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A Level H1 Physics Electricity Magnetism Quiz

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Questions

A-Level Physics H1 Quiz - Electricity Magnetism

Name: __________________________
Class: __________________________
Date: __________________________
Score: ______ / 50

Duration: 45 minutes
Total Marks: 50

Instructions:

Answer all questions.
Write your answers in the spaces provided.
Show all working clearly. Numerical answers should be given to an appropriate number of significant figures.
The use of a scientific calculator is allowed.

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

1. A copper wire has a length of 2.0 m and a cross-sectional area of $1.5 \times 10^{-6} \text{ m}^2$ . The resistivity of copper is $1.7 \times 10^{-8} \Omega \text{ m}$ . (a) Calculate the resistance of the wire.

Resistance = ____________________ $\Omega$ [2]

(b) The wire is stretched uniformly to twice its original length. Assuming the volume of the copper remains constant, state and explain the effect on its resistance.

_________________________________________________________________________ [2]

2. Define the electromotive force (e.m.f.) of a battery.

_________________________________________________________________________ [2]

3. A lamp is rated at 12 V, 24 W. (a) Calculate the current flowing through the lamp when it is operating at its rated power.

Current = ____________________ A [1]

(b) Calculate the resistance of the filament at this operating temperature.

Resistance = ____________________ $\Omega$ [1]

4. The graph below shows the variation of current $I$ with potential difference $V$ for a semiconductor diode.

Image pending generation for this question.

Explain why the diode does not obey Ohm’s Law.

_________________________________________________________________________ [2]

5. A potential divider circuit consists of a battery of e.m.f. 12 V and negligible internal resistance, connected in series with two resistors $R_1 = 4.0 \text{ k}\Omega$ and $R_2 = 2.0 \text{ k}\Omega$ . Calculate the potential difference across $R_2$ .

Potential Difference = ____________________ V [2]

Section B: D.C. Circuits and Kirchhoff’s Laws (Questions 6–10)

6. State Kirchhoff’s First Law.

_________________________________________________________________________ [1]

7. State Kirchhoff’s Second Law.

_________________________________________________________________________ [1]

8. In the circuit shown, a battery of e.m.f. $E$ and internal resistance $r$ is connected to a variable resistor $R$ .

(Diagram: Battery in series with Ammeter and Variable Resistor R. Voltmeter connected across the battery terminals.)

The terminal potential difference $V$ is measured for different values of current $I$ . The relationship is given by $V = E - Ir$ .

Explain how the e.m.f. $E$ and internal resistance $r$ can be determined from a graph of $V$ against $I$ .

_________________________________________________________________________ [2]

9. Two resistors of resistance $6.0 \Omega$ and $3.0 \Omega$ are connected in parallel. This combination is connected in series with a $4.0 \Omega$ resistor and a battery of e.m.f. 12 V and internal resistance $1.0 \Omega$ . (a) Calculate the total resistance of the external circuit.

Total Resistance = ____________________ $\Omega$ [2]

(b) Calculate the current supplied by the battery.

Current = ____________________ A [1]

10. In the circuit in Question 9, calculate the power dissipated in the $4.0 \Omega$ resistor.

Power = ____________________ W [2]

Section C: Electric Fields and Forces (Questions 11–15)

11. Define electric field strength at a point.

_________________________________________________________________________ [2]

12. An electron is placed in a uniform electric field of strength $5.0 \times 10^4 \text{ V m}^{-1}$ . (a) Calculate the magnitude of the electric force acting on the electron. (Charge of electron $e = 1.6 \times 10^{-19} \text{ C}$ )

Force = ____________________ N [1]

(b) State the direction of the force relative to the electric field lines.

Direction: _________________________________________________________ [1]

13. Two parallel metal plates are separated by a distance of 4.0 cm. A potential difference of 200 V is applied across them. Calculate the electric field strength between the plates, assuming it is uniform.

Electric Field Strength = ____________________ V m⁻¹ [2]

14. An oil drop of mass $3.0 \times 10^{-15} \text{ kg}$ is held stationary between two horizontal charged plates. The electric field strength between the plates is $2.0 \times 10^5 \text{ V m}^{-1}$ . (a) State the condition required for the oil drop to remain stationary.

_________________________________________________________________________ [1]

(b) Calculate the charge on the oil drop. ( $g = 9.81 \text{ m s}^{-2}$ )

Charge = ____________________ C [2]

15. Sketch the electric field pattern around a positive point charge. Include at least four field lines with arrows indicating direction.

[2]

Section D: Magnetic Fields and Electromagnetism (Questions 16–20)

16. Define magnetic flux density $B$ .

_________________________________________________________________________ [2]

17. A straight wire of length 0.50 m carries a current of 3.0 A. The wire is placed perpendicular to a uniform magnetic field of flux density 0.20 T. Calculate the magnitude of the force acting on the wire.

Force = ____________________ N [2]

18. An electron moves with a velocity of $2.0 \times 10^6 \text{ m s}^{-1}$ perpendicular to a magnetic field of flux density 0.10 T. (a) Calculate the magnitude of the magnetic force on the electron.

Force = ____________________ N [1]

(b) Describe the path of the electron in this magnetic field.

Path: _______________________________________________________________ [1]

19. Two long, straight, parallel wires X and Y carry currents in the same direction. (a) State whether the force between the wires is attractive or repulsive.

Force is: __________________________ [1]

(b) Explain this interaction in terms of magnetic fields.

_________________________________________________________________________ [2]

20. A rectangular coil rotates in a uniform magnetic field. State the orientation of the coil relative to the magnetic field when the magnetic flux linkage through the coil is: (a) Maximum

_________________________________________________________________________ [1]

(b) Zero

_________________________________________________________________________ [1]

End of Quiz

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Answers

A-Level Physics H1 Quiz - Electricity Magnetism (Answer Key)

1. (a) $R = \frac{\rho L}{A}$ [M1] $R = \frac{1.7 \times 10^{-8} \times 2.0}{1.5 \times 10^{-6}} = 0.0227 \Omega$ [A1] Answer: $0.023 \Omega$ (2 s.f.)

(b) Resistance increases by a factor of 4. [B1] Explanation: $R = \frac{\rho L}{A}$ . If length $L$ doubles, area $A$ halves (constant volume). New $R' = \frac{\rho (2L)}{(A/2)} = 4 \frac{\rho L}{A} = 4R$ . [B1]

2. E.m.f. is the energy converted from non-electrical forms (chemical, etc.) to electrical energy per unit charge passing through the source. [B1] Alternatively: The work done by the source in driving a unit charge around a complete circuit. [B1]

3. (a) $P = VI \Rightarrow I = P/V = 24/12 = 2.0$ A [A1] (b) $R = V/I = 12/2.0 = 6.0 \Omega$ [A1] (Or $R = V^2/P$ )

4. Ohm’s Law states that current is directly proportional to potential difference ( $I \propto V$ ) provided physical conditions (like temperature) remain constant, resulting in a constant resistance. [B1] For a diode, the graph is not a straight line through the origin; the resistance changes with voltage (non-linear). [B1]

5. $V_{out} = V_{in} \times \frac{R_2}{R_1 + R_2}$ [M1] $V_{out} = 12 \times \frac{2.0}{4.0 + 2.0} = 12 \times \frac{2}{6} = 4.0$ V [A1]

6. The sum of currents entering a junction is equal to the sum of currents leaving the junction. [B1] (Or: The algebraic sum of currents at a junction is zero.)

7. In any closed loop, the sum of the e.m.f.s is equal to the sum of the potential differences (voltage drops). [B1] (Or: The algebraic sum of potential differences around any closed loop is zero.)

8. The graph of $V$ against $I$ is a straight line with a negative gradient. [B1] The y-intercept represents the e.m.f. $E$ . [B1] The magnitude of the gradient represents the internal resistance $r$ . [B1] (Note: Award 2 marks total. 1 for intercept=e.m.f, 1 for gradient=r)

9. (a) Parallel combination $R_p$ : $\frac{1}{R_p} = \frac{1}{6} + \frac{1}{3} = \frac{1}{6} + \frac{2}{6} = \frac{3}{6} = \frac{1}{2}$ . So $R_p = 2.0 \Omega$ . [M1] Total external resistance $R_{ext} = R_p + R_{series} = 2.0 + 4.0 = 6.0 \Omega$ . [A1]

(b) Total circuit resistance $R_{tot} = R_{ext} + r = 6.0 + 1.0 = 7.0 \Omega$ . [M1] Current $I = \frac{E}{R_{tot}} = \frac{12}{7.0} = 1.71$ A. [A1] Answer: 1.7 A (2 s.f.)

10. Power $P = I^2 R$ [M1] $P = (1.714)^2 \times 4.0 = 11.75$ W. [A1] Answer: 12 W (2 s.f.)

11. Electric field strength is the electric force experienced per unit positive charge placed at that point. [B1] Formula: $E = F/Q$ . [B1]

12. (a) $F = EQ = (5.0 \times 10^4) \times (1.6 \times 10^{-19}) = 8.0 \times 10^{-15}$ N. [A1] (b) Opposite to the direction of the electric field. [B1] (Since electron is negative).

13. $E = \frac{V}{d}$ [M1] $d = 4.0 \text{ cm} = 0.04 \text{ m}$ . $E = \frac{200}{0.04} = 5000$ V m⁻¹. [A1] Answer: $5.0 \times 10^3$ V m⁻¹

14. (a) The upward electric force equals the downward gravitational force (weight). [B1] (b) $QE = mg \Rightarrow Q = \frac{mg}{E}$ [M1] $Q = \frac{3.0 \times 10^{-15} \times 9.81}{2.0 \times 10^5} = 1.47 \times 10^{-19}$ C. [A1] Answer: $1.5 \times 10^{-19}$ C (2 s.f.)

15. Radial lines originating from the center. [B1] Arrows pointing outwards (away from the positive charge). [B1]

16. Magnetic flux density $B$ is defined by the force acting on a current-carrying conductor per unit current per unit length, when the conductor is placed perpendicular to the magnetic field. [B1] Formula: $F = BIL \sin \theta$ (where $\theta=90^\circ$ ). [B1]

17. $F = BIL \sin \theta$ [M1] Since perpendicular, $\sin 90^\circ = 1$ . $F = 0.20 \times 3.0 \times 0.50 = 0.30$ N. [A1]

18. (a) $F = BQv \sin \theta$ [M1] $F = 0.10 \times (1.6 \times 10^{-19}) \times (2.0 \times 10^6) = 3.2 \times 10^{-14}$ N. [A1] (b) Circular path. [B1] (Because force is always perpendicular to velocity).

19. (a) Attractive. [B1] (b) Wire X creates a magnetic field that passes through Wire Y. Using Fleming’s Left Hand Rule (or Right Hand Grip Rule + F=BIL), the force on Y is towards X. By Newton’s 3rd Law, the force on X is towards Y. [B1 for field concept, B1 for direction/attraction explanation].

20. (a) Plane of the coil is perpendicular to the magnetic field lines. [B1] (b) Plane of the coil is parallel to the magnetic field lines. [B1]