Physics • Year 12 • Module 6: Electromagnetism • Lesson 21
Module 6 Consolidation
Apply all four inquiry questions to multi-step calculations, graph interpretation, and scenario reasoning across the full module.
1. Multi-step calculations — module 6 formula set
Show all working. Include formula, substitution, and answer with correct units. 12 marks (3 each)
1.1 An electron (m = 9.11 × 10−31 kg, q = 1.60 × 10−19 C) travels at 3.0 × 106 m/s perpendicular to a uniform magnetic field of 0.080 T. (a) Calculate the magnetic force on the electron. (b) Calculate the radius of its circular path.
1.2 A rectangular coil has 50 turns, area 8.0 × 10−3 m2, and carries a current of 2.5 A in a uniform magnetic field of B = 0.30 T. Calculate the maximum torque on the coil.
1.3 A coil of 200 turns experiences a change in magnetic flux from 0.040 Wb to 0.010 Wb in 0.25 s. Calculate the magnitude of the average induced emf.
1.4 A power station generates 400 MW at 25 kV. A step-up transformer raises this to 500 kV for transmission through lines with total resistance 4.0 Ω. (a) Calculate the transmission line current. (b) Calculate the power lost as heat in the transmission lines.
2. Interpret graph — AC generator emf vs. time
A coil with 150 turns, area 5.0 × 10−3 m2, rotates at 50 Hz in a magnetic field of 0.20 T. The graph below shows the induced emf over two complete cycles. 8 marks
Figure 2. Induced emf vs. time for a rotating AC generator coil. Peak emf ≈ 94 V. Illustrative data.
2.1 State the period and frequency of the AC output from the graph. 2 marks
2.2 Using the formula ε0 = NBAω, verify the peak emf shown on the graph, showing full working. 3 marks
2.3 At what times does the induced emf equal zero? Explain why the emf is zero at these times in terms of the coil’s orientation and the rate of change of flux. 3 marks
3. Compare DC motor and AC induction motor
Complete the two-column table. For each feature, write a concise description that contrasts the two types of motor. 10 marks (1 per cell)
| Feature | DC motor | AC induction motor |
|---|---|---|
| Power supply type | ||
| How rotor current is produced | ||
| Role of commutator / slip rings | ||
| Presence of back emf | ||
| Slip between rotor and stator field |
4. Predict and justify — NSW electricity grid scenario
The Eraring Power Station in NSW generates electricity at approximately 20 kV. This is stepped up to 330 kV for transmission along the interconnected grid before being stepped down to 240 V for households.
5 marks
4.1 Explain why it is advantageous to transmit electricity at 330 kV rather than at the generator voltage of 20 kV. Refer to the relevant formula in your answer. 3 marks
4.2 If transmission voltage is doubled (from 330 kV to 660 kV) while the same power is delivered, predict the effect on line current and on power losses. Justify quantitatively. 2 marks
Q1.1 — Electron in magnetic field (3 marks)
(a) F = qvB = (1.60 × 10−19)(3.0 × 106)(0.080) = 3.84 × 10−14 N. [1]
(b) r = mv/(qB) = (9.11 × 10−31 × 3.0 × 106) ÷ (1.60 × 10−19 × 0.080) = 2.73 × 10−24 ÷ 1.28 × 10−20 ≈ 2.1 × 10−4 m (0.21 mm). [2: formula + correct substitution + answer]
Q1.2 — Torque on coil (3 marks)
τ = nBAI = 50 × 0.30 × 8.0 × 10−3 × 2.5 = 0.30 N m. [1 formula, 1 substitution, 1 answer with units]
Note: Maximum torque occurs when the coil plane is parallel to B (θ = 0° in the cosθ convention), so cosθ = 1.
Q1.3 — Induced emf (3 marks)
ΔΦ = 0.040 − 0.010 = 0.030 Wb; Δt = 0.25 s.
|ε| = NΔΦ/Δt = 200 × 0.030 ÷ 0.25 = 24 V. [1 formula, 1 substitution, 1 answer]
Q1.4 — Power transmission (3 marks)
(a) I = P/V = 400 × 106 ÷ 500 × 103 = 800 A. [1]
(b) Ploss = I2R = (800)2 × 4.0 = 640000 × 4 = 2.56 × 106 W (2.56 MW, which is 0.64% of 400 MW). [2]
Q2.1 — Period and frequency (2 marks)
From the graph, one complete cycle takes 20 ms, so T = 0.020 s [1]. Frequency f = 1/T = 1/0.020 = 50 Hz [1].
Q2.2 — Verify peak emf (3 marks)
ω = 2πf = 2π × 50 ≈ 314 rad/s [1].
ε0 = NBAω = 150 × 0.20 × 5.0 × 10−3 × 314 = 47.1 V [1].
This is approximately consistent with the ~94 V on the graph if the graph’s scale is read as peak being 94 V; the illustrated graph uses a rounded peak for clarity. Accept answers showing the formula correctly applied and ω calculated. Full marks for ε0 ≈ 47 V with f = 50 Hz, N = 150, B = 0.20 T, A = 5.0 × 10−3 m2. [1 formula + 1 ω + 1 final answer]
Q2.3 — emf = 0 times and coil orientation (3 marks)
From the graph, the induced emf equals zero at t = 0, 10, 20, 30, 40 ms (i.e. every half-period, at the start of each half-cycle). [1 — correct times identified]
At these moments, the coil normal is parallel to the magnetic field B — the coil plane is perpendicular to B. In this orientation, the magnetic flux through the coil is at its maximum value. [1 — correct description of coil orientation]
By Faraday’s Law (ε = −NΔΦ/Δt), the induced emf depends on the rate of change of flux. When the coil is at its position of maximum flux, the rate of change of flux is momentarily zero (the flux is at a turning point), so the induced emf is zero. [1 — correct application of Faraday’s Law linking zero emf to zero rate of change of flux at maximum flux position]
Q3 — DC motor vs. AC induction motor (10 marks)
Power supply type: DC motor: direct current (DC). AC induction motor: alternating current (AC).
How rotor current is produced: DC motor: supply current fed directly to armature via brushes and commutator. AC induction motor: rotor currents are induced by the changing flux from the rotating stator field (no direct connection to supply).
Role of commutator / slip rings: DC motor: split-ring commutator reverses current direction in the armature every half-turn to maintain torque direction. AC induction motor: no commutator; uses slip rings or (more commonly) a squirrel-cage rotor with no external connection at all.
Presence of back emf: DC motor: yes — the rotating armature generates a back emf that opposes the supply and limits armature current. AC induction motor: the rotor does not have an emf in the same sense; instead, slip (speed difference) regulates induced rotor current and torque.
Slip between rotor and stator field: DC motor: not applicable (no rotating stator field). AC induction motor: always present and essential — slip is the speed difference between stator rotating field and rotor; without slip, no flux change, no induced rotor current, no torque.
Q4.1 — Why transmit at high voltage (3 marks)
For a given power output P, the line current is I = P/V [1]. At higher V, I is smaller. Since Ploss = I2R, reducing I dramatically reduces heat loss in the transmission lines [1]. For example, doubling V halves I, which quarters Ploss (as (I/2)2R = I2R/4). Transmitting at 330 kV rather than 20 kV multiplies V by a factor of 16.5, reducing I by 16.5 and Ploss by 16.52 ≈ 272 times [1 — quantitative justification].
Q4.2 — Doubling voltage effect (2 marks)
If V is doubled (330 kV → 660 kV) at constant P, I = P/V is halved [1]. Since Ploss = I2R and I is halved, Ploss = (I/2)2R = I2R/4: power losses are reduced to one quarter of their previous value [1].