = \frac{3}{R}$
$R_{\text{eff}} = \frac{R}{3}$
Mark breakdown: 1 mark for correct formula application and answer.

4. Answer: A (Upwards)
Explanation: Using Fleming's Left-Hand Rule:

• First finger (Field): Points from N to S (left to right)
• Second finger (Current): Points left to right (given)
• Thumb (Force): Points upwards
  Note: The force is perpendicular to both the magnetic field and the current direction.

5. Answer: B
Working:
Energy = Power × Time
$E = 2000 \ \text{W} \times (15 \times 60) \ \text{s} = 2000 \times 900 = 1,800,000 \ \text{J} = 1800 \ \text{kJ}$
Alternative: $E = 2 \ \text{kW} \times 0.25 \ \text{h} = 0.5 \ \text{kWh} = 1800 \ \text{kJ}$
Mark breakdown: 1 mark for correct unit conversion and calculation.

6. Answer: D
Explanation: The magnitude of induced e.m.f. depends on the rate of change of magnetic flux linkage (Faraday's Law): $|\mathcal{E}| = N \frac{\Delta \Phi}{\Delta t}$. This depends on:

• Speed of magnet (rate of flux change)
• Number of turns $N$
• Strength of magnet (flux density)
  The resistance of the galvanometer affects the induced current ($I = \mathcal{E}/R$), not the induced e.m.f. itself.

7. Answer: A (Maximum)
Explanation: For an a.c. generator, induced e.m.f. $\mathcal{E} = \mathcal{E}_0 \sin(\omega t)$ where $\mathcal{E}_0 = N B A \omega$. When the coil is horizontal, its plane is parallel to the magnetic field, so the rate of change of magnetic flux through the coil is maximum. Hence induced e.m.f. is maximum.
Common mistake: Confusing flux (maximum when coil is vertical) with rate of change of flux (maximum when coil is horizontal).

8. Answer: A
Working:
For an ideal transformer: $\frac{V_s}{V_p} = \frac{N_s}{N_p}$
$V_s = V_p \times \frac{N_s}{N_p} = 240 \times \frac{100}{500} = 240 \times 0.2 = 48 \ \text{V}$
Mark breakdown: 1 mark for correct formula and calculation.

9. Answer: D
Explanation: Magnetic field lines form continuous closed loops. Outside the magnet, they point from North to South. Inside the magnet, they continue from South back to North to complete the loop. They do not "start" at North and "end" at South inside the magnet.
Mark breakdown: 1 mark for identifying the incorrect statement.

10. Answer: A
Working:
Force on a current-carrying conductor: $F = B I L \sin\theta$
Since wire is perpendicular to field, $\theta = 90^\circ$, $\sin\theta = 1$
$F = 0.5 \times 3 \times 0.2 = 0.3 \ \text{N}$
Mark breakdown: 1 mark for correct formula, substitution, and answer with unit.

Section B: Structured Questions (30 marks)

11. (a) Answer: $8 \ \Omega$
Working:
$R_{\text{total}} = \frac{V}{I} = \frac{12}{1.5} = 8 \ \Omega$
Mark breakdown: 1 mark for correct formula, 1 mark for correct answer with unit.

(b) Answer: $4 \ \Omega$
Working:
$R_{\text{variable}} = R_{\text{total}} - R_{\text{fixed}} = 8 - 4 = 4 \ \Omega$
Mark breakdown: 1 mark for correct answer with unit.

(c) Answer: $6 \ \text{V}$
Working:
$V_{\text{fixed}} = I R = 1.5 \times 4 = 6 \ \text{V}$
Mark breakdown: 1 mark for correct answer with unit.

(d) Answer: The brightness of the lamp decreases.
Explanation: When the variable resistor's resistance increases, the total circuit resistance increases, so the total current from the battery decreases. The potential difference across the parallel combination (fixed resistor and lamp) decreases because the variable resistor takes a larger share of the battery voltage. With lower p.d. across the lamp, the current through it decreases ($I = V/R_{\text{lamp}}$), so its power ($P = VI$) decreases and it becomes dimmer.
Mark breakdown: 1 mark for "brightness decreases", 1 mark for correct explanation linking resistance increase → current decrease → p.d. across parallel branch decrease → lamp power decrease.

12. (a) Answer: On the diagram, the compass needle should point towards the North (i.e., tangent to the circular field line, pointing North at a position East of the wire with upward current).
Explanation: Using the Right-Hand Grip Rule: thumb points up (current), fingers curl around wire. At a point East of the wire, the field direction is North. The compass needle aligns with the field, pointing North.
Mark breakdown: 1 mark for correct needle direction on diagram.

(b) Answer: North (or towards geographic North)
Mark breakdown: 1 mark for correct direction.

(c) Answer: The compass needle reverses direction (now points South).
Explanation: Reversing the current reverses the magnetic field direction (Right-Hand Grip Rule). The compass needle aligns with the new field direction, so it points in the opposite direction.
Mark breakdown: 1 mark for correct description of reversal.

13. (a) Answer: $1256.6 \ \text{V}$ (or $400\pi \ \text{V} \approx 1257 \ \text{V}$)
Working:
Maximum e.m.f. for a rotating coil: $\mathcal{E}_0 = N B A \omega$
$N = 200$, $B = 0.4 \ \text{T}$, $A = 5 \times 10^{-3} \ \text{m}^2$
$\omega = 2\pi f = 2\pi \times 50 = 100\pi \ \text{rad/s}$
$\mathcal{E}_0 = 200 \times 0.4 \times (5 \times 10^{-3}) \times 100\pi$
$\mathcal{E}_0 = 200 \times 0.4 \times 0.005 \times 100\pi$
$\mathcal{E}_0 = 40 \times 100\pi = 4000\pi \ \text{V}$? Wait, recalculate:
$200 \times 0.4 = 80$
$80 \times 5 \times 10^{-3} = 0.4$
$0.4 \times 100\pi = 40\pi \approx 125.7 \ \text{V}$
Let me recalculate carefully:
$\mathcal{E}_0 = N B A \omega = 200 \times 0.4 \times 5 \times 10^{-3} \times (2\pi \times 50)$
$= 200 \times 0.4 \times 0.005 \times 100\pi$
$= 200 \times 0.4 \times 0.5\pi$
$= 200 \times 0.2\pi$
$= 40\pi \approx 125.7 \ \text{V}$
Corrected Answer: $40\pi \ \text{V} \approx 126 \ \text{V}$
Mark breakdown: 1 mark for correct formula $\mathcal{E}_0 = N B A \omega$, 1 mark for correct $\omega = 2\pi f$, 1 mark for correct calculation and unit.

(b) Answer: See graph description below.
Graph description:

• Sinusoidal waveform (sine wave)
• Period $T = \frac{1}{f} = \frac{1}{50} = 0.02 \ \text{s}$
• Two complete cycles shown (0 to 0.04 s)
• Amplitude = $40\pi \ \text{V} \approx 126 \ \text{V}$
• Axes labelled: Time (s) on x-axis, Induced e.m.f. (V) on y-axis
• Key points: Zero at $t=0, 0.01, 0.02, 0.03, 0.04$ s; Maximum at $t=0.005, 0.025$ s; Minimum at $t=0.015, 0.035$ s
  Mark breakdown: 1 mark for sinusoidal shape, 1 mark for correct period and two cycles, 1 mark for correct amplitude and labelled axes.

14. (a) Answer: $12 \ \text{V}$
Working:
$\frac{V_s}{V_p} = \frac{N_s}{N_p}$
$V_s = 240 \times \frac{60}{1200} = 240 \times 0.05 = 12 \ \text{V}$
Mark breakdown: 1 mark for correct formula, 1 mark for correct answer with unit.

(b) Answer: $0.06 \ \text{A}$ (or $60 \ \text{mA}$)
Working:
For 100% efficient transformer: $V_p I_p = V_s I_s$
$I_p = \frac{V_s I_s}{V_p} = \frac{12 \times 1.2}{240} = \frac{14.4}{240} = 0.06 \ \text{A}$
Alternative: Using turns ratio: $\frac{I_p}{I_s} = \frac{N_s}{N_p} = \frac{60}{1200} = \frac{1}{20}$
$I_p = \frac{1.2}{20} = 0.06 \ \text{A}$
Mark breakdown: 1 mark for correct principle (power conservation or current ratio), 1 mark for correct answer with unit.

(c) Answer: Any two of:

1. Eddy current losses in the soft iron core (induced currents in the core cause heating)
2. Hysteresis losses (energy needed to magnetise/demagnetise the core each cycle)
3. Copper losses ($I^2R$ heating in the primary and secondary windings due to their resistance)
4. Flux leakage (not all magnetic flux from primary links with secondary)
   Mark breakdown: 1 mark each for any two valid reasons, correctly stated.

15. (a) Answer: On the diagram: Current in side AB flows into the page (or away from viewer), current in side CD flows out of the page (or towards viewer).
Explanation: Current enters from left brush, flows through commutator into coil. Trace the path: left brush → left commutator half → side AB (downwards/into page) → side CD (upwards/out of page) → right commutator half → right brush.
Mark breakdown: 1 mark for both directions correctly labelled.

(b) Answer: Force on AB: Downwards (or towards bottom of page). Force on CD: Upwards (or towards top of page).
Explanation: Using Fleming's Left-Hand Rule:

• Field: Left to right (N to S)
• Current in AB: Into page → Force: Downwards
• Current in CD: Out of page → Force: Upwards
  These forces form a couple producing clockwise rotation (viewed from left).
  Mark breakdown: 1 mark for both forces correctly stated.

(c) Answer: The split-ring commutator reverses the current direction in the coil every half-turn. As the coil rotates past the vertical position, the commutator halves swap contact with the brushes. This reverses the current in the coil sides, so the forces on AB and CD also reverse. The torque direction remains the same, allowing continuous rotation in one direction.
Mark breakdown: 1 mark for "reverses current every half-turn", 1 mark for "maintains torque direction for continuous rotation".

16. (a) Answer: The iron filings form concentric circles centred on the wire.
Mark breakdown: 1 mark for "concentric circles" or "circular pattern around the wire".

(b) Answer: The circles become closer together (or the pattern becomes denser), indicating a stronger magnetic field.
Mark breakdown: 1 mark for correct description of change.

(c) Answer: The direction of the field lines reverses (circles go the opposite way), but the pattern (concentric circles) remains the same.
Mark breakdown: 1 mark for correct description.

(d) Answer: Right-Hand Grip Rule (or Right-Hand Thumb Rule): If you grip the wire with your right hand such that your thumb points in the direction of conventional current, your curled fingers show the direction of the magnetic field lines.
Mark breakdown: 1 mark for naming the rule correctly.

17. (a) Answer: $3 \ \text{kWh}$
Working:
Energy = Power × Time
$E = 1.5 \ \text{kW} \times 2 \ \text{h} = 3 \ \text{kWh}$
Mark breakdown: 1 mark for power in kW, 1 mark for correct answer with unit.

(b) Answer: $25.20$
Working:
Energy for 30 days = $3 \times 30 = 90 \ \text{kWh}$
Cost = 90 \times \0.28 = $25.20$**Mark breakdown:** 1 mark for total energy, 1 mark for correct cost with$ sign.

(c) Answer: Verification shown below.
Working:
Method 1: Using $P = VI$
$I = \frac{P}{V} = \frac{1500}{240} = 6.25 \ \text{A}$ ✓
Method 2: Using $P = I^2 R$
$I = \sqrt{\frac{P}{R}} = \sqrt{\frac{1500}{38.4}} = \sqrt{39.0625} = 6.25 \ \text{A}$ ✓
Method 3: Using $V = IR$
$I = \frac{V}{R} = \frac{240}{38.4} = 6.25 \ \text{A}$ ✓
Mark breakdown: 1 mark for correct formula, 1 mark for correct calculation showing 6.25 A.

(d) Answer: As the potential difference increases, the current increases, causing the filament temperature to rise. The higher temperature increases the lattice ion vibration, which increases collisions with conduction electrons, thus increasing the resistance of the metal filament.
Mark breakdown: 1 mark for linking temperature rise to increased resistance via increased lattice vibrations/collisions.

18. (a) Answer: As the magnet falls, the magnetic flux through the copper tube changes. This induces eddy currents in the tube walls (Faraday's Law). These eddy currents create their own magnetic field which opposes the change producing them (Lenz's Law). The induced magnetic field repels the falling magnet (or attracts it from below), creating an upward magnetic force that opposes gravity. This reduces the net downward force, so the magnet falls slower than in a plastic tube (where no eddy currents are induced).
Mark breakdown: 1 mark for "eddy currents induced", 1 mark for "oppose the change (Lenz's Law)", 1 mark for "upward magnetic force opposes gravity/reduces acceleration".

(b) Answer: Gravitational potential energy → Kinetic energy + Electrical energy (in eddy currents) → Heat energy (dissipated in the tube).
Or simply: Gravitational potential energy → Heat (thermal energy)
Mark breakdown: 1 mark for correct energy conversion chain.

19. (a) Answer: $0.45 \ \text{A}$
Explanation: From the graph, at $V = 6 \ \text{V}$, the corresponding current is $0.45 \ \text{A}$.
Mark breakdown: 1 mark for correct reading from graph.

(b) Answer: $2.7 \ \text{V}$
Working:
$V_R = I R = 0.45 \times 6 = 2.7 \ \text{V}$
Mark breakdown: 1 mark for correct calculation with unit.

(c) Answer: $8.7 \ \text{V}$
Working:
In a series circuit, e.m.f. = sum of p.d.s across components
$\mathcal{E} = V_{\text{lamp}} + V_R = 6 + 2.7 = 8.7 \ \text{V}$
Note: The question states the battery is 12 V, but asks to "determine the e.m.f. of the battery" from the graph data. This is a consistency check - the graph data implies a battery of 8.7 V, not 12 V. Students should use the graph data, not the stated 12 V.
Mark breakdown: 1 mark for correct principle and answer.

(d) **Answer: The filament lamp's resistance increases because as the voltage increases, the current increases, heating the filament. The higher temperature causes the metal ions in the filament to vibrate more vigorously, impeding the flow of electrons (more frequent collisions), which increases the resistance.
Mark breakdown: 1 mark for explanation linking temperature → lattice vibrations → increased collisions → increased resistance.

20. (a) Answer: Any two of:

1. Increase the number of turns in the coil
2. Use a soft iron core with higher magnetic permeability (or larger cross-sectional area)
3. Decrease the length of the core (more compact coil)
4. Use a core material with higher relative permeability
   Mark breakdown: 1 mark each for any two valid methods.

(b) Answer: The graph passes through the origin because when the current $I = 0$, there is no current to produce a magnetic field, so the magnetic field strength $B = 0$. The relationship $B \propto I$ (for a given electromagnet geometry and core) is linear, so the line passes through the origin.
Mark breakdown: 1 mark for explaining that zero current gives zero magnetic field.

(c) Answer: With a steel core, the electromagnet retains its magnetism (becomes a permanent magnet) when the current is switched off. With a soft iron core, the magnetism is lost almost completely when the current is switched off.
Explanation: Steel is a "hard" magnetic material with high retentivity (high coercivity) - it retains magnetic alignment. Soft iron is a "soft" magnetic material with low retentivity - domains easily return to random orientation when the magnetising current is removed.
Mark breakdown: 1 mark for stating steel retains magnetism / soft iron loses it, 1 mark for explanation using retentivity/coercivity or domain theory.

End of Answer Key