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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: _______ / 50

Duration: 60 minutes
Total Marks: 50

Instructions:

Answer all questions.
Write your answers in the spaces provided.
Show all working clearly. Marks may be awarded for correct working even if the final answer is incorrect.
Use $g = 9.81 \text{ m s}^{-2}$ where applicable.
The use of an approved scientific calculator is expected.

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

1. Two point charges, $+Q$ and $-4Q$ , are fixed at a distance $d$ apart in a vacuum. (a) Sketch the electric field lines for this system. Indicate the direction of the field. [2]

(b) State the location, relative to the charges, where the electric potential is zero. Explain your reasoning. [2]

2. An electron is accelerated from rest through a potential difference of $2.5 \text{ kV}$ . (a) Calculate the kinetic energy gained by the electron in Joules. [2]

(b) Hence, calculate the final speed of the electron. (Mass of electron $m_e = 9.11 \times 10^{-31} \text{ kg}$ ) [2]

3. Define electric field strength $E$ at a point in an electric field. [1]

4. Two parallel horizontal metal plates are separated by a distance of $4.0 \text{ cm}$ . A potential difference of $1.2 \text{ kV}$ is applied across them. (a) Calculate the magnitude of the uniform electric field between the plates. [1]

(b) A charged oil drop of mass $3.0 \times 10^{-15} \text{ kg}$ remains stationary between the plates. Determine the magnitude of the charge on the oil drop. [3]

5. The electric potential $V$ at a distance $r$ from a point charge $Q$ is given by $V = \frac{Q}{4\pi\epsilon_0 r}$ . Explain why the electric field strength $E$ is equal to the negative gradient of the potential, i.e., $E = -\frac{dV}{dr}$ . [2]

Section B: Current Electricity and D.C. Circuits (Questions 6–12)

6. A copper wire of length $L$ and cross-sectional area $A$ has a resistance $R$ . If the wire is stretched to twice its original length ( $2L$ ) while maintaining constant volume, determine the new resistance in terms of $R$ . [2]

7. State Kirchhoff’s Second Law. [1]

8. A battery has an electromotive force (e.m.f.) of $12.0 \text{ V}$ and an internal resistance of $2.0 \, \Omega$ . It is connected to a variable resistor $R$ . (a) Calculate the current in the circuit when $R = 4.0 \, \Omega$ . [2]

(b) Calculate the power dissipated in the variable resistor $R$ for this value. [2]

9. The graph below shows the variation of terminal potential difference $V$ with current $I$ for a real battery.

Image pending generation for this question.

(Imagine a linear graph starting at $V=12$ V on the y-axis and ending at $I=6$ A on the x-axis)

(a) Determine the e.m.f. of the battery. [1]

(b) Determine the internal resistance of the battery. [2]

10. In the circuit shown, a thermistor $T$ is connected in series with a fixed resistor $R_1 = 1.0 \text{ k}\Omega$ across a $10 \text{ V}$ supply. The output voltage $V_{out}$ is taken across the thermistor.

(a) The resistance of the thermistor decreases as temperature increases. State and explain the effect on $V_{out}$ as the temperature rises. [2]

  



(b) At $25^\circ\text{C}$, the resistance of the thermistor is $2.0 \text{ k}\Omega$. Calculate $V_{out}$. [2]

11. A potential divider circuit is used to provide a variable voltage from $0$ to $6.0 \text{ V}$ using a $6.0 \text{ V}$ supply. Explain why a potentiometer (rheostat connected as a potential divider) is preferred over a fixed resistor in series with a variable resistor for this purpose. [2]

12. Two resistors, $R_1 = 6.0 \, \Omega$ and $R_2 = 3.0 \, \Omega$ , are connected in parallel. This combination is connected in series with a $4.0 \, \Omega$ resistor and a $12 \text{ V}$ battery of negligible internal resistance. (a) Calculate the total resistance of the circuit. [2]

(b) Calculate the current flowing through the $3.0 \, \Omega$ resistor. [3]

Section C: Electromagnetism and Induction (Questions 13–20)

13. A proton moves with velocity $v$ into a uniform magnetic field of flux density $B$ . The velocity is perpendicular to the magnetic field. (a) State the direction of the magnetic force acting on the proton relative to $v$ and $B$ . [1]

(b) Explain why the kinetic energy of the proton remains constant while it is in the magnetic field. [2]

14. A straight wire of length $0.50 \text{ m}$ carries a current of $3.0 \text{ A}$ . It is placed in a uniform magnetic field of flux density $0.40 \text{ T}$ . Calculate the magnitude of the force on the wire when the angle between the wire and the magnetic field is: (a) $90^\circ$ [1]

(b) $30^\circ$ [2]

15. State Faraday’s Law of Electromagnetic Induction. [2]

16. A coil of area $0.02 \text{ m}^2$ with $50$ turns is placed in a magnetic field of flux density $0.5 \text{ T}$ . The plane of the coil is perpendicular to the field. (a) Calculate the magnetic flux linkage through the coil. [2]

(b) The coil is rotated through $90^\circ$ in $0.1 \text{ s}$ so that its plane is parallel to the field. Calculate the average induced e.m.f. during this rotation. [3]

17. A transformer has a primary coil with $1000$ turns and a secondary coil with $50$ turns. (a) If the primary voltage is $240 \text{ V}$ (a.c.), calculate the secondary voltage. [2]

(b) State one reason why real transformers are not $100\%$ efficient. [1]

18. An electron beam enters a region where there is both a uniform electric field $E$ and a uniform magnetic field $B$ . The fields are perpendicular to each other and to the initial velocity of the electrons. Derive the expression for the velocity $v$ of electrons that pass through this region undeflected. [3]

19. A magnet is dropped through a long vertical copper tube. Explain, using Lenz’s Law, why the magnet reaches a terminal velocity rather than accelerating continuously under gravity. [3]

20. A rectangular coil rotates with constant angular velocity $\omega$ in a uniform magnetic field. Sketch a graph of the induced e.m.f. $\varepsilon$ against time $t$ for one complete rotation, starting from the position where the plane of the coil is perpendicular to the magnetic field. [2]

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Answers

A-Level Physics H2 Quiz - Electricity Magnetism (Answer Key)

1. (a) Sketch: Lines originate from $+Q$ and terminate on $-4Q$ . Lines are denser near $-4Q$ indicating stronger field. At least 4-6 lines drawn. Arrows pointing away from $+Q$ and towards $-4Q$ . [2] (b) Location: On the line joining the charges, closer to $+Q$ (outside the region between them). [1] Reasoning: Potential is a scalar. $V_{total} = V_+ + V_- = \frac{kQ}{r_1} - \frac{k(4Q)}{r_2}$ . For $V=0$ , $\frac{Q}{r_1} = \frac{4Q}{r_2} \Rightarrow r_2 = 4r_1$ . The point must be closer to the smaller magnitude charge to balance the potential. [1]

2. (a) $KE = qV = (1.60 \times 10^{-19} \text{ C})(2500 \text{ V}) = 4.0 \times 10^{-16} \text{ J}$ . [2] (b) $KE = \frac{1}{2}mv^2 \Rightarrow v = \sqrt{\frac{2KE}{m}}$ $v = \sqrt{\frac{2(4.0 \times 10^{-16})}{9.11 \times 10^{-31}}} = \sqrt{8.78 \times 10^{14}} \approx 2.96 \times 10^7 \text{ m s}^{-1}$ . [2]

3. Electric field strength is the electric force experienced per unit positive charge placed at that point. ( $E = F/q$ ) [1]

4. (a) $E = \frac{V}{d} = \frac{1200}{0.04} = 30,000 \text{ V m}^{-1}$ (or $\text{N C}^{-1}$ ). [1] (b) For stationary drop, Electric Force = Weight. $qE = mg \Rightarrow q = \frac{mg}{E}$ $q = \frac{(3.0 \times 10^{-15})(9.81)}{30000} = \frac{2.943 \times 10^{-14}}{30000} \approx 9.81 \times 10^{-19} \text{ C}$ . [3]

5. Work done moving a unit positive charge against the field is equal to the increase in potential. $dW = -F dx = -qE dx$ . Also $dW = q dV$ . Therefore, $q dV = -qE dx \Rightarrow E = -\frac{dV}{dx}$ . The negative sign indicates the field points in the direction of decreasing potential. [2]

6. $R = \rho \frac{L}{A}$ . Volume $V_{ol} = L \times A$ is constant. New length $L' = 2L$ . New area $A' = \frac{V_{ol}}{2L} = \frac{A}{2}$ . $R' = \rho \frac{2L}{A/2} = 4 \rho \frac{L}{A} = 4R$ . [2]

7. The sum of the electromotive forces in any closed loop is equal to the sum of the potential differences (voltage drops) across the components in that loop. (Or: The algebraic sum of changes in potential around any closed circuit loop is zero). [1]

8. (a) $I = \frac{\mathcal{E}}{R + r} = \frac{12.0}{4.0 + 2.0} = \frac{12.0}{6.0} = 2.0 \text{ A}$ . [2] (b) $P = I^2 R = (2.0)^2 (4.0) = 16 \text{ W}$ . [2]

9. (a) Intercept on V-axis (when $I=0$ ) is the e.m.f. $\mathcal{E} = 12 \text{ V}$ . [1] (b) Gradient magnitude $= \frac{\Delta V}{\Delta I} = \frac{12 - 0}{6 - 0} = 2.0 \, \Omega$ . Internal resistance $r = 2.0 \, \Omega$ . [2]

10. (a) As temperature rises, $R_T$ decreases. The total resistance decreases, current increases. However, using the potential divider rule $V_{out} = V_{in} \frac{R_T}{R_T + R_1}$ , as $R_T$ decreases relative to $R_1$ , the fraction decreases. Thus, $V_{out}$ decreases. [2] (b) $V_{out} = 10 \times \frac{2000}{2000 + 1000} = 10 \times \frac{2}{3} = 6.67 \text{ V}$ . [2]

11. A potentiometer allows the output voltage to be varied continuously from $0 \text{ V}$ (when the slider is at the ground end) to the full supply voltage. A series variable resistor cannot reduce the voltage across the load to zero (unless the load resistance is zero) and has a non-linear control range depending on load. [2]

12. (a) Parallel combination $R_p$ : $\frac{1}{R_p} = \frac{1}{6} + \frac{1}{3} = \frac{1}{6} + \frac{2}{6} = \frac{3}{6} \Rightarrow R_p = 2.0 \, \Omega$ . Total Resistance $R_{tot} = R_p + 4.0 = 2.0 + 4.0 = 6.0 \, \Omega$ . [2] (b) Total Current $I_{tot} = \frac{V}{R_{tot}} = \frac{12}{6.0} = 2.0 \text{ A}$ . Voltage across parallel section $V_p = I_{tot} R_p = 2.0 \times 2.0 = 4.0 \text{ V}$ . Current through $3.0 \, \Omega$ resistor: $I_3 = \frac{V_p}{3.0} = \frac{4.0}{3.0} = 1.33 \text{ A}$ . [3]

13. (a) Perpendicular to both the velocity vector and the magnetic field vector (determined by Fleming’s Left Hand Rule). [1] (b) The magnetic force is always perpendicular to the direction of motion (velocity). Therefore, the work done by the magnetic force is zero ( $W = F \cdot d \cos 90^\circ = 0$ ). Since no work is done, the kinetic energy does not change. [2]

14. (a) $F = BIL \sin 90^\circ = 0.40 \times 3.0 \times 0.50 \times 1 = 0.60 \text{ N}$ . [1] (b) $F = BIL \sin 30^\circ = 0.40 \times 3.0 \times 0.50 \times 0.5 = 0.30 \text{ N}$ . [2]

15. The magnitude of the induced e.m.f. is proportional to the rate of change of magnetic flux linkage through the circuit. ( $\mathcal{E} = -\frac{d(N\Phi)}{dt}$ ). [2]

16. (a) Flux $\Phi = BA = 0.5 \times 0.02 = 0.01 \text{ Wb}$ . Flux Linkage $= N\Phi = 50 \times 0.01 = 0.50 \text{ Wb turns}$ . [2] (b) Change in flux linkage $\Delta(N\Phi) = N\Phi_{final} - N\Phi_{initial}$ . Final position (parallel): $\Phi = 0$ . Initial: $0.50 \text{ Wb turns}$ . $\Delta(N\Phi) = 0 - 0.50 = -0.50 \text{ Wb turns}$ . Magnitude of induced e.m.f. $|\mathcal{E}| = \left| \frac{\Delta(N\Phi)}{\Delta t} \right| = \frac{0.50}{0.1} = 5.0 \text{ V}$ . [3]

17. (a) $\frac{V_s}{V_p} = \frac{N_s}{N_p} \Rightarrow V_s = 240 \times \frac{50}{1000} = 240 \times 0.05 = 12 \text{ V}$ . [2] (b) Energy losses due to: heating of coils (resistance), eddy currents in the core, hysteresis in the core, or magnetic flux leakage. (Any one). [1]

18. For undeflected motion, the net force is zero. Electric force balances Magnetic force. $F_E = F_B$ $qE = Bqv$ (since $v \perp B$ ) $v = \frac{E}{B}$ . [3]

19. As the magnet falls, the changing magnetic flux through the copper tube induces eddy currents in the tube (Faraday's Law). [1] According to Lenz's Law, the direction of these induced currents creates a magnetic field that opposes the change causing it (the motion of the magnet). [1] This results in an upward magnetic force on the falling magnet. As speed increases, this opposing force increases until it equals the weight of the magnet, resulting in zero net force and constant terminal velocity. [1]

20. Graph: Sinusoidal wave. Starts at $\varepsilon = 0$ at $t=0$ (since flux is max, rate of change is zero). Reaches maximum positive peak at $T/4$ . Crosses zero at $T/2$ . Reaches maximum negative peak at $3T/4$ . Returns to zero at $T$ . [2]