Module 6: Electromagnetism — Worked Solutions (HSC Physics)
Created with Intu AI Reviewed by Intuition's expert tutors
Studying this? See our HSC Physics course →
Worked examples for HSC Physics Module 6: Electromagnetism. Each shows where the marks are awarded, the key idea, and the full solution explained by your choice of tutor — Stella, Ella or Cassie.
In short: full step-by-step worked solutions for HSC Physics — Module 6: Electromagnetism. Every question is worked through with the method and reasoning shown, in NESA exam style, so you can check how to reach the answer, not just the final result.
How to use these
Try each question first, then check your working. Use the tutor tabs to read the full solution in the style that suits you: Stella is direct and challenging, Ella is warm and explains the why, and Cassie is concise and analytical.
Watch your units: forces in newtons (N), magnetic flux density in tesla (T), flux in webers (Wb) and EMF in volts (V).
Example 1 — The motor effect
Question
A straight conductor of length $0.25\ \text{m}$ carries a current of $4.0\ \text{A}$ and lies at an angle of $60^\circ$ to a uniform magnetic field of strength $0.30\ \text{T}$. Calculate the magnitude of the force experienced by the conductor.
Solution
Use the motor-effect formula directly: $F = BIL\sin\theta$, where $\theta$ is the angle between the conductor and the field.
$$F = (0.30)(4.0)(0.25)\sin 60^\circ$$
$$F = 0.30 \times 4.0 \times 0.25 \times 0.8660 = 0.26\ \text{N}$$
So the force is $0.26\ \text{N}$. Don't forget the $\sin\theta$ — leaving it out is the most common way to lose this mark.
A current-carrying wire in a magnetic field feels a force — that's the motor effect, and the formula is $F = BIL\sin\theta$.
Here $\theta$ is the angle between the wire and the field lines, which is $60^\circ$. The $\sin\theta$ term matters because only the component of the field perpendicular to the current produces a force; if the wire were parallel to the field ($\theta = 0$) there'd be no force at all.
$$F = BIL\sin\theta = (0.30)(4.0)(0.25)\sin 60^\circ$$
$$F = 0.30 \times 4.0 \times 0.25 \times 0.8660 = 0.26\ \text{N}$$
So the conductor experiences a force of $0.26\ \text{N}$. The key idea is that the angle dramatically affects the force — maximum when perpendicular, zero when parallel.
Motor effect: $F = BIL\sin\theta$.
- $B = 0.30\ \text{T}$, $I = 4.0\ \text{A}$, $L = 0.25\ \text{m}$, $\theta = 60^\circ$
- $F = (0.30)(4.0)(0.25)\sin 60^\circ$
- $F = 0.30 \times 4.0 \times 0.25 \times 0.8660$
- $F = 0.26\ \text{N}$
Where the marks go
- 1 mark: States the correct formula $F = BIL\sin\theta$
- 1 mark: Substitutes all values correctly including $\sin 60^\circ$
- 1 mark: Correct force $F = 0.26\ \text{N}$ with units
Key idea
The force on a current-carrying conductor is $F = BIL\sin\theta$, where $\theta$ is the angle between the conductor and the field — the force is maximum at $90^\circ$ and zero when parallel.
Example 2 — Induction and transformers
Question
An ideal transformer has 200 turns on its primary coil and 1500 turns on its secondary coil. The primary coil is connected to a $240\ \text{V}$ AC supply and the secondary supplies a current of $0.50\ \text{A}$. Calculate the secondary voltage and the current drawn by the primary coil.
Solution
Use the transformer ratio: $\dfrac{V_p}{V_s} = \dfrac{N_p}{N_s}$.
$$V_s = V_p \times \frac{N_s}{N_p} = 240 \times \frac{1500}{200} = 1800\ \text{V}$$
This is a step-up transformer. For an ideal transformer, power in equals power out, so $V_p I_p = V_s I_s$.
$$I_p = \frac{V_s I_s}{V_p} = \frac{1800 \times 0.50}{240} = 3.75\ \text{A}$$
Secondary voltage $1800\ \text{V}$, primary current $3.75\ \text{A}$. Note the voltage steps up while the current steps down — power is conserved.
A transformer changes voltage using the ratio of turns on its two coils. The relationship is $\dfrac{V_p}{V_s} = \dfrac{N_p}{N_s}$, so:
$$V_s = V_p \times \frac{N_s}{N_p} = 240 \times \frac{1500}{200} = 1800\ \text{V}$$
Because there are more turns on the secondary, the voltage goes up — this is a step-up transformer.
Now for the primary current. An ideal transformer wastes no energy, so the power going in must equal the power coming out: $V_p I_p = V_s I_s$. That's why a step-up in voltage forces a step-down in current. Rearranging:
$$I_p = \frac{V_s I_s}{V_p} = \frac{1800 \times 0.50}{240} = 3.75\ \text{A}$$
So the secondary delivers $1800\ \text{V}$ and the primary draws $3.75\ \text{A}$. The big idea is conservation of power: you can trade voltage for current, but not get something for nothing.
Turns ratio for voltage; power conservation for current.
- $V_s = V_p \dfrac{N_s}{N_p} = 240 \times \dfrac{1500}{200} = 1800\ \text{V}$
Ideal transformer: $V_p I_p = V_s I_s$.
- $I_p = \dfrac{V_s I_s}{V_p} = \dfrac{1800 \times 0.50}{240} = 3.75\ \text{A}$
$V_s = 1800\ \text{V}$, $I_p = 3.75\ \text{A}$.
Where the marks go
- 1 mark: Uses turns ratio $\dfrac{V_p}{V_s} = \dfrac{N_p}{N_s}$
- 1 mark: Correct secondary voltage $V_s = 1800\ \text{V}$
- 1 mark: Applies power conservation $V_p I_p = V_s I_s$
- 1 mark: Correct primary current $I_p = 3.75\ \text{A}$ with units
Key idea
In an ideal transformer the voltage follows the turns ratio $\dfrac{V_p}{V_s} = \dfrac{N_p}{N_s}$, and power is conserved $V_p I_p = V_s I_s$, so stepping voltage up steps current down.
Frequently asked questions
Step-by-step solutions to exam-style questions on Electromagnetism in HSC Physics, with the full method shown for each — so you can follow the reasoning, not just the final answer.
Attempt each question yourself first, then compare your method — not just your answer — against the worked solution. The aim is to learn the approach so you can handle unfamiliar questions in the exam, rather than memorising answers.
Yes — they follow the NESA HSC Physics syllabus for Electromagnetism, using the methods and notation expected in the exam.