Worked Solutions
Module 6: Electromagnetism — Worked Solutions (HSC Physics)
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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.
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.