A Level H2 Physics Electricity Magnetism Quiz

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: $\varepsilon = -\frac{d(N\Phi)}{dt}$) [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 $B$.
[3]
Answer:
Magnetic force provides centripetal force:
$Bqv = \frac{mv^2}{r} \Rightarrow B = \frac{mv}{qr}$
$B = \frac{(1.67 \times 10^{-27}) \times (2.0 \times 10^6)}{(1.60 \times 10^{-19}) \times 0.15}$
$B = \frac{3.34 \times 10^{-21}}{2.4 \times 10^{-20}}$
$B \approx 0.14 \text{ T}$
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
$\frac{1}{2}mv^2 = qV \Rightarrow v = \sqrt{\frac{2qV}{m}}$
$v = \sqrt{\frac{2 \times (1.60 \times 10^{-19}) \times 250}{9.11 \times 10^{-31}}}$
$v = \sqrt{8.78 \times 10^{13}}$
$v = 9.37 \times 10^6 \text{ m s}^{-1}$
Marking Note: 1 mark for formula, 1 mark for substitution, 1 mark for answer ($9.4 \times 10^6 \text{ m s}^{-1}$).

4. Electric field between plates.
(a) Electric field strength.
[2]
Answer:
$E = \frac{V}{d}$
$E = \frac{120}{0.040}$
$E = 3000 \text{ V m}^{-1}$ (or $\text{N C}^{-1}$)
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:
$B = \frac{\mu_0 I}{2\pi r}$
$B = \frac{(4\pi \times 10^{-7}) \times 5.0}{2\pi \times 0.020}$
$B = 5.0 \times 10^{-5} \text{ T}$
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 $R$.
[3]
Answer:
$V_T = 3.0 \text{ V}$, so $V_R = 9.0 - 3.0 = 6.0 \text{ V}$.
Current $I = \frac{V_T}{R_T} = \frac{3.0}{2000} = 1.5 \times 10^{-3} \text{ A}$.
$R = \frac{V_R}{I} = \frac{6.0}{1.5 \times 10^{-3}} = 4000 \, \Omega$ ($4.0 \text{ k}\Omega$).
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 $V/I$ is not constant [1].

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

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

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

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

13. Maximum induced e.m.f.
(a) Determine $\varepsilon_{max}$.
[3]
Answer:
$\varepsilon_{max} = BAN\omega$
$\varepsilon_{max} = 0.040 \times 100\pi$
$\varepsilon_{max} = 4\pi \approx 12.6 \text{ V}$.
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 $v = \sqrt{\frac{2qV}{m}}$.
[2]
Answer:
Work done by electric field = Kinetic Energy gained
$qV = \frac{1}{2}mv^2$ [1]
$v^2 = \frac{2qV}{m} \Rightarrow v = \sqrt{\frac{2qV}{m}}$ [1].

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

18. Isotope radius ratio.
(a) Calculate ratio.
[2]
Answer:
$r \propto \sqrt{m}$ (since $B, V, q$ constant).
$\frac{r_{13}}{r_{12}} = \sqrt{\frac{13}{12}} \approx 1.04$.
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 ($F=Bqv$) [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:
$\frac{V_s}{V_p} = \frac{N_s}{N_p}$
$V_s = 240 \times \frac{50}{1000} = 240 \times 0.05 = 12 \text{ V}$.

(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).