## ERQ · 14 marks · Topics: D.4 Induction (HL) + B.5 Current and circuits
**Stem.** A research team studies induction heating using a single rectangular copper loop of area A = 0.50 m² placed horizontally inside a large solenoid. The solenoid produces a spatially uniform but time-varying magnetic field directed vertically through the loop, given by B(t) = B₀(1 + αt), with B₀ = 0.80 T and α = 0.020 s⁻¹. The plane of the loop is perpendicular to the field at all times. The loop (of internal resistance r = 0.50 Ω) is connected through fine leads to an external variable resistor of resistance R_ext = 1.50 Ω and an ideal ammeter, forming a closed series circuit. The team records the current and uses it to estimate the thermal energy delivered to the resistor between t = 0 and t = 10 s. Edge effects and self-inductance of the loop are assumed negligible.
### Part (a) State [2 marks] · AO1 · Topic: D.4
State an expression for the magnetic flux Φ through the loop at time t, and define induced electromotive force (emf).
### Part (b)(i) Derive [3 marks] · AO2/AO3 · Topic: D.4
Using Faraday's law, derive an expression for the magnitude of the induced emf ε in the loop as a function of the given constants. Comment briefly on its time dependence.
### Part (b)(ii) Calculate [3 marks] · AO2/AO3 · Topic: D.4 + B.5
Calculate the current registered by the ammeter at t = 5.0 s.
### Part (c) Determine [4 marks] · AO3 · Topic: D.4 + B.5
Determine the total electrical energy dissipated in the external resistor R_ext between t = 0 and t = 10 s.
### Part (d) Suggest [2 marks] · AO3 · ASSUMPTIONS DISCRIMINATOR
The external resistor is replaced by a thermistor whose resistance falls as it heats up. Suggest how this would affect the validity of your answer in (c), and state one consequence for the design of an induction-heating control circuit.
---
## Mark Scheme
### Part (a) [2 marks]
- M1: Φ = B(t)·A = B₀A(1 + αt) (must include area; symbolic form acceptable) [ECF: no]
- M2: Induced emf is the rate of change of magnetic flux linkage through a circuit / work done per unit charge by the non-electrostatic field around the loop [ECF: no]
### Part (b)(i) [3 marks]
- M1: Statement of Faraday's law ε = −dΦ/dt (negative sign or magnitude both acceptable) [ECF: no]
- M2: Substitution dΦ/dt = d/dt [B₀A(1 + αt)] = B₀Aα [ECF: yes from a]
- M3: |ε| = B₀Aα ; constant in time / independent of t [ECF: yes]
### Part (b)(ii) [3 marks]
- M1: |ε| = (0.80)(0.50)(0.020) = 8.0 × 10⁻³ V [ECF: yes from b(i)]
- M2: I = ε/(r + R_ext) = 8.0 × 10⁻³ / (0.50 + 1.50) [ECF: yes]
- M3: I = 4.0 × 10⁻³ A = 4.0 mA (2 s.f., correct unit) [ECF: yes]
### Part (c) [4 marks]
- M1: Recognises emf (hence current) is constant in time, so P is constant and E = P·Δt [ECF: yes]
- M2: Power in R_ext: P_ext = I²·R_ext = (4.0 × 10⁻³)²·(1.50) = 2.4 × 10⁻⁵ W [ECF: yes from b(ii)]
- M3: Substitution E = (2.4 × 10⁻⁵)(10) [ECF: yes]
- M4: E = 2.4 × 10⁻⁴ J (2 s.f., correct unit) [ECF: yes]
### Part (d) [2 marks]
- M1: As thermistor heats, R_ext decreases ⇒ total circuit resistance falls ⇒ current rises (since ε is fixed by Faraday's law and independent of R) ⇒ the constant-R calculation in (c) underestimates / no longer correctly partitions energy between r and R_ext; relation P = I²R is still valid instantaneously but must be integrated over time [ECF: no]
- M2: One valid design consequence, e.g.: thermal runaway risk (positive-feedback heating) requires a current-limiting / ballast resistor; or a feedback controller is needed to stabilise the operating point; or fuse protection because steady-state current is no longer predictable from initial R [ECF: no]
### Marker notes
- (b)(i): Award full marks for arriving at ε = B₀Aα by any equivalent route (e.g. ε = A·dB/dt). The comment "constant / time-independent" is required for M3.
- (c) alternative: integrating ∫₀¹⁰ I²R_ext dt with constant I gives the same result; explicit integral notation acceptable in lieu of M1.
- (c) common error: using (r + R_ext) instead of R_ext for the energy in the *external* resistor — penalise M2 only; M3, M4 ECF.
- (d) discriminator: accept any of {non-ohmic behaviour invalidates constant-P assumption; current rises as R falls causing more heating; need for stabilising element; calculation in (c) is an underestimate of energy delivered to thermistor over 10 s}. Do NOT accept vague answers such as "the resistor gets hot" without linking to R changing.