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A Level H1 Physics Electricity Magnetism Quiz

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Questions

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A-Level Physics H1 Quiz - Electricity Magnetism

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

Duration: 45 minutes
Total Marks: 50

Instructions:

  1. Answer all questions.
  2. Write your answers in the spaces provided.
  3. Show all working clearly. Numerical answers should be given to an appropriate number of significant figures.
  4. 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 1.5×106 m21.5 \times 10^{-6} \text{ m}^2. The resistivity of copper is 1.7×108Ω m1.7 \times 10^{-8} \Omega \text{ m}. (a) Calculate the resistance of the wire.

<br><br><br> Resistance = ____________________ Ω\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.

<br><br><br>


_________________________________________________________________________ [2]

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

<br><br>


_________________________________________________________________________ [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.

<br><br> Current = ____________________ A [1]

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

<br><br> Resistance = ____________________ Ω\Omega [1]

4. The graph below shows the variation of current II with potential difference VV for a semiconductor diode.

(Imagine a graph where I is near zero for negative V, and rises exponentially for positive V > 0.6V)

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

<br><br><br>


_________________________________________________________________________ [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 R1=4.0 kΩR_1 = 4.0 \text{ k}\Omega and R2=2.0 kΩR_2 = 2.0 \text{ k}\Omega. Calculate the potential difference across R2R_2.

<br><br><br> Potential Difference = ____________________ V [2]


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

6. State Kirchhoff’s First Law.

<br><br>


_________________________________________________________________________ [1]

7. State Kirchhoff’s Second Law.

<br><br>


_________________________________________________________________________ [1]

8. In the circuit shown, a battery of e.m.f. EE and internal resistance rr is connected to a variable resistor RR.

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

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

Explain how the e.m.f. EE and internal resistance rr can be determined from a graph of VV against II.

<br><br><br>



_________________________________________________________________________ [2]

9. Two resistors of resistance 6.0Ω6.0 \Omega and 3.0Ω3.0 \Omega are connected in parallel. This combination is connected in series with a 4.0Ω4.0 \Omega resistor and a battery of e.m.f. 12 V and internal resistance 1.0Ω1.0 \Omega. (a) Calculate the total resistance of the external circuit.

<br><br><br> Total Resistance = ____________________ Ω\Omega [2]

(b) Calculate the current supplied by the battery.

<br><br> Current = ____________________ A [1]

10. In the circuit in Question 9, calculate the power dissipated in the 4.0Ω4.0 \Omega resistor.

<br><br><br> Power = ____________________ W [2]


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

11. Define electric field strength at a point.

<br><br>


_________________________________________________________________________ [2]

12. An electron is placed in a uniform electric field of strength 5.0×104 V m15.0 \times 10^4 \text{ V m}^{-1}. (a) Calculate the magnitude of the electric force acting on the electron. (Charge of electron e=1.6×1019 Ce = 1.6 \times 10^{-19} \text{ C})

<br><br> Force = ____________________ N [1]

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

<br> 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.

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

14. An oil drop of mass 3.0×1015 kg3.0 \times 10^{-15} \text{ kg} is held stationary between two horizontal charged plates. The electric field strength between the plates is 2.0×105 V m12.0 \times 10^5 \text{ V m}^{-1}. (a) State the condition required for the oil drop to remain stationary.

<br><br> _________________________________________________________________________ [1]

(b) Calculate the charge on the oil drop. (g=9.81 m s2g = 9.81 \text{ m s}^{-2})

<br><br><br> Charge = ____________________ C [2]

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

<br><br><br><br><br> [2]


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

16. Define magnetic flux density BB.

<br><br>


_________________________________________________________________________ [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.

<br><br><br> Force = ____________________ N [2]

18. An electron moves with a velocity of 2.0×106 m s12.0 \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.

<br><br> Force = ____________________ N [1]

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

<br> 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.

<br> Force is: __________________________ [1]

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

<br><br><br>


_________________________________________________________________________ [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

<br> _________________________________________________________________________ [1]

(b) Zero

<br> _________________________________________________________________________ [1]

End of Quiz

Answers

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

1. (a) R=ρLAR = \frac{\rho L}{A} [M1] R=1.7×108×2.01.5×106=0.0227ΩR = \frac{1.7 \times 10^{-8} \times 2.0}{1.5 \times 10^{-6}} = 0.0227 \Omega [A1] Answer: 0.023Ω0.023 \Omega (2 s.f.)

(b) Resistance increases by a factor of 4. [B1] Explanation: R=ρLAR = \frac{\rho L}{A}. If length LL doubles, area AA halves (constant volume). New R=ρ(2L)(A/2)=4ρLA=4RR' = \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=VII=P/V=24/12=2.0P = VI \Rightarrow I = P/V = 24/12 = 2.0 A [A1] (b) R=V/I=12/2.0=6.0ΩR = V/I = 12/2.0 = 6.0 \Omega [A1] (Or R=V2/PR = V^2/P)

4. Ohm’s Law states that current is directly proportional to potential difference (IVI \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. Vout=Vin×R2R1+R2V_{out} = V_{in} \times \frac{R_2}{R_1 + R_2} [M1] Vout=12×2.04.0+2.0=12×26=4.0V_{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 VV against II is a straight line with a negative gradient. [B1] The y-intercept represents the e.m.f. EE. [B1] The magnitude of the gradient represents the internal resistance rr. [B1] (Note: Award 2 marks total. 1 for intercept=e.m.f, 1 for gradient=r)

9. (a) Parallel combination RpR_p: 1Rp=16+13=16+26=36=12\frac{1}{R_p} = \frac{1}{6} + \frac{1}{3} = \frac{1}{6} + \frac{2}{6} = \frac{3}{6} = \frac{1}{2}. So Rp=2.0ΩR_p = 2.0 \Omega. [M1] Total external resistance Rext=Rp+Rseries=2.0+4.0=6.0ΩR_{ext} = R_p + R_{series} = 2.0 + 4.0 = 6.0 \Omega. [A1]

(b) Total circuit resistance Rtot=Rext+r=6.0+1.0=7.0ΩR_{tot} = R_{ext} + r = 6.0 + 1.0 = 7.0 \Omega. [M1] Current I=ERtot=127.0=1.71I = \frac{E}{R_{tot}} = \frac{12}{7.0} = 1.71 A. [A1] Answer: 1.7 A (2 s.f.)

10. Power P=I2RP = I^2 R [M1] P=(1.714)2×4.0=11.75P = (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/QE = F/Q. [B1]

12. (a) F=EQ=(5.0×104)×(1.6×1019)=8.0×1015F = 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=VdE = \frac{V}{d} [M1] d=4.0 cm=0.04 md = 4.0 \text{ cm} = 0.04 \text{ m}. E=2000.04=5000E = \frac{200}{0.04} = 5000 V m⁻¹. [A1] Answer: 5.0×1035.0 \times 10^3 V m⁻¹

14. (a) The upward electric force equals the downward gravitational force (weight). [B1] (b) QE=mgQ=mgEQE = mg \Rightarrow Q = \frac{mg}{E} [M1] Q=3.0×1015×9.812.0×105=1.47×1019Q = \frac{3.0 \times 10^{-15} \times 9.81}{2.0 \times 10^5} = 1.47 \times 10^{-19} C. [A1] Answer: 1.5×10191.5 \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 BB 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=BILsinθF = BIL \sin \theta (where θ=90\theta=90^\circ). [B1]

17. F=BILsinθF = BIL \sin \theta [M1] Since perpendicular, sin90=1\sin 90^\circ = 1. F=0.20×3.0×0.50=0.30F = 0.20 \times 3.0 \times 0.50 = 0.30 N. [A1]

18. (a) F=BQvsinθF = BQv \sin \theta [M1] F=0.10×(1.6×1019)×(2.0×106)=3.2×1014F = 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]