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A Level H2 Physics Electricity Magnetism Quiz

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A Level H2 Physics From Real Exams Generated by Qwen3.6 Plus Updated 2026-06-03

Questions

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A-Level Physics H2 Quiz - Electricity Magnetism

Name: __________________________
Class: __________________________
Date: __________________________
Score: ________ / 45

Duration: 60 minutes
Total Marks: 45

Instructions:

  1. Answer all questions.
  2. Write your answers in the spaces provided.
  3. Show all working clearly. Marks may be awarded for correct working even if the final answer is incorrect.
  4. Use g=9.81 m s2g = 9.81 \text{ m s}^{-2} where appropriate.
  5. The use of an approved scientific calculator is expected.

Section A: Fundamentals of Fields and Forces (Questions 1-5)

1. State Faraday’s law of electromagnetic induction.
[2]

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2. A proton moves with a constant speed of 2.0×106 m s12.0 \times 10^6 \text{ m s}^{-1} in a circular path of radius 0.15 m0.15 \text{ m} in a uniform magnetic field. The plane of the circle is perpendicular to the magnetic field.

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

<br> <br>

(b) Calculate the magnitude of the magnetic flux density BB.
(Mass of proton mp=1.67×1027 kgm_p = 1.67 \times 10^{-27} \text{ kg}, charge q=1.60×1019 Cq = 1.60 \times 10^{-19} \text{ C})
[3]

<br> <br> <br> <br> <br>

3. An electron is accelerated from rest through a potential difference of 250 V250 \text{ V}.

(a) Calculate the speed of the electron as it exits the acceleration region.
(Mass of electron me=9.11×1031 kgm_e = 9.11 \times 10^{-31} \text{ kg}, charge e=1.60×1019 Ce = 1.60 \times 10^{-19} \text{ C})
[3]

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4. The electron from Question 3 enters a region of uniform electric field between two parallel horizontal plates separated by 4.0 cm4.0 \text{ cm}. The potential difference across the plates is 120 V120 \text{ V}.

(a) Determine the magnitude of the electric field strength between the plates.
[2]

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5. A long straight wire carries a current of 5.0 A5.0 \text{ A} upwards (out of the page).

(a) Sketch the magnetic field pattern around the wire as viewed from above. Indicate the direction of the field lines.
[2]

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(b) Calculate the magnetic flux density at a distance of 2.0 cm2.0 \text{ cm} from the wire.
[2]

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Section B: Circuits and Components (Questions 6-10)

6. In a circuit designed to investigate the characteristics of a thermistor, a student connects the thermistor in series with a fixed resistor RR and a battery of e.m.f. 9.0 V9.0 \text{ V} and negligible internal resistance. A voltmeter is connected in parallel with the thermistor.

(a) Explain why the voltmeter reading changes when the temperature of the NTC thermistor increases.
[2]

<br> <br> <br> <br>

7. The resistance of the thermistor in Question 6 is 2.0 kΩ2.0 \text{ k}\Omega at 20C20^\circ\text{C}.

(a) Calculate the resistance of the fixed resistor RR if the voltmeter reads 3.0 V3.0 \text{ V} at this temperature.
[3]

<br> <br> <br> <br> <br>

8. A student investigates the relationship between the current II in a filament lamp and the potential difference VV across it. The results are plotted on a graph of II against VV.

(a) Explain why the graph is not a straight line through the origin.
[2]

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9. For the filament lamp in Question 8, at a certain operating point, V=6.0 VV = 6.0 \text{ V} and I=0.50 AI = 0.50 \text{ A}.

(a) Calculate the resistance of the filament at this point.
[1]

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10. The student connects a second identical lamp in series with the first lamp and the same power supply (6.0 V6.0 \text{ V} source for simplicity, or same supply as before).

(a) State and explain the effect on the total power dissipated by the two lamps compared to the single lamp connected to the same supply.
[3]

<br> <br> <br> <br> <br> <br>

Section C: Electromagnetic Induction (Questions 11-15)

11. A rectangular coil of area 4.0×103 m24.0 \times 10^{-3} \text{ m}^2 with 50 turns rotates in a uniform magnetic field of flux density 0.20 T0.20 \text{ T}.

(a) Calculate the maximum magnetic flux linkage through the coil.
[2]

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12. The coil in Question 11 rotates at a constant frequency of 50 Hz50 \text{ Hz}. The axis of rotation is perpendicular to the magnetic field.

(a) Determine the angular frequency ω\omega of the rotation.
[1]

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13. Using the values from Questions 11 and 12:

(a) Determine the maximum e.m.f. induced in the coil.
[3]

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14. Consider the orientation of the rotating coil in Question 11.

(a) State the orientation of the coil relative to the magnetic field when the induced e.m.f. is zero.
[1]

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15. A bar magnet is dropped through a vertical copper tube.

(a) Explain why the magnet falls slower than it would in free fall, referencing Lenz’s Law.
[3]

<br> <br> <br> <br> <br> <br>

Section D: Advanced Applications (Questions 16-20)

16. In a mass spectrometer, ions of charge +q+q and mass mm are accelerated from rest through a potential difference VV.

(a) Show that the speed vv of the ions upon entering the magnetic field is given by v=2qVmv = \sqrt{\frac{2qV}{m}}.
[2]

<br> <br> <br> <br> <br>

17. The ions from Question 16 enter a uniform magnetic field BB perpendicular to their velocity.

(a) Show that the radius rr of the circular path is given by r=1B2mVqr = \frac{1}{B} \sqrt{\frac{2mV}{q}}.
[2]

<br> <br> <br> <br> <br> <br>

18. Two isotopes of carbon, 12C^{12}\text{C} and 13C^{13}\text{C}, are ionized to have charge +e+e. They are accelerated through the same potential difference and enter the same magnetic field.

(a) Calculate the ratio r13r12\frac{r_{13}}{r_{12}}.
[2]

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19. A Hall probe is used to measure magnetic flux density.

(a) Explain the physical mechanism that generates the Hall voltage in a conductor carrying current in a magnetic field.
[3]

<br> <br> <br> <br> <br> <br>

20. A transformer has a primary coil with 1000 turns and a secondary coil with 50 turns. The primary is connected to a 240 V240 \text{ V} a.c. supply.

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

<br> <br> <br> <br>

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

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Answers

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

1. State Faraday’s law of electromagnetic induction.
[2]
Answer:
The induced e.m.f. is proportional to the rate of change of magnetic flux linkage [1].
(Mathematically: ε=d(NΦ)dt\varepsilon = -\frac{d(N\Phi)}{dt}) [1].
Marking Note: Must mention "rate of change" and "flux linkage".

2. Proton in magnetic field.
(a) Direction of magnetic force.
[1]
Answer:
Perpendicular to the velocity (and perpendicular to the magnetic field).
Marking Note: "Centripetal" is acceptable in this context.

(b) Calculate magnetic flux density BB.
[3]
Answer:
Magnetic force provides centripetal force:
Bqv=mv2rB=mvqrBqv = \frac{mv^2}{r} \Rightarrow B = \frac{mv}{qr}
B=(1.67×1027)×(2.0×106)(1.60×1019)×0.15B = \frac{(1.67 \times 10^{-27}) \times (2.0 \times 10^6)}{(1.60 \times 10^{-19}) \times 0.15}
B=3.34×10212.4×1020B = \frac{3.34 \times 10^{-21}}{2.4 \times 10^{-20}}
B0.14 TB \approx 0.14 \text{ T}
Marking Note: 1 mark for formula, 1 mark for substitution, 1 mark for answer.

3. Electron accelerated.
(a) Speed of electron.
[3]
Answer:
Gain in KE = Work done by electric field
12mv2=qVv=2qVm\frac{1}{2}mv^2 = qV \Rightarrow v = \sqrt{\frac{2qV}{m}}
v=2×(1.60×1019)×2509.11×1031v = \sqrt{\frac{2 \times (1.60 \times 10^{-19}) \times 250}{9.11 \times 10^{-31}}}
v=8.78×1013v = \sqrt{8.78 \times 10^{13}}
v=9.37×106 m s1v = 9.37 \times 10^6 \text{ m s}^{-1}
Marking Note: 1 mark for formula, 1 mark for substitution, 1 mark for answer (9.4×106 m s19.4 \times 10^6 \text{ m s}^{-1}).

4. Electric field between plates.
(a) Electric field strength.
[2]
Answer:
E=VdE = \frac{V}{d}
E=1200.040E = \frac{120}{0.040}
E=3000 V m1E = 3000 \text{ V m}^{-1} (or N C1\text{N C}^{-1})
Marking Note: 1 mark for formula, 1 mark for answer.

5. Straight wire magnetic field.
(a) Sketch field pattern.
[2]
Answer:
Concentric circles centered on the wire [1].
Arrows indicating counter-clockwise direction (Right-Hand Grip Rule) [1].

(b) Calculate magnetic flux density.
[2]
Answer:
B=μ0I2πrB = \frac{\mu_0 I}{2\pi r}
B=(4π×107)×5.02π×0.020B = \frac{(4\pi \times 10^{-7}) \times 5.0}{2\pi \times 0.020}
B=5.0×105 TB = 5.0 \times 10^{-5} \text{ T}
Marking Note: 1 mark for formula, 1 mark for answer.

6. Thermistor circuit explanation.
(a) Explain voltmeter reading change.
[2]
Answer:
As temperature increases, the resistance of the NTC thermistor decreases [1].
This causes the potential difference across the thermistor to decrease (smaller share of e.m.f. in potential divider) [1].

7. Thermistor calculation.
(a) Calculate resistance of fixed resistor RR.
[3]
Answer:
VT=3.0 VV_T = 3.0 \text{ V}, so VR=9.03.0=6.0 VV_R = 9.0 - 3.0 = 6.0 \text{ V}.
Current I=VTRT=3.02000=1.5×103 AI = \frac{V_T}{R_T} = \frac{3.0}{2000} = 1.5 \times 10^{-3} \text{ A}.
R=VRI=6.01.5×103=4000ΩR = \frac{V_R}{I} = \frac{6.0}{1.5 \times 10^{-3}} = 4000 \, \Omega (4.0 kΩ4.0 \text{ k}\Omega).
Marking Note: 1 mark for voltage/current logic, 1 mark for substitution, 1 mark for answer.

8. Filament lamp characteristics.
(a) Explain non-linear graph.
[2]
Answer:
As current/voltage increases, the temperature of the filament increases [1].
Resistance increases due to increased lattice vibrations, so V/IV/I is not constant [1].

9. Filament lamp resistance.
(a) Calculate resistance.
[1]
Answer:
R=VI=6.00.50=12ΩR = \frac{V}{I} = \frac{6.0}{0.50} = 12 \, \Omega.

10. Two lamps in series.
(a) Effect on total power.
[3]
Answer:
Total resistance increases [1].
Current decreases (I=V/RtotalI = V/R_{total}) [1].
Total power P=V2/RtotalP = V^2/R_{total} decreases because RtotalR_{total} increases [1].

11. Rotating coil flux linkage.
(a) Maximum magnetic flux linkage.
[2]
Answer:
Max Flux Linkage =NBA= NBA
=50×0.20×(4.0×103)= 50 \times 0.20 \times (4.0 \times 10^{-3})
=0.040 Wb turns= 0.040 \text{ Wb turns}.
Marking Note: 1 mark for NBANBA, 1 mark for answer.

12. Angular frequency.
(a) Calculate ω\omega.
[1]
Answer:
ω=2πf=2π(50)=100π rad s1\omega = 2\pi f = 2\pi(50) = 100\pi \text{ rad s}^{-1} (314 rad s1\approx 314 \text{ rad s}^{-1}).

13. Maximum induced e.m.f.
(a) Determine εmax\varepsilon_{max}.
[3]
Answer:
εmax=BANω\varepsilon_{max} = BAN\omega
εmax=0.040×100π\varepsilon_{max} = 0.040 \times 100\pi
εmax=4π12.6 V\varepsilon_{max} = 4\pi \approx 12.6 \text{ V}.
Marking Note: 1 mark for formula, 1 mark for substitution, 1 mark for answer.

14. Orientation for zero e.m.f.
(a) State orientation.
[1]
Answer:
The plane of the coil is perpendicular to the magnetic field (flux linkage is maximum, rate of change is zero).

15. Magnet in copper tube.
(a) Explain slower fall.
[3]
Answer:
Changing magnetic flux induces eddy currents in the copper tube (Faraday's Law) [1].
By Lenz's Law, these currents create a magnetic field that opposes the motion of the magnet [1].
This creates an upward magnetic force that reduces the net downward acceleration [1].

16. Mass spectrometer speed.
(a) Show v=2qVmv = \sqrt{\frac{2qV}{m}}.
[2]
Answer:
Work done by electric field = Kinetic Energy gained
qV=12mv2qV = \frac{1}{2}mv^2 [1]
v2=2qVmv=2qVmv^2 = \frac{2qV}{m} \Rightarrow v = \sqrt{\frac{2qV}{m}} [1].

17. Mass spectrometer radius.
(a) Show r=1B2mVqr = \frac{1}{B} \sqrt{\frac{2mV}{q}}.
[2]
Answer:
Magnetic force = Centripetal force: Bqv=mv2rr=mvBqBqv = \frac{mv^2}{r} \Rightarrow r = \frac{mv}{Bq} [1].
Substitute vv: r=mBq2qVm=1Bm22qVq2m=1B2mVqr = \frac{m}{Bq}\sqrt{\frac{2qV}{m}} = \frac{1}{B}\sqrt{\frac{m^2 2qV}{q^2 m}} = \frac{1}{B}\sqrt{\frac{2mV}{q}} [1].

18. Isotope radius ratio.
(a) Calculate ratio.
[2]
Answer:
rmr \propto \sqrt{m} (since B,V,qB, V, q constant).
r13r12=13121.04\frac{r_{13}}{r_{12}} = \sqrt{\frac{13}{12}} \approx 1.04.
Marking Note: 1 mark for proportionality, 1 mark for answer.

19. Hall Effect.
(a) Explain mechanism.
[3]
Answer:
Charge carriers (electrons) moving in a magnetic field experience a magnetic Lorentz force (F=BqvF=Bqv) [1].
This force deflects carriers to one side of the conductor, creating a charge separation [1].
This separation creates an electric field (Hall voltage) that balances the magnetic force [1].

20. Transformer.
(a) Calculate secondary voltage.
[2]
Answer:
VsVp=NsNp\frac{V_s}{V_p} = \frac{N_s}{N_p}
Vs=240×501000=240×0.05=12 VV_s = 240 \times \frac{50}{1000} = 240 \times 0.05 = 12 \text{ V}.

(b) Reason for inefficiency.
[1]
Answer:
Energy loss due to heating in coils (resistance) OR eddy currents in the core OR hysteresis in the core OR flux leakage. (Any one).