IGCSE · 4 · extended response
Induced current opposes the change, not the field
Electricity and magnetism · Original GioPhysics question with a detailed, mark-by-mark answer guide.
- Demand
- discriminating
- Marks
- 6
- Topics
- 1
- Answer
- Complete
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.
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
Ready to self-mark?Reveal the detailed answer guide
(a)
Calculate the signed induced emf and current during each interval, using ε = −N dΦ/dt.
Build a chain of claim, evidence and physics reasoning. Address more than one relevant factor, identify a limitation or assumption, and finish with a conclusion that is conditional on the evidence rather than absolute.
- 1
With the field normal to the coil, flux per turn is Φ = BA and flux linkage is NBA.
- 2
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.
- 3
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.
(b)
Find total resistive energy dissipated and explain the energy source and current direction during the decreasing-field interval.
Build a chain of claim, evidence and physics reasoning. Address more than one relevant factor, identify a limitation or assumption, and finish with a conclusion that is conditional on the evidence rather than absolute.
- 1
During each 0.20 s ramp, I²Rt = (0.40)²(5.0)(0.20) = 0.16 J; total heating is 0.32 J.
- 2
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.
- 3
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.
Complete marking guidance
Use these worked steps alongside the part-by-part answers. They explain the method and mark allocation; they do not add extra marks.
- 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.
Private on this device