Physics • Year 12 • Module 6 • Lesson 13
Faraday's Law of Induction
Apply Faraday's Law to calculate induced emf, interpret data, compare scenarios, and evaluate the effect of changing flux parameters.
1. Calculate the induced emf
Use Faraday's Law (ε = −NΔΦ/Δt) to solve each problem. Show all working. 12 marks (3 each)
1.1 A coil of 100 turns experiences a change in magnetic flux from 0.020 Wb to 0.050 Wb in 0.10 s. Calculate the magnitude of the average induced emf. 3 marks
1.2 The same coil from Q1.1 undergoes the same ΔΦ but in only 0.010 s. Calculate the new average induced emf and explain why it differs from Q1.1. 3 marks
1.3 A coil of 200 turns and area 5.0 × 10−3 m2 is placed perpendicular to a uniform magnetic field of 0.40 T. The field is reduced to zero in 0.20 s. Calculate (a) the initial flux and (b) the average induced emf. 3 marks
1.4 An average emf of 12 V is induced in a coil of 60 turns over an interval of 0.050 s. Calculate the change in magnetic flux (ΔΦ) through the coil. 3 marks
2. Interpret a flux-vs-time graph
The graph below shows the magnetic flux through a 50-turn coil as a function of time. Use it to answer the questions. 9 marks
Figure 2. Magnetic flux through a 50-turn coil vs time. Illustrative data.
2.1 Calculate the average induced emf in Region A (0 to 0.2 s). Show your working. 2 marks
2.2 State the induced emf in Region B (0.2 to 0.5 s) and explain why. 2 marks
2.3 Calculate the average induced emf in Region C (0.5 to 0.8 s) and compare its magnitude with Region A. Account for any difference. 3 marks
2.4 Sketch a rough emf-vs-time graph for this coil for the interval 0 to 1.0 s (no axes values needed — indicate zero and non-zero regions only). 2 marks
3. Compare scenarios
Complete the table by calculating the missing emf values and explaining any patterns. 9 marks
| Scenario | N (turns) | ΔΦ (mWb) | Δt (ms) | |ε| (V) — calculate |
|---|---|---|---|---|
| A | 50 | 20 | 100 | |
| B | 100 | 20 | 100 | |
| C | 50 | 40 | 100 | |
| D | 50 | 20 | 50 | |
| E | 200 | 20 | 200 |
3.1 Compare Scenarios A and B. What does the result tell you about the effect of increasing N? 2 marks
3.2 Compare Scenarios A and D. What does the result tell you about the effect of halving Δt? 2 marks
3.3 In Scenario E, N is quadrupled but Δt is also doubled compared to Scenario A. Predict whether |ε| will be larger, smaller, or equal to A, and verify with your calculation. 1 mark
Q1 — Calculations
1.1 ΔΦ = 0.050 − 0.020 = 0.030 Wb; ε = 100 × 0.030 / 0.10 = 30 V. (1 mark each: ΔΦ correct, substitution correct, final answer with unit.)
1.2 ΔΦ unchanged = 0.030 Wb; Δt = 0.010 s; ε = 100 × 0.030 / 0.010 = 300 V. The emf is 10 times larger because Δt is 10 times smaller, so the rate of flux change is 10 times greater. (1 mark calculation, 1 mark comparison/explanation, 1 mark linking to rate of change.)
1.3 (a) Φ = BA = 0.40 × 5.0 × 10−3 = 2.0 × 10−3 Wb = 2.0 mWb. (b) ΔΦ = 0 − 2.0 × 10−3 = −2.0 × 10−3 Wb; |ε| = 200 × 2.0 × 10−3 / 0.20 = 2.0 V. (1 mark flux calculation, 1 mark ΔΦ, 1 mark final emf.)
1.4 Rearrange: ΔΦ = |ε| × Δt / N = 12 × 0.050 / 60 = 0.60 / 60 = 0.010 Wb. (1 mark rearrangement, 1 mark substitution, 1 mark answer with unit.)
Q2 — Graph interpretation
2.1 Region A (0 to 0.2 s): ΔΦ = 40 − 0 = 40 mWb = 0.040 Wb; Δt = 0.2 s; ε = 50 × 0.040 / 0.2 = 10 V. (1 mark correct ΔΦ/Δt, 1 mark final answer.)
2.2 Region B (0.2 to 0.5 s): Induced emf = 0 V. The flux is constant in this interval (ΔΦ = 0), so ΔΦ/Δt = 0 and Faraday's Law gives zero emf. (1 mark answer, 1 mark explanation referencing constant flux.)
2.3 Region C (0.5 to 0.8 s): ΔΦ = 0 − 40 = −40 mWb; Δt = 0.3 s; |ε| = 50 × 0.040 / 0.3 ≈ 6.7 V. Magnitude is smaller than Region A (10 V) because the same ΔΦ change occurs over a longer time interval (0.3 s vs 0.2 s), giving a smaller rate of flux change. (1 mark ε calculation, 1 mark comparison, 1 mark reason referencing Δt.)
2.4 Sketch should show: non-zero positive constant emf in Region A; zero emf in Region B; non-zero negative (or opposite direction) constant emf in Region C (smaller magnitude than A); zero emf in Region D. (1 mark correct zero/nonzero pattern, 1 mark indicating regions B and D are zero.)
Q3 — Compare scenarios
Scenario emf values (convert mWb → Wb, ms → s):
A: 50 × 0.020/0.100 = 10 V
B: 100 × 0.020/0.100 = 20 V
C: 50 × 0.040/0.100 = 20 V
D: 50 × 0.020/0.050 = 20 V
E: 200 × 0.020/0.200 = 20 V
3.1 Doubling N from 50 to 100 doubles the emf from 10 V to 20 V. Increasing N increases flux linkage, so the total induced emf is proportionally larger. (1 mark correct comparison, 1 mark explanation.)
3.2 Halving Δt from 100 ms to 50 ms doubles the emf from 10 V to 20 V. A shorter Δt means a greater rate of flux change (ΔΦ/Δt is doubled), so Faraday's Law gives a doubled emf. (1 mark correct comparison, 1 mark explanation.)
3.3 Quadrupling N (factor of 4) and doubling Δt (factor of 2) gives a net factor of 4/2 = 2 → emf = 2 × 10 = 20 V, which is larger than Scenario A. The quadrupling of N outweighs the halving effect of the doubled Δt. (1 mark correct prediction with verification.)