A Level H2 Physics Electricity Magnetism Quiz

A-Level Physics H2 Quiz - Electricity Magnetism — Answer Key

Total Marks: 50

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Section A: Electric Fields and Current Electricity (Questions 1–7)

1. State Coulomb's law for the force between two point charges. [2 marks]

Answer: The electrostatic force between two point charges is directly proportional to the product of the charges and inversely proportional to the square of the distance between them. [1 mark]

The force acts along the line joining the two charges. [1 mark]

Accept: $F = \frac{kQ_1Q_2}{r^2}$ or $F = \frac{Q_1Q_2}{4\pi\varepsilon_0 r^2}$ with correct identification of symbols.

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2. Two point charges, $Q_1 = +3.0 \times 10^{-6}$ C and $Q_2 = -4.0 \times 10^{-6}$ C, are placed 0.20 m apart in a vacuum.

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

Answer: $F = \frac{1}{4\pi\varepsilon_0} \frac{|Q_1 Q_2|}{r^2}$ [1 mark for formula]

$F = (8.99 \times 10^9) \times \frac{(3.0 \times 10^{-6})(4.0 \times 10^{-6})}{(0.20)^2}$

$F = 8.99 \times 10^9 \times \frac{1.2 \times 10^{-11}}{0.040}$

$F = 2.697 \approx 2.7$ N [1 mark for correct answer with unit]

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

Answer: Attractive. [0.5 mark] The charges have opposite signs (one positive, one negative). [0.5 mark]

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3. Define electric field strength at a point. [2 marks]

Answer: Electric field strength at a point is the force per unit positive charge [1 mark] acting on a small stationary test charge placed at that point. [1 mark]

Accept: $E = \frac{F}{q}$ with correct explanation of symbols.

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4. A small charged sphere of mass $2.0 \times 10^{-3}$ kg is suspended in equilibrium by a light insulating thread in a uniform horizontal electric field of strength $5.0 \times 10^{4}$ N C$^{-1}$. The thread makes an angle of $30^\circ$ with the vertical.

(a) Draw a labelled free-body diagram showing all the forces acting on the sphere. [2 marks]

Answer: Diagram should show:

• Weight $mg$ acting vertically downwards [0.5 mark]
• Tension $T$ acting along the thread [0.5 mark]
• Electric force $F_E = qE$ acting horizontally [0.5 mark]
• Correct labelling of all three forces [0.5 mark]

(b) Calculate the magnitude of the charge on the sphere. [3 marks]

Answer: Resolving forces horizontally and vertically: $T\sin 30^\circ = qE$ [1 mark] $T\cos 30^\circ = mg$ [1 mark]

Dividing: $\tan 30^\circ = \frac{qE}{mg}$

$q = \frac{mg \tan 30^\circ}{E} = \frac{(2.0 \times 10^{-3})(9.81)(\tan 30^\circ)}{5.0 \times 10^4}$

$q = \frac{(2.0 \times 10^{-3})(9.81)(0.577)}{5.0 \times 10^4} = 2.26 \times 10^{-7}$ C $\approx 2.3 \times 10^{-7}$ C [1 mark for correct answer with unit]

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5. A copper wire of length 2.0 m and cross-sectional area $1.5 \times 10^{-6}$ m$^2$ has a resistance of 0.023 $\Omega$. Calculate the resistivity of copper. [2 marks]

Answer: $R = \frac{\rho L}{A}$ so $\rho = \frac{RA}{L}$ [1 mark for formula]

$\rho = \frac{0.023 \times 1.5 \times 10^{-6}}{2.0}$

$\rho = 1.725 \times 10^{-8} \approx 1.7 \times 10^{-8}$ $\Omega$ m [1 mark for correct answer with unit]

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6. A filament lamp is rated at 60 W, 240 V.

(a) Calculate the current flowing through the lamp when it operates at its rated voltage. [1 mark]

Answer: $P = VI$ so $I = \frac{P}{V} = \frac{60}{240} = 0.25$ A [1 mark]

(b) Explain why the resistance of the filament lamp increases as the current through it increases. [2 marks]

Answer: As current increases, the filament temperature increases. [1 mark]

The increased thermal vibrations of the metal ions cause more frequent collisions with conduction electrons, increasing resistance. [1 mark]

Accept: Reference to increased lattice vibrations impeding electron flow.

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7. The current $I$ through a component varies with the potential difference $V$ across it according to the relationship $I = kV^2$, where $k$ is a constant.

(a) State whether the component obeys Ohm's law. Explain your answer. [1 mark]

Answer: No, it does not obey Ohm's law. [0.5 mark] Ohm's law requires current to be directly proportional to potential difference ($I \propto V$), but here $I \propto V^2$. [0.5 mark]

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Section B: D.C. Circuits (Questions 8–13)

8. A student sets up the circuit shown in Fig. 8.1 to investigate the characteristics of a component X.

(a) Identify the component labelled R and explain its function in this circuit. [2 marks]

Answer: R is a rheostat (or variable resistor/potential divider). [1 mark]

Its function is to vary the current in the circuit and/or the potential difference across component X, allowing measurements at different values. [1 mark]

(b) State how the voltmeters should be connected to measure the potential difference across component X and across component R. [1 mark]

Answer: Each voltmeter should be connected in parallel with (across) the respective component. [1 mark]

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9. Three resistors of resistances 4.0 $\Omega$, 6.0 $\Omega$, and 12.0 $\Omega$ are connected in parallel across a 12 V battery of negligible internal resistance.

(a) Calculate the total resistance of the parallel combination. [2 marks]

Answer: $\frac{1}{R_T} = \frac{1}{4.0} + \frac{1}{6.0} + \frac{1}{12.0}$ [1 mark for formula]

$\frac{1}{R_T} = 0.25 + 0.1667 + 0.0833 = 0.50$

$R_T = \frac{1}{0.50} = 2.0$ $\Omega$ [1 mark for correct answer with unit]

(b) Calculate the current supplied by the battery. [1 mark]

Answer: $I = \frac{V}{R_T} = \frac{12}{2.0} = 6.0$ A [1 mark]

(c) Determine the current flowing through the 6.0 $\Omega$ resistor. [2 marks]

Answer: In parallel, the potential difference across each resistor is 12 V. [1 mark]

$I_{6\Omega} = \frac{V}{R} = \frac{12}{6.0} = 2.0$ A [1 mark for correct answer with unit]

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10. A battery of e.m.f. 9.0 V and internal resistance 0.50 $\Omega$ is connected to an external resistor of resistance 4.0 $\Omega$.

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

Answer: Total resistance $R_T = 4.0 + 0.50 = 4.5$ $\Omega$ [0.5 mark]

Current $I = \frac{\mathcal{E}}{R_T} = \frac{9.0}{4.5} = 2.0$ A [0.5 mark]

Terminal p.d. $V = \mathcal{E} - Ir = 9.0 - (2.0 \times 0.50) = 8.0$ V [1 mark]

Alternative: $V = IR = 2.0 \times 4.0 = 8.0$ V

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

Answer: $P = I^2 R$ [1 mark for formula]

$P = (2.0)^2 \times 4.0 = 16$ W [1 mark for correct answer with unit]

Accept: $P = VI = 8.0 \times 2.0 = 16$ W or $P = \frac{V^2}{R} = \frac{8.0^2}{4.0} = 16$ W

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11. In the circuit shown in Fig. 11.1, the ammeter has negligible resistance and the voltmeter has infinite resistance.

(a) Calculate the reading on the ammeter. [2 marks]

Answer: The 3.0 $\Omega$ and 6.0 $\Omega$ resistors are in parallel. [0.5 mark]

$\frac{1}{R_P} = \frac{1}{3.0} + \frac{1}{6.0} = \frac{2}{6.0} + \frac{1}{6.0} = \frac{3}{6.0}$, so $R_P = 2.0$ $\Omega$ [0.5 mark]

$I = \frac{V}{R_P} = \frac{12}{2.0} = 6.0$ A [1 mark for correct answer with unit]

(b) A voltmeter is now connected across the 6.0 $\Omega$ resistor. Calculate the voltmeter reading. [2 marks]

Answer: The potential difference across the parallel combination is 12 V. [1 mark]

Therefore, the voltmeter reading = 12 V. [1 mark]

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Section C: Electromagnetism and Electromagnetic Induction (Questions 12–20)

12. State Faraday's law of electromagnetic induction. [2 marks]

Answer: The magnitude of the induced e.m.f. in a circuit is directly proportional to the rate of change of magnetic flux linkage through the circuit. [2 marks]

Accept: $\mathcal{E} = -\frac{d(N\Phi)}{dt}$ or $\mathcal{E} = -\frac{d\Phi}{dt}$ for a single turn, with correct identification of symbols. Award 1 mark for mentioning rate of change of flux; 2 marks for complete statement including flux linkage or number of turns.

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13. A straight wire of length 0.50 m moves with a constant velocity of 4.0 m s$^{-1}$ perpendicular to a uniform magnetic field of flux density 0.80 T. Calculate the magnitude of the e.m.f. induced across the ends of the wire. [2 marks]

Answer: $\mathcal{E} = BLv$ [1 mark for formula]

$\mathcal{E} = 0.80 \times 0.50 \times 4.0 = 1.6$ V [1 mark for correct answer with unit]

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14. State Lenz's law of electromagnetic induction. [2 marks]

Answer: The direction of the induced e.m.f. (or induced current) is such that it opposes the change in magnetic flux that produced it. [2 marks]

Award 1 mark for mentioning opposition; 2 marks for complete statement linking opposition to the change in flux.

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15. A bar magnet is dropped vertically through a horizontal circular coil connected to a sensitive centre-zero galvanometer.

(a) Describe and explain the deflection of the galvanometer as the magnet approaches the coil from above. [2 marks]

Answer: The galvanometer deflects in one direction (e.g., to the right). [1 mark]

As the magnet approaches, the magnetic flux through the coil increases. By Lenz's law, the induced current creates a magnetic field that opposes this increase, so the coil's upper face becomes a north pole to repel the approaching north pole. [1 mark]

(b) Describe and explain the deflection of the galvanometer as the magnet leaves the coil from below. [2 marks]

Answer: The galvanometer deflects in the opposite direction (e.g., to the left). [1 mark]

As the magnet leaves, the magnetic flux through the coil decreases. By Lenz's law, the induced current creates a magnetic field that opposes this decrease, so the coil's lower face becomes a south pole to attract the receding north pole. [1 mark]

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16. A proton moves with a speed of $3.0 \times 10^{6}$ m s$^{-1}$ perpendicular to a uniform magnetic field of flux density 0.50 T.

(a) State the direction of the magnetic force on the proton relative to its velocity and the magnetic field. [1 mark]

Answer: The magnetic force is perpendicular to both the velocity of the proton and the magnetic field. [1 mark]

Accept: The force is mutually perpendicular to v and B / given by Fleming's left-hand rule.

(b) Calculate the magnitude of the magnetic force on the proton. [2 marks]

Answer: $F = Bqv$ [1 mark for formula]

$F = 0.50 \times (1.60 \times 10^{-19}) \times (3.0 \times 10^6)$

$F = 2.4 \times 10^{-13}$ N [1 mark for correct answer with unit]

(c) Explain why the proton moves in a circular path. [2 marks]

Answer: The magnetic force is always perpendicular to the velocity of the proton. [1 mark]

This force provides the centripetal force required for circular motion, causing the proton to move in a circular path at constant speed. [1 mark]

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17. A coil of 200 turns and area $4.0 \times 10^{-3}$ m$^2$ is placed in a uniform magnetic field of flux density 0.60 T. The plane of the coil is perpendicular to the magnetic field. Calculate the magnetic flux linkage through the coil. [2 marks]

Answer: Flux through one turn: $\Phi = BA = 0.60 \times 4.0 \times 10^{-3} = 2.4 \times 10^{-3}$ Wb [1 mark]

Flux linkage: $N\Phi = 200 \times 2.4 \times 10^{-3} = 0.48$ Wb [1 mark for correct answer with unit]

Accept: Direct calculation $N\Phi = NBA = 200 \times 0.60 \times 4.0 \times 10^{-3} = 0.48$ Wb

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18. The magnetic flux through a coil of 50 turns changes uniformly from $8.0 \times 10^{-3}$ Wb to $2.0 \times 10^{-3}$ Wb in 0.30 s. Calculate the magnitude of the induced e.m.f. in the coil. [2 marks]

Answer: Change in flux linkage: $\Delta(N\Phi) = N(\Phi_f - \Phi_i) = 50 \times (2.0 \times 10^{-3} - 8.0 \times 10^{-3}) = 50 \times (-6.0 \times 10^{-3}) = -0.30$ Wb [1 mark for method]

Magnitude of induced e.m.f.: $|\mathcal{E}| = \left|\frac{\Delta(N\Phi)}{\Delta t}\right| = \frac{0.30}{0.30} = 1.0$ V [1 mark for correct answer with unit]

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19. Explain why the core of a transformer is laminated. [2 marks]

Answer: Lamination means the core is made of thin sheets of iron insulated from each other. [1 mark]

This reduces eddy currents induced in the core, thereby reducing energy losses due to heating. [1 mark]

Accept: Laminations increase the resistance to eddy currents, minimising power loss.

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20. A step-down transformer has 500 turns on its primary coil and 50 turns on its secondary coil. The primary coil is connected to a 240 V a.c. supply.

(a) Calculate the output voltage across the secondary coil, assuming the transformer is ideal. [2 marks]

Answer: $\frac{V_s}{V_p} = \frac{N_s}{N_p}$ [1 mark for formula]

$V_s = V_p \times \frac{N_s}{N_p} = 240 \times \frac{50}{500} = 24$ V [1 mark for correct answer with unit]

(b) The secondary coil supplies a current of 2.0 A to a load. Calculate the current in the primary coil, assuming the transformer is 100% efficient. [2 marks]

Answer: For an ideal transformer: $V_p I_p = V_s I_s$ [1 mark for formula]

$I_p = \frac{V_s I_s}{V_p} = \frac{24 \times 2.0}{240} = 0.20$ A [1 mark for correct answer with unit]

Alternative: $\frac{I_p}{I_s} = \frac{N_s}{N_p}$, so $I_p = 2.0 \times \frac{50}{500} = 0.20$ A

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END OF ANSWER KEY