AP Physics · topic 13
Electromagnetic Induction
Connect the ideas, make your working visible, and test the result. Use the hints when you need a starting point; open the full solution after your attempt.
Extra challenging questions
Work through the linked stages and explain what each result means.
Induced current opposes the change, not the field
A 100-turn closed coil of area 0.010 m² and resistance 5.0 Ω has its normal parallel to a uniform magnetic field. The field changes linearly through the supplied points. Positive flux is defined along the coil normal; neglect self-inductance for the current calculation. Supplied data (Time / s; Magnetic field / T): (0, 0), (0.2, 0.4), (0.5, 0.4), (0.7, 0). (a) Calculate the signed induced emf and current during each interval, using ε = −N dΦ/dt. [3 marks] (b) Find total resistive energy dissipated and explain the energy source and current direction during the decreasing-field interval. [3 marks]
Applied field through the coil
Supplied question data. Read the labelled axes and use the observations to test the physical model.
Vertical axis (y): Magnetic field / T
Horizontal axis (x): Time / s
Read the exact data as a table
| Time / s | Magnetic field / T |
|---|---|
| 0 | 0 |
| 0.2 | 0.4 |
| 0.5 | 0.4 |
| 0.7 | 0 |
These are the supplied values, not a worked answer. Lines join the supplied values; they are not a fitted model.
Show a hint
Begin with part (a): Calculate the signed induced emf and current during each interval, using ε = −N dΦ/dt.
Reveal the worked solution
- With the field normal to the coil, flux per turn is Φ = BA and flux linkage is NBA.
- The initial field gradient is +2.0 T s⁻¹, so ε = −100(0.010)(2.0) = −2.0 V and I = ε/R = −0.40 A.
- Constant field gives zero flux-change emf. The final gradient is −2.0 T s⁻¹, reversing both emf and current to +2.0 V and +0.40 A.
- During each 0.20 s ramp, I²Rt = (0.40)²(5.0)(0.20) = 0.16 J; total heating is 0.32 J.
- When the applied field decreases, the induced current creates a field in the original positive direction, opposing the decrease rather than always opposing the existing field.
- Lenz's law gives a reaction opposing the flux change. The apparatus maintaining that change must supply energy that is dissipated in the coil; induction is not a free source of energy.
Answer (a) 0–0.2 s: ε = −2.0 V, I = −0.40 A; 0.2–0.5 s: zero; 0.5–0.7 s: ε = +2.0 V, I = +0.40 A. (b) Total heating = 0.32 J; the external apparatus changing the field supplies the energy.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.