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A Level H2 Physics Electricity Magnetism Quiz

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Questions

A-Level Physics H2 Quiz - Electricity Magnetism

Name: _______________________
Class: _______________________
Date: _______________________
Score: _______ / 60

Duration: 60 minutes
Total Marks: 60

Instructions

Answer ALL questions in the spaces provided.
Show all working clearly. Answers without working may not receive full credit.
The number of marks for each question is given in brackets [ ].
You may use a calculator.
Take $g = 9.81 \text{ m s}^{-2}$ where needed.
Fundamental constants are provided where required.

Section A: Electric Fields and Current Electricity (Questions 1–5)

1. State Coulomb's Law in words and write its mathematical expression.
Define all symbols used.

[3]

2. Two point charges $Q_1 = +4.0 \text{ \mu C}$ and $Q_2 = -6.0 \text{ \mu C}$ are placed $0.30 \text{ m}$ apart in a vacuum.

(a) Calculate the magnitude of the electrostatic force between the two charges.
[2]

(b) State whether the force is attractive or attractive, and explain your reasoning.
[1]

(c) A third charge $Q_3 = +2.0 \text{ \mu C}$ is placed on the line joining $Q_1$ and $Q_2$ , exactly midway between them. Calculate the magnitude and direction of the resultant force on $Q_3$ .
[4]

3. The figure below shows a uniform electric field between two parallel plates.

<image_placeholder> id: Q3-fig1 type: diagram linked_question: Q3 description: Two horizontal parallel plates separated by distance d, with uniform electric field lines pointing vertically downward from positive upper plate to negative lower plate. A charged particle of mass m and charge +q is shown between the plates. labels: Upper plate (+), Lower plate (-), Electric field lines (downward), Separation = d, Particle with charge +q and mass m values: Plate separation d = 0.040 m, Potential difference V = 200 V, Charge q = 3.2 × 10⁻¹⁹ C, Mass m = 6.6 × 10⁻²⁶ kg must_show: Parallel plates with polarity labels, uniform field lines, particle position, separation distance d, direction of field </image_placeholder>

(a) Calculate the electric field strength between the plates.
[2]

(b) Calculate the force on the charged particle due to the electric field.
[2]

(c) The particle is released from rest at the positive plate. Calculate its speed just before it reaches the negative plate.
[3]

4. Define electric potential at a point in an electric field.
Explain how it differs from electric potential energy.

[3]

5. A student sets up the circuit shown below to investigate the $I$ - $V$ characteristics of a filament lamp.

<image_placeholder> id: Q5-fig1 type: experimental_setup linked_question: Q5 description: Circuit diagram with a battery (emf E, internal resistance r) connected in series with a variable resistor (rheostat), an ammeter, and a filament lamp. A voltmeter is connected in parallel across the filament lamp. Switch is in series. labels: Battery (E, r), Variable resistor (rheostat), Ammeter (A), Filament lamp, Voltmeter (V) across lamp, Switch (S) values: Battery emf = 12.0 V, Internal resistance r = 0.50 Ω must_show: Complete circuit with all components labelled, ammeter in series, voltmeter in parallel across lamp, variable resistor in series </image_placeholder>

(a) Explain why the variable resistor is necessary in this circuit.
[2]

(b) The student records the following data:

$V$ / V	0.0	1.0	2.0	3.0	4.0	5.0	6.0
$I$ / A	0.00	0.12	0.22	0.30	0.37	0.43	0.48

(i) Plot a graph of $I$ against $V$ on the grid provided.
[3]

(ii) Use your graph to determine the resistance of the lamp at $V = 4.0 \text{ V}$ .
[2]

(iii) Explain why the resistance of the filament lamp changes with voltage.
[2]

Section B: D.C. Circuits (Questions 6–10)

6. State Kirchhoff's First Law (Junction Rule).
State the physical principle on which it is based.

[2]

7. For the circuit shown below, calculate the current through the $6.0 \text{ \Omega}$ resistor.

<image_placeholder> id: Q7-fig1 type: diagram linked_question: Q7 description: Circuit with a 12.0 V battery (negligible internal resistance) connected to a network of resistors: R1 = 4.0 Ω in series with a parallel combination of R2 = 6.0 Ω and R3 = 3.0 Ω. labels: Battery = 12.0 V, R1 = 4.0 Ω, R2 = 6.0 Ω, R3 = 3.0 Ω values: EMF = 12.0 V, R1 = 4.0 Ω, R2 = 6.0 Ω, R3 = 3.0 Ω must_show: Complete circuit diagram with battery, all resistor values labelled, current direction indicated </image_placeholder>

[4]

8. A cell of emf $1.50 \text{ V}$ and internal resistance $0.20 \text{ \Omega}$ is connected to an external resistor of $2.80 \text{ \Omega}$ .

(a) Calculate the terminal potential difference across the cell.
[2]

(b) Calculate the power dissipated in the external resistor.
[2]

9. A potential divider circuit is shown below.

<image_placeholder> id: Q9-fig1 type: diagram linked_question: Q9 description: Potential divider circuit with a battery of emf 9.0 V and negligible internal resistance, connected across two resistors R1 = 3.0 kΩ and R2 = 6.0 kΩ in series. The output voltage is taken across R2. labels: Battery = 9.0 V, R1 = 3.0 kΩ, R2 = 6.0 kΩ, Output terminals across R2 values: EMF = 9.0 V, R1 = 3.0 kΩ, R2 = 6.0 kΩ must_show: Complete potential divider circuit with battery, two resistors in series, output terminals clearly marked across R2 </image_placeholder>

(a) Calculate the output voltage across $R_2$ .
[2]

(b) A load resistor of $12.0 \text{ k\Omega}$ is now connected in parallel with $R_2$ . Calculate the new output voltage.
[3]

10. A battery of emf $12.0 \text{ V}$ and internal resistance $1.0 \text{ \Omega}$ is connected to a network of resistors as shown.

<image_placeholder> id: Q10-fig1 type: diagram linked_question: Q10 description: Circuit with a 12.0 V battery (internal resistance 1.0 Ω) connected to a network: R1 = 2.0 Ω in series with a parallel combination of R2 = 4.0 Ω and R3 = 4.0 Ω. labels: Battery = 12.0 V, r = 1.0 Ω, R1 = 2.0 Ω, R2 = 4.0 Ω, R3 = 4.0 Ω values: EMF = 12.0 V, r = 1.0 Ω, R1 = 2.0 Ω, R2 = 4.0 Ω, R3 = 4.0 Ω must_show: Complete circuit with battery, internal resistance, all resistor values labelled </image_placeholder>

(a) Calculate the total resistance of the external circuit.
[2]

(b) Calculate the current drawn from the battery.
[2]

Section C: Electromagnetism (Questions 11–15)

11. Define magnetic flux density.
State its SI unit and give its symbol.

[2]

12. A straight wire of length $0.50 \text{ m}$ carrying a current of $4.0 \text{ A}$ is placed perpendicular to a uniform magnetic field of flux density $0.20 \text{ T}$ .

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

(b) The wire is now rotated so that it makes an angle of $30^\circ$ with the magnetic field. Calculate the new force on the wire.
[2]

13. A rectangular coil of 50 turns, dimensions $0.040 \text{ m} \times 0.060 \text{ m}$ , carries a current of $2.0 \text{ A}$ . The coil is placed in a uniform magnetic field of flux density $0.15 \text{ T}$ such that the plane of the coil is parallel to the field.

<image_placeholder> id: Q13-fig1 type: diagram linked_question: Q13 description: Rectangular coil (50 turns) in a uniform magnetic field. The plane of the coil is parallel to the magnetic field direction. Dimensions 0.040 m × 0.060 m. Current I = 2.0 A flowing in the coil. Magnetic field B = 0.15 T directed horizontally. labels: N = 50 turns, a = 0.040 m, b = 0.060 m, I = 2.0 A, B = 0.15 T, Plane of coil parallel to B values: N = 50, a = 0.040 m, b = 0.060 m, I = 2.0 A, B = 0.15 T must_show: Rectangular coil with turns, dimensions, current direction, magnetic field direction, angle between plane and field </image_placeholder>

(a) Calculate the magnetic flux through one turn of the coil.
[2]

(b) Calculate the torque on the coil.
[3]

14. A charged particle of mass $9.11 \times 10^{-31} \text{ kg}$ and charge $-1.60 \times 10^{-19} \text{ C}$ moves perpendicular to a uniform magnetic field of flux density $0.050 \text{ T}$ with a speed of $3.0 \times 10^6 \text{ m s}^{-1}$ .

(a) Calculate the magnetic force on the particle.
[2]

(b) Show that the particle moves in a circular path and calculate the radius of the path.
[3]

15. Two long straight parallel wires P and Q are placed $0.10 \text{ m}$ apart in a vacuum. Wire P carries a current of $5.0 \text{ A}$ and wire Q carries a current of $8.0 \text{ A}$ in the same direction.

<image_placeholder> id: Q15-fig1 type: diagram linked_question: Q15 description: Two long straight parallel wires P and Q, separated by distance r = 0.10 m. Current IP = 5.0 A flowing upward in wire P, current IQ = 8.0 A flowing upward in wire Q. Magnetic field lines around each wire shown. labels: Wire P, Wire Q, IP = 5.0 A (upward), IQ = 8.0 A (upward), Separation r = 0.10 m values: IP = 5.0 A, IQ = 8.0 A, r = 0.10 m must_show: Two parallel wires with current directions, separation distance, magnetic field direction around each wire </image_placeholder>

(a) Calculate the magnetic flux density at the position of wire Q due to the current in wire P.
[2]

(b) Calculate the force per unit length on wire Q due to the magnetic field from wire P.
[2]

Section D: Electromagnetic Induction (Questions 16–20)

16. State Faraday's Law of Electromagnetic Induction.
State Lenz's Law and explain how it relates to Faraday's Law.

[4]

17. A coil of 200 turns and area $0.020 \text{ m}^2$ is placed in a uniform magnetic field of flux density $0.30 \text{ T}$ . The coil is rotated from a position where its plane is perpendicular to the field to a position where its plane is parallel to the field in $0.10 \text{ s}$ .

(a) Calculate the initial magnetic flux linkage through the coil.
[2]

(b) Calculate the average emf induced in the coil during this rotation.
[3]

18. A metal rod of length $0.40 \text{ m}$ moves at a constant speed of $5.0 \text{ m s}^{-1}$ perpendicular to a uniform magnetic field of flux density $0.25 \text{ T}$ .

<image_placeholder> id: Q18-fig1 type: diagram linked_question: Q18 description: A metal rod of length L = 0.40 m moving with velocity v = 5.0 m/s perpendicular to a uniform magnetic field B = 0.25 T directed into the page. The rod is on two parallel conducting rails. labels: Rod length L = 0.40 m, Velocity v = 5.0 m/s (right), Magnetic field B = 0.25 T (into page), Conducting rails values: L = 0.40 m, v = 5.0 m/s, B = 0.25 T must_show: Rod on rails, velocity direction, magnetic field direction (into page), length of rod </image_placeholder>

(a) Calculate the emf induced across the ends of the rod.
[2]

(b) Explain, in terms of the motion of charges in the rod, how this emf is produced.
[3]

(c) The rod is now moved at the same speed but at an angle of $60^\circ$ to the magnetic field. Calculate the new induced emf.
[2]

19. A transformer has 1000 turns on its primary coil and 200 turns on its secondary coil. The primary coil is connected to a $240 \text{ V}$ a.c. supply.

(a) Calculate the secondary voltage.
[2]

(b) The transformer is $90\%$ efficient. If the primary current is $0.50 \text{ A}$ , calculate the secondary current.
[3]

20. A small circular coil of 100 turns and radius $0.020 \mathrm{~m}$ is placed at the centre of a long solenoid of 500 turns per metre. The current in the solenoid increases uniformly from $0$ to $2.0 \mathrm{~A}$ in $0.50 \mathrm{~s}$ .

<image_placeholder> id: Q20-fig1 type: diagram linked_question: Q20 description: A small circular coil (N = 100 turns, radius r = 0.020 m) placed at the centre of a long solenoid (n = 500 turns per metre). The solenoid carries a current I that increases from 0 to 2.0 A. labels: Small coil: N = 100, r = 0.020 m; Solenoid: n = 500 turns/m, I = 0 → 2.0 A in 0.50 s values: N_coil = 100, r = 0.020 m, n = 500 turns/m, ΔI = 2.0 A, Δt = 0.50 s must_show: Solenoid with turns, small coil at centre, dimensions labelled, current direction </image_placeholder>

(a) Calculate the magnetic flux density inside the solenoid when the current is $2.0 \mathrm{~A}$ .
(Take $\mu_0 = 4\pi \times 10^{-7} \mathrm{~T\,m\,A^{-1}}$ )
[2]

(b) Calculate the magnetic flux through the small coil.
[2]

(d) State the direction of the induced current in the small coil, explaining your answer using Lenz's Law.
[2]

END OF QUIZ

Answers

A-Level Physics H2 Quiz - Electricity Magnetism

Answer Key

Section A: Electric Fields and Current Electricity

Question 1 [3 marks]

Coulomb's Law states that the magnitude of the electrostatic force between two point charges is directly proportional to the product of the magnitudes of the charges and inversely proportional to the square of the distance between them.

Mathematical expression: $F = \frac{1}{4\pi\varepsilon_0} \frac{Q_1 Q_2}{r^2}$

Symbols:

$F$ = electrostatic force (N)
$Q_1, Q_2$ = magnitudes of the point charges (C)
$r$ = distance between the charges (m)
$\varepsilon_0$ = permittivity of free space ( $8.85 \times 10^{-12} \text{ F m}^{-1}$ )

Marking: 1 mark for correct statement, 1 mark for correct equation, 1 mark for defining all symbols.

Question 2 [7 marks]

(a) [2 marks]

Using Coulomb's Law: $F = \frac{1}{4\pi\varepsilon_0} \frac{|Q_1 Q_2|}{r^2}$

$F = \frac{(8.99 \times 10^9) \times (4.0 \times 10^{-6}) \times (6.0 \times 10^{-6})}{(0.30)^2}$

$F = \frac{(8.99 \times 10^9) \times (2.4 \times 10^{-11})}{0.090}$

$F = \frac{0.21576}{0.090}$

$\boxed{F = 2.4 \text{ N}}$

Marking: 1 mark for correct substitution, 1 mark for correct answer.

(b) [1 mark]

The force is attractive because the charges are of opposite sign (one positive, one negative). Opposite charges attract.

(c) [4 marks]

At the midpoint, $r = 0.15 \text{ m}$ from each charge.

Force on $Q_3$ due to $Q_1$ (repulsive, away from $Q_1$ ): $F_1 = \frac{(8.99 \times 10^9) \times (4.0 \times 10^{-6}) \times (2.0 \times 10^{-6})}{(0.15)^2} = \frac{0.07192}{0.0225} = 3.20 \text{ N (towards } Q_2\text{)}$

Force on $Q_3$ due to $Q_2$ (attractive, towards $Q_2$ ): $F_2 = \frac{(8.99 \times 10^9) \times (6.0 \times 10^{-6}) \times (2.0 \times 10^{-6})}{(0.15)^2} = \frac{0.10788}{0.0225} = 4.79 \text{ N (towards } Q_2\text{)}$

Both forces act in the same direction (towards $Q_2$ ): $F_{\text{resultant}} = F_1 + F_2 = 3.20 + 4.79 = \boxed{7.99 \text{ N towards } Q_2}$

Marking: 1 mark for each force calculation, 1 mark for correct resultant direction and magnitude.

Question 3 [7 marks]

(a) [2 marks]

$E = \frac{V}{d} = \frac{200}{0.040} = \boxed{5000 \text{ V m}^{-1}}$

(b) [2 marks]

$F = qE = (3.2 \times 10^{-19}) \times 5000 = \boxed{1.6 \times 10^{-15} \text{ N}}$

Direction: downward (same direction as field, since charge is positive).

(c) [3 marks]

Using conservation of energy: $\text{Work done by field} = \text{Kinetic energy gained}$

$qV = \frac{1}{2}mv^2$

$(3.2 \times 10^{-19}) \times 200 = \frac{1}{2} \times (6.6 \times 10^{-26}) \times v^2$

$6.4 \times 10^{-17} = 3.3 \times 10^{-26} \times v^2$

$v^2 = \frac{6.4 \times 10^{-17}}{3.3 \times 10^{-26}} = 1.94 \times 10^9$

$\boxed{v = 4.4 \times 10^4 \text{ m s}^{-1}}$

Marking: 1 mark for energy equation, 1 mark for substitution, 1 mark for correct answer.

Question 4 [3 marks]

Electric potential at a point in an electric field is defined as the work done per unit positive charge in bringing a small test charge from infinity to that point.

$V = \frac{W}{q}$

Difference from electric potential energy:

Electric potential energy ( $U$ ) is the total energy a charge possesses due to its position in an electric field. It depends on both the field and the charge: $U = qV$ .
Electric potential ( $V$ ) is a property of the field itself at a point, independent of any test charge placed there. It is the potential energy per unit charge.

Marking: 1 mark for definition of potential, 1 mark for definition of potential energy, 1 mark for clear distinction.

Question 5 [9 marks]

(a) [2 marks]

The variable resistor is necessary to:

Vary the current through the lamp and the voltage across it, allowing multiple readings to be taken.
Limit the current to prevent damage to the lamp or components.

(b)(i) [3 marks]

Graph should show:

Correctly labelled axes ( $I$ on y-axis, $V$ on x-axis)
Appropriate scale
All 7 points plotted correctly
A smooth curve (not a straight line) showing increasing gradient

(b)(ii) [2 marks]

At $V = 4.0 \text{ V}$ , $I = 0.37 \text{ A}$ :

$R = \frac{V}{I} = \frac{4.0}{0.37} = \boxed{10.8 \text{ \Omega}}$

(Alternatively, from gradient of tangent at that point: approximately $11 \text{ \Omega}$ )

(b)(iii) [2 marks]

As voltage increases, the current increases, causing the filament to heat up. The increased temperature causes the metal ions in the filament to vibrate more, increasing the frequency of collisions between conduction electrons and the lattice. This increases the resistance of the filament.

Marking: 1 mark for temperature effect, 1 mark for explanation of resistance increase mechanism.

Section B: D.C. Circuits

Question 6 [2 marks]

Kirchhoff's First Law (Junction Rule): The algebraic sum of currents at a junction in a circuit is zero.

$\sum I_{\text{in}} = \sum I_{\text{out}}$

Physical principle: Conservation of electric charge. Charge cannot accumulate at a junction; whatever charge flows in must flow out.

Marking: 1 mark for statement, 1 mark for principle.

Question 7 [4 marks]

First, find the equivalent resistance of the parallel combination: $\frac{1}{R_{\text{parallel}}} = \frac{1}{6.0} + \frac{1}{3.0} = \frac{1}{6.0} + \frac{2}{6.0} = \frac{3}{6.0}$

$R_{\text{parallel}} = 2.0 \text{ \Omega}$

Total resistance: $R_{\text{total}} = R_1 + R_{\text{parallel}} = 4.0 + 2.0 = 6.0 \text{ \Omega}$

Total current: $I_{\text{total}} = \frac{V}{R_{\text{total}}} = \frac{12.0}{6.0} = 2.0 \text{ A}$

Voltage across parallel combination: $V_{\text{parallel}} = I_{\text{total}} \times R_{\text{parallel}} = 2.0 \times 2.0 = 4.0 \text{ V}$

Current through $6.0 \text{ \Omega}$ resistor: $I_{6\Omega} = \frac{V_{\text{parallel}}}{6.0} = \frac{4.0}{6.0} = \boxed{0.67 \text{ A}}$

Marking: 1 mark for parallel resistance, 1 mark for total current, 1 mark for voltage across parallel, 1 mark for final answer.

Question 8 [6 marks]

(a) [2 marks]

Total resistance: $R_{\text{total}} = 0.20 + 2.80 = 3.00 \text{ \Omega}$

Current: $I = \frac{\mathcal{E}}{R_{\text{total}}} = \frac{1.50}{3.00} = 0.50 \text{ A}$

Terminal p.d.: $V = \mathcal{E} - Ir = 1.50 - (0.50 \times 0.20) = 1.50 - 0.10 = \boxed{1.40 \text{ V}}$

(b) [2 marks]

$P = I^2 R = (0.50)^2 \times 2.80 = 0.25 \times 2.80 = \boxed{0.70 \text{ W}}$

(c) [2 marks]

When external resistance decreases:

Total current increases
Greater voltage drop across internal resistance ( $Ir$ increases)
Terminal p.d. ( $V = \mathcal{E} - Ir$ ) decreases

Marking: 1 mark for each correct explanation point.

Question 9 [7 marks]

(a) [2 marks]

$V_{\text{out}} = \frac{R_2}{R_1 + R_2} \times \mathcal{E} = \frac{6.0}{3.0 + 6.0} \times 9.0 = \frac{6.0}{9.0} \times 9.0 = \boxed{6.0 \text{ V}}$

(b) [3 marks]

Equivalent resistance of $R_2$ and $R_L$ in parallel: $\frac{1}{R_{\text{eq}}} = \frac{1}{6.0} + \frac{1}{12.0} = \frac{2}{12.0} + \frac{1}{12.0} = \frac{3}{12.0}$

$R_{\text{eq}} = 4.0 \text{ k\Omega}$

New output voltage: $V_{\text{out}} = \frac{R_{\text{eq}}}{R_1 + R_{\text{eq}}} \times \mathcal{E} = \frac{4.0}{3.0 + 4.0} \times 9.0 = \frac{4.0}{7.0} \times 9.0 = \boxed{5.14 \text{ V}}$

(c) [2 marks]

The load resistor reduces the effective resistance of the lower part of the potential divider. This causes a greater proportion of the voltage to be dropped across $R_1$ , reducing the output voltage.

Marking: 1 mark for equivalent resistance, 1 mark for correct calculation, 1 mark for explanation.

Question 10 [6 marks]

(a) [2 marks]

Parallel combination of $R_2$ and $R_3$ : $R_{\text{parallel}} = \frac{4.0 \times 4.0}{4.0 + 4.0} = \frac{16.0}{8.0} = 2.0 \text{ \Omega}$

Total external resistance: $R_{\text{ext}} = R_1 + R_{\text{parallel}} = 2.0 + 2.0 = \boxed{4.0 \text{ \Omega}}$

(b) [2 marks]

Total circuit resistance: $R_{\text{total}} = 4.0 + 1.0 = 5.0 \text{ \Omega}$

$I = \frac{\mathcal{E}}{R_{\text{total}}} = \frac{12.0}{5.0} = \boxed{2.4 \text{ A}}$

(c) [2 marks]

$P = I^2 r = (2.4)^2 \times 1.0 = 5.76 \times 1.0 = \boxed{5.76 \text{ W}}$

Section C: Electromagnetism

Question 11 [2 marks]

Magnetic flux density ( $B$ ) is defined as the force per unit length per unit current on a straight conductor placed perpendicular to a uniform magnetic field.

$B = \frac{F}{IL}$

SI unit: Tesla (T), where $1 \text{ T} = 1 \text{ N A}^{-1} \text{ m}^{-1}$

Marking: 1 mark for definition, 1 mark for unit.

Question 12 [6 marks]

(a) [2 marks]

$F = BIL\sin\theta = 0.20 \times 4.0 \times 0.50 \times \sin 90^\circ = 0.20 \times 4.0 \times 0.50 \times 1 = \boxed{0.40 \text{ N}}$

(b) [2 marks]

$F = BIL\sin\theta = 0.20 \times 4.0 \times 0.50 \times \sin 30^\circ = 0.40 \times 0.50 = \boxed{0.20 \text{ N}}$

(c) [2 marks]

The force can be increased by:

Increasing the current in the wire
Increasing the magnetic flux density
Increasing the length of the wire in the field
Orienting the wire perpendicular to the field ( $\theta = 90^\circ$ )

(Any two valid methods, 1 mark each)

Question 13 [6 marks]

(a) [2 marks]

Area of one turn: $A = 0.040 \times 0.060 = 2.4 \times 10^{-3} \text{ m}^2$

Since the plane is parallel to the field, the normal to the coil is perpendicular to $B$ : $\Phi = BA\cos\theta = 0.15 \times (2.4 \times 10^{-3}) \times \cos 90^\circ = \boxed{0 \text{ Wb}}$

(b) [3 marks]

$\tau = BANI\sin\theta$

where $\theta$ is the angle between the normal to the coil and the field. Since the plane is parallel to $B$ , the normal is perpendicular to $B$ , so $\theta = 90^\circ$ :

$\tau = 0.15 \times (2.4 \times 10^{-3}) \times 50 \times 2.0 \times \sin 90^\circ$

$\tau = 0.15 \times 2.4 \times 10^{-3} \times 50 \times 2.0 \times 1 = \boxed{0.036 \text{ N m}}$

(c) [1 mark]

If the number of turns is doubled, the torque is also doubled (since $\tau \propto N$ ).

Question 14 [7 marks]

(a) [2 marks]

$F = qvB\sin\theta = (1.60 \times 10^{-19}) \times (3.0 \times 10^6) \times 0.050 \times \sin 90^\circ$

$F = 1.60 \times 10^{-19} \times 3.0 \times 10^6 \times 0.050 = \boxed{2.4 \times 10^{-14} \text{ N}}$

(b) [3 marks]

The magnetic force provides the centripetal force for circular motion: $qvB = \frac{mv^2}{r}$

$r = \frac{mv}{qB} = \frac{(9.11 \times 10^{-31}) \times (3.0 \times 10^6)}{(1.60 \times 10^{-19}) \times 0.050}$

$r = \frac{2.733 \times 10^{-24}}{8.0 \times 10^{-21}} = \boxed{3.42 \times 10^{-4} \text{ m}}$

(c) [2 marks]

$T = \frac{2\pi r}{v} = \frac{2\pi \times 3.42 \times 10^{-4}}{3.0 \times 10^6} = \frac{2.148 \times 10^{-3}}{3.0 \times 10^6} = \boxed{7.16 \times 10^{-10} \text{ s}}$

Question 15 [6 marks]

(a) [2 marks]

$B = \frac{\mu_0 I_P}{2\pi r} = \frac{(4\pi \times 10^{-7}) \times 5.0}{2\pi \times 0.10} = \frac{2.0 \times 10^{-6}}{0.10} = \boxed{2.0 \times 10^{-5} \text{ T}}$

(b) [2 marks]

$\frac{F}{L} = B I_Q = (2.0 \times 10^{-5}) \times 8.0 = \boxed{1.6 \times 10^{-4} \text{ N m}^{-1}}$

(c) [2 marks]

The force is attractive. Parallel currents in the same direction produce attractive forces between wires. This can be verified using Fleming's Left-Hand Rule: the magnetic field from wire P at wire Q is directed into the page, and with current upward in wire Q, the force is towards wire P.

Section D: Electromagnetic Induction

Question 16 [4 marks]

Faraday's Law: The magnitude of the induced emf in a circuit is directly proportional to the rate of change of magnetic flux linkage through the circuit.

$\mathcal{E} = -\frac{d(N\Phi)}{dt}$

Lenz's Law: The direction of the induced current is such that it opposes the change in magnetic flux that produced it.

Relation: The negative sign in Faraday's Law represents Lenz's Law. It indicates that the induced emf (and hence induced current) acts in a direction that opposes the change in flux linkage, ensuring conservation of energy.

Marking: 1 mark for Faraday's Law statement, 1 mark for equation, 1 mark for Lenz's Law, 1 mark for explanation of relationship.

Question 17 [6 marks]

(a) [2 marks]

Initial flux linkage (plane perpendicular to field, so $\theta = 0^\circ$ ): $N\Phi = NBA\cos\theta = 200 \times 0.30 \times 0.020 \times \cos 0^\circ = 200 \times 0.30 \times 0.020 \times 1 = \boxed{1.2 \text{ Wb}}$

(b) [3 marks]

Final flux linkage (plane parallel to field, so $\theta = 90^\circ$ ): $N\Phi_{\text{final}} = NBA\cos 90^\circ = 0$

Average emf: $\mathcal{E} = -\frac{\Delta(N\Phi)}{\Delta t} = -\frac{0 - 1.2}{0.10} = \frac{1.2}{0.10} = \boxed{12 \text{ V}}$

(c) [1 mark]

If the rotation is carried out more slowly, the rate of change of flux linkage is smaller, so the induced emf would be smaller (since $\mathcal{E} \propto \frac{d\Phi}{dt}$ ).

Question 18 [7 marks]

(a) [2 marks]

$\mathcal{E} = BLv = 0.25 \times 0.40 \times 5.0 = \boxed{0.50 \text{ V}}$

(b) [3 marks]

As the rod moves through the magnetic field, the free electrons in the rod also move with velocity $v$ . These moving charges experience a magnetic force $F = qvB$ (Fleming's Left-Hand Rule). The force on positive charges is in one direction, causing a separation of charges. This creates an electric field within the rod. Equilibrium is reached when the electric force equals the magnetic force, resulting in a potential difference (emf) across the ends of the rod.

(c) [2 marks]

$\mathcal{E} = BLv\sin\theta = 0.25 \times 0.40 \times 5.0 \times \sin 60^\circ = 0.50 \times 0.866 = \boxed{0.43 \text{ V}}$

Question 19 [8 marks]

(a) [2 marks]

$\frac{V_S}{V_P} = \frac{N_S}{N_P}$

$V_S = V_P \times \frac{N_S}{N_P} = 240 \times \frac{200}{1000} = 240 \times 0.20 = \boxed{48 \text{ V}}$

(b) [3 marks]

Input power: $P_{\text{in}} = V_P I_P = 240 \times 0.50 = 120 \text{ W}$

Output power: $P_{\text{out}} = \eta \times P_{\text{in}} = 0.90 \times 120 = 108 \text{ W}$

$I_S = \frac{P_{\text{out}}}{V_S} = \frac{108}{48} = \boxed{2.25 \text{ A}}$

(c) [3 marks]

Sources of energy loss:

Copper losses ( $I^2R$ losses): Resistance of the windings causes heating. Minimised by using thick copper wire with low resistance.
Eddy current losses: Changing magnetic flux induces eddy currents in the core, causing heating. Minimised by using a laminated core (thin insulated sheets).
Hysteresis losses: Repeated magnetisation and demagnetisation of the core causes energy loss. Minimised by using soft magnetic materials (e.g., soft iron) with low hysteresis.

(Any two sources with minimisation methods, 1.5 marks each)

Question 20 [9 marks]

(a) [2 marks]

$B = \mu_0 n I = (4\pi \times 10^{-7}) \times 500 \times 2.0 = 4\pi \times 10^{-7} \times 1000 = \boxed{1.26 \times 10^{-3} \text{ T}}$

(b) [2 marks]

Area of small coil: $A = \pi r^2 = \pi \times (0.020)^2 = 1.257 \times 10^{-3} \text{ m}^2$

$\Phi = BA = (1.26 \times 10^{-3}) \times (1.257 \times 10^{-3}) = \boxed{1.58 \times 10^{-6} \text{ Wb}}$

(c) [3 marks]

Initial flux linkage: $N\Phi_{\text{initial}} = 0$ (since $I = 0$ )

Final flux linkage: $N\Phi_{\text{final}} = 100 \times 1.58 \times 10^{-6} = 1.58 \times 10^{-4} \text{ Wb}$

$\mathcal{E} = -\frac{\Delta(N\Phi)}{\Delta t} = -\frac{1.58 \times 10^{-4} - 0}{0.50} = \boxed{3.16 \times 10^{-4} \text{ V}}$

(d) [2 marks]

The current in the solenoid is increasing, so the magnetic flux through the small coil is increasing. By Lenz's Law, the induced current in the small coil will flow in a direction that opposes this increase. Therefore, the induced current will create a magnetic field in the opposite direction to the solenoid's field. Using the right-hand grip rule, the induced current flows in the opposite direction to the solenoid current.

END OF ANSWER KEY