IB Diploma Physics · topic D.4
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.
Reducing grid losses while respecting transformer limits
An ideal transformer raises an AC supply from 250 V to 10.0 kV. It delivers 20.0 kW to a transmission line with total loop resistance 5.0 Ω. Compare this with delivering the same 20.0 kW into the line at 250 V. Treat 20.0 kW as power at the sending end and use rms values. The graph supplies transformer voltage against turns ratio. Supplied model: For an ideal transformer use Vs/Vp = Ns/Np and VpIp = VsIs; wire heating is I²R. Supplied data (Secondary/primary turns ratio; Secondary voltage / V): (1, 250), (10, 2500), (20, 5000), (40, 10000). (a) Find turns ratio, secondary and primary currents, line loss and receiving-end power for the 10 kV case. [3 marks] (b) Evaluate the direct 250 V comparison and explain why a transformer does not work continuously with steady DC. [3 marks]
Ideal transformer voltage ratio
Supplied question data. Read the labelled axes and use the observations to test the physical model.
Vertical axis (y): Secondary voltage / V
Horizontal axis (x): Secondary/primary turns ratio
Read the exact data as a table
| Secondary/primary turns ratio | Secondary voltage / V |
|---|---|
| 1 | 250 |
| 10 | 2500 |
| 20 | 5000 |
| 40 | 10000 |
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): Find turns ratio, secondary and primary currents, line loss and receiving-end power for the 10 kV case.
Reveal the worked solution
- Vs/Vp = Ns/Np = 10000/250 = 40. At the sending end Is = P/Vs = 20000/10000 = 2.0 A.
- For an ideal transformer input power equals output power, so Ip = 20000/250 = 80 A.
- Line heating is I²R = 2²(5) = 20 W, leaving 20000−20 = 19980 W at the receiving end; the line voltage drop is IR = 10 V.
- Using sending current P/V = 20000/250 = 80 A would give a 400 V line drop and I²R = 32000 W, exceeding the sending power.
- This contradicts a passive receiving load at the assumed operating point; the calculation demonstrates why voltage drop and the load must be solved together, and why high-voltage transmission is useful.
- A steady DC current produces steady flux after the switching transient, so it induces no sustained secondary emf. Real AC transformers also have core and winding losses and insulation limits.
Answer (a) Turns ratio = 40; secondary current = 2.0 A; primary current = 80 A; line loss = 20 W; received power = 19.98 kW. (b) The naive 250 V case demands 32 kW of line loss for only 20 kW sent, revealing that the assumed delivered operating point is impossible with a passive load.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.