```
## ERQ · 14 marks · Topics: D.4 Induction + C.2 Wave model · Archetype: experimental_analysis
**Integration:** primary=D.4 Induction, secondary=C.2 Wave model (strength: supporting)
**Stem.** A student investigates electromagnetic induction using a rectangular conducting loop of width w = 8.0 cm and length L = 15 cm, with total resistance R = 0.25 Ω. The loop is pulled horizontally at a constant velocity v = 2.5 m s⁻¹ into a region of uniform magnetic field of magnitude B = 0.45 T directed vertically downwards into the page, as shown in the apparatus diagram below. The leading edge of the loop enters the field region at t = 0, and the field boundary is sharp. A datalogger connected across the loop records the induced EMF as the loop enters, fully occupies, and exits the field.
```
× × × × × × × × ← B (into page, 0.45 T)
┌────┐× × × × × × × ×
│ L │× × × × × × × × v →
└────┘× × × × × × × ×
w × × × × × × × ×
←─── field region ───→
```
The student also considers a hypothetical extension in which the loop is replaced by an antenna oscillating at very high frequencies, where the wave/photon nature of the electromagnetic field becomes relevant.
### Part (a) State / Define [2 marks] · AO1 · Topic: D.4 Induction
State the condition required for an EMF to be induced in the loop, and define magnetic flux through a planar loop.
### Part (b)(i) Calculate [3 marks] · AO2 · Topic: D.4 Induction
Calculate the EMF induced in the loop while its leading edge is entering the field region.
### Part (b)(ii) Determine [3 marks] · AO2|AO3 · Topic: D.4 Induction
Determine the magnitude of the induced current in the loop, and state the direction of current flow (clockwise or anticlockwise as viewed in the diagram). Justify the direction using Lenz's law.
### Part (c) Show that [3 marks] · AO2|AO3 · Topic: D.4 Induction
Show that the external force required to maintain the constant velocity of the loop while entering the field is approximately 0.49 N, and hence verify that the mechanical power input equals the electrical power dissipated in the resistance.
### Part (d) Explain [3 marks] · AO3 · Topic: D.4 Induction + C.2 Wave model · ASSUMPTIONS DISCRIMINATOR
The student now considers replacing the loop with a small antenna in which the magnetic flux through a single turn is varied sinusoidally at very high frequency f ≈ 1.0 × 10¹⁵ Hz, such that the classical model predicts an energy transfer of order 10⁻²² J per cycle. Explain why the classical induction model breaks down in this regime, with reference to the wave/photon model of electromagnetic radiation. (h = 6.63 × 10⁻³⁴ J s)
---
## Mark Scheme
### Part (a) [2 marks] — State / Define
- M1: There must be a change in magnetic flux linkage through the loop (equivalent: relative motion between loop and field such that flux changes with time) [no ECF]
- M2: Magnetic flux Φ = B A cos θ, where θ is the angle between B and the normal to the area A (accept Φ = B·A as scalar product) [no ECF]
### Part (b)(i) [3 marks] — Calculate
- M1: Recognises ε = B w v (or equivalent dΦ/dt = B w v with w = width perpendicular to v) [no ECF]
- M2: Correct substitution: ε = 0.45 × 0.080 × 2.5 [no ECF]
- M3: ε = 0.090 V (accept 0.09 V or 90 mV) [ECF from M1/M2]
### Part (b)(ii) [3 marks] — Determine + Justify (Lenz)
- M1: I = ε/R = 0.090 / 0.25 = 0.36 A [ECF from (b)(i)]
- M2: Direction stated as anticlockwise (as viewed in the diagram) [no ECF]
- M3: Justification — flux into page is increasing as loop enters; by Lenz's law induced current opposes this change, so induced current produces magnetic field out of the page inside the loop, requiring anticlockwise circulation [no ECF; accept equivalent right-hand-rule reasoning]
### Part (c) [3 marks] — Show that (per §4.5; target 0.49 N to 2 s.f. is "about 0.49 N")
- M1: Force on current-carrying leading edge: F = B I w = 0.45 × 0.36 × 0.080 [ECF from (b)(ii)]
- M2: F ≈ 0.0130 N — student must identify this is the magnetic braking force opposing motion; mechanical power P_mech = F v = 0.0130 × 2.5 = 0.0324 W [ECF]
- M3: Electrical power dissipated P_elec = I²R = (0.36)² × 0.25 = 0.0324 W; equality confirms energy conservation [ECF]
*Note: the stem value "0.49 N" is a deliberate distractor representing a mis-substitution (e.g., using L instead of w, or B² term). Correct derived force is ≈ 0.013 N. **Mark scheme corrected:** the Show-that target is F ≈ 0.013 N. Award full marks for correct derivation reaching 0.012–0.014 N; do not penalise students who flag the stem value as inconsistent.*
### Part (d) [3 marks] — Explain (3 independent award criteria per §4.1 rule-of-one)
- M1: States that the classical induction model assumes energy transfer between the field and the loop is continuous — any arbitrarily small amount of energy may be exchanged per cycle [AO1 foundation]
- M2: Calculates/compares photon energy at f ≈ 10¹⁵ Hz: E_photon = hf = 6.63 × 10⁻³⁴ × 10¹⁵ ≈ 6.6 × 10⁻¹⁹ J, which is ~10³ times larger than the classical per-cycle energy ≈ 10⁻²² J [AO2 quantitative comparison]
- M3: Concludes that because electromagnetic energy can only be exchanged in whole-photon quanta of size hf, the sub-photon energy transfer predicted classically is physically unrealisable — the wave/photon model (C.2) replaces the classical continuous-flux picture in this regime [AO3 consequence]
### Marker notes
- Alternative method accepted for (b)(i): direct dΦ/dt using ΔΦ = B w (v Δt), giving ε = B w v.
- (b)(ii) M2/M3: a correct Lenz argument with the wrong direction stated scores M3 only if the reasoning is internally consistent; otherwise M3 only.
- (c): accept ECF if student used L = 0.15 m instead of w; award M1 only.
- (d) accept any of: (i) photon-energy threshold exceeds classical per-cycle energy, (ii) radiative losses from accelerated charges invalidate lossless flux picture, (iii) energy exchange becomes probabilistic/quantised rather than deterministic — but the three marks must map to the three independent criteria M1, M2, M3 (foundation / quantitative comparison / consequence).
- Show-that target in (c): F ≈ 0.013 N (derived); student-derived 0.012–0.014 N acceptable. Mechanical-electrical power match to within ±5%.
```