A Level H2 Physics Electricity Magnetism Quiz

A-Level Physics H2 Quiz - Electricity Magnetism

Answer Key

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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.

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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.

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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.

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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.

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Question 5 [9 marks]

(a) [2 marks]

The variable resistor is necessary to:

1. Vary the current through the lamp and the voltage across it, allowing multiple readings to be taken.
2. 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.

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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.

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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.

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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.

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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.

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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]

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

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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.

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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:

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

(Any two valid methods, 1 mark each)

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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 $\tau \propto N$).

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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]

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

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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.

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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.

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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 $\mathcal{E} \propto \frac{d\Phi}{dt}$).

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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]

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

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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:

1. Copper losses ($I^2R$ losses): Resistance of the windings causes heating. Minimised by using thick copper wire with low resistance.

2. Eddy current losses: Changing magnetic flux induces eddy currents in the core, causing heating. Minimised by using a laminated core (thin insulated sheets).

3. 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)

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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.

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END OF ANSWER KEY