MYP Physics · topic MYP-14.04
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
Optional enrichment: this challenge may extend beyond the usual syllabus depth and assumes confidence with algebra and graph interpretation. 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.