Module 6: Electromagnetism Quiz

Test Your Understanding

Instructions

This quiz covers Module 6: Electromagnetism. Answer all questions, then check your answers at the bottom.

Topics covered:

  • Charged particles in fields
  • Motor effect
  • Electromagnetic induction
  • Transformers

Section A: Multiple Choice (1 mark each)

Question 1

A proton enters a uniform magnetic field perpendicular to its velocity. The resulting motion is:

A. Straight line at constant speed B. Parabolic path C. Circular path D. Stationary

Question 2

A current-carrying wire experiences maximum force in a magnetic field when the angle between current and field is:

A. 0° B. 45° C. 90° D. 180°

Question 3

According to Lenz’s Law, the direction of induced current:

A. Is always clockwise B. Opposes the change in flux C. Is the same as the applied field D. Depends on the wire material

Question 4

An ideal step-up transformer with turns ratio 1:10 has input voltage 12 V. The output voltage is:

A. 1.2 V B. 12 V C. 120 V D. 1200 V

Question 5

The force between two parallel wires carrying currents in the same direction is:

A. Attractive B. Repulsive C. Zero D. Depends on current magnitude

Section B: Short Answer (3-4 marks each)

Question 6

An electron travels at \(2.0 \times 10^6\) m/s perpendicular to a 0.5 T magnetic field.

  1. Calculate the magnetic force on the electron. (2 marks)

  2. Calculate the radius of the circular path. (2 marks)

Data: \(e = 1.6 \times 10^{-19}\) C, \(m_e = 9.11 \times 10^{-31}\) kg

Use: \(F = qvB\), \(r = \frac{mv}{qB}\)

Question 7

A rectangular coil with 50 turns and area 0.02 m² is perpendicular to a magnetic field that decreases from 0.8 T to 0.2 T in 0.1 s.

  1. Calculate the change in magnetic flux. (2 marks)

  2. Calculate the induced EMF. (2 marks)

Use: \(\Phi = BA\), \(\varepsilon = -N\frac{\Delta\Phi}{\Delta t}\)

Question 8

Two parallel wires 0.1 m apart carry currents of 5 A and 10 A in the same direction.

Calculate the force per unit length between them.

Data: \(\mu_0 = 4\pi \times 10^{-7}\) T·m/A

Use: \(\frac{F}{l} = \frac{\mu_0 I_1 I_2}{2\pi d}\)

Section C: Extended Response (5-6 marks each)

Question 9

A transformer has 200 primary turns and 1000 secondary turns. The primary voltage is 240 V and the secondary current is 2 A.

  1. Calculate the secondary voltage. (2 marks)

  2. Calculate the primary current (assuming 100% efficiency). (2 marks)

  3. If the transformer is 95% efficient, calculate the actual primary current. (2 marks)

Question 10

Explain how a DC motor converts electrical energy to mechanical energy. Include in your answer:

  • The role of the magnetic field
  • How force is produced on the coil
  • The function of the split-ring commutator
  • Why back EMF limits motor speed (5 marks)

Answers

  1. C - Magnetic force is perpendicular to velocity, causing circular motion

  2. C - \(F = BIl\sin\theta\) is maximum when \(\theta = 90°\)

  3. B - Lenz’s Law: induced current opposes change in flux (conservation of energy)

  4. C - \(V_s = V_p \times \frac{N_s}{N_p} = 12 \times 10 = 120\) V

  5. A - Currents in same direction attract (right-hand rule)

Q6: (a) \(F = qvB = 1.6 \times 10^{-19} \times 2.0 \times 10^6 \times 0.5 = 1.6 \times 10^{-13}\) N (b) \(r = \frac{mv}{qB} = \frac{9.11 \times 10^{-31} \times 2.0 \times 10^6}{1.6 \times 10^{-19} \times 0.5} = 2.3 \times 10^{-5}\) m

Q7: (a) \(\Delta\Phi = \Delta B \times A = (0.2 - 0.8) \times 0.02 = -0.012\) Wb (b) \(\varepsilon = -N\frac{\Delta\Phi}{\Delta t} = -50 \times \frac{-0.012}{0.1} = 6.0\) V

Q8: \(\frac{F}{l} = \frac{4\pi \times 10^{-7} \times 5 \times 10}{2\pi \times 0.1} = 1.0 \times 10^{-4}\) N/m (attractive)

Q9: (a) \(V_s = 240 \times \frac{1000}{200} = 1200\) V (b) Ideal: \(I_p = \frac{V_s I_s}{V_p} = \frac{1200 \times 2}{240} = 10\) A (c) Efficiency = \(\frac{P_{out}}{P_{in}}\): \(P_{in} = \frac{2400}{0.95} = 2526\) W, so \(I_p = \frac{2526}{240} = 10.5\) A

Q10: Key points: - Permanent magnets create uniform field across coil - Current in coil → force on sides (\(F = BIl\)) due to motor effect - Forces create torque causing rotation - Commutator reverses current every half rotation to maintain unidirectional torque - Rotating coil generates back EMF opposing supply (Lenz’s Law) - Back EMF reduces net voltage → limits current → limits speed - At max speed: back EMF ≈ supply voltage

Score Guide

Score Performance
90-100% Excellent - Ready for HSC
75-89% Good - Minor revision needed
60-74% Satisfactory - Review weak areas
Below 60% More study required