MYP Physics · unit 14
Magnetism and electromagnetism
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
A selection from this pathway. Open a topic below for its complete practice and challenge set.
Mapping an electromagnet without mistaking a field line for a path
Optional enrichment: this challenge may extend beyond the usual syllabus depth and assumes confidence with algebra and graph interpretation. A long solenoid is tested at its centre. Current and magnetic flux-density measurements are supplied. A compass at the right-hand end points away from the coil when current is positive. For a long air-core solenoid use B = μ₀nI, with μ₀ = 4π×10⁻⁷ T m A⁻¹. Ignore Earth's field after background subtraction. Supplied data (Current / A; Flux density / mT): (-2, -2), (-1, -1), (0, 0), (1, 1), (2, 2). (a) Identify the right-end pole for positive current, estimate turns per metre, and predict the field after reversing a 1.5 A current. [3 marks] (b) Design a fair comparison of air and iron cores and explain a limitation of field-line maps. [3 marks]
Solenoid field after background subtraction
Supplied question data. Read the labelled axes and use the observations to test the physical model.
Vertical axis (y): Flux density / mT
Horizontal axis (x): Current / A
Read the exact data as a table
| Current / A | Flux density / mT |
|---|---|
| -2 | -2 |
| -1 | -1 |
| 0 | 0 |
| 1 | 1 |
| 2 | 2 |
These are the supplied values, not a worked answer. No fitted trend is drawn.
Show a hint
Begin with part (a): Identify the right-end pole for positive current, estimate turns per metre, and predict the field after reversing a 1.5 A current.
Reveal the worked solution
- Outside a magnet, a compass's north-seeking end follows the field direction from north to south. Pointing away from the right end identifies that end as north.
- The measured gradient is 1.0×10⁻³ T A⁻¹ = μ₀n, so n = 0.001/(4π×10⁻⁷)≈796 m⁻¹.
- Reversing current reverses field direction. At −1.5 A the linear calibration predicts −1.5 mT; its magnitude is unchanged.
- Control current rather than supply voltage, because resistance and heating may change. Keep probe orientation and centre position fixed, repeat measurements and subtract the ambient field.
- Measure a current-up and current-down sequence to detect hysteresis or remanence in the iron core; the air-core proportional model need not describe a saturating core.
- Field lines represent local directions and relative density. They are not physical wires or trajectories that isolated magnetic poles follow; a complete magnetic field map forms loops.
Answer (a) Right end is north; n≈796 turns m⁻¹; reversed field is −1.5 mT. (b) Vary core material at fixed coil geometry and current; measure signed field at fixed locations and check for residual magnetism.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.
A motor coil that must keep turning
Optional enrichment: this challenge may extend beyond the usual syllabus depth and assumes confidence with algebra and graph interpretation. A rectangular coil has 20 turns and area 0.010 m² in a uniform 0.20 T field. It carries 2.0 A. The supplied graph gives torque versus sinθ, where θ is the angle between the coil's area normal and the magnetic field. Use τ = NBIA sinθ. Supplied data (sin θ; Torque / N m): (0, 0), (0.5, 0.04), (1, 0.08). (a) Find the maximum torque, torque at θ = 30°, and mechanical power when the instantaneous torque is 0.060 N m at 50 rad s⁻¹. [3 marks] (b) Explain why a DC motor needs current reversal and assess one energy-loss mechanism and one useful application trade-off. [3 marks]
Coil torque calibration
Supplied question data. Read the labelled axes and use the observations to test the physical model.
Vertical axis (y): Torque / N m
Horizontal axis (x): sin θ
Read the exact data as a table
| sin θ | Torque / N m |
|---|---|
| 0 | 0 |
| 0.5 | 0.04 |
| 1 | 0.08 |
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 the maximum torque, torque at θ = 30°, and mechanical power when the instantaneous torque is 0.060 N m at 50 rad s⁻¹.
Reveal the worked solution
- The torque factor is NBIA = 20(0.20)(2.0)(0.010) = 0.080 N m.
- At θ = 30°, multiply by sin30° = 0.50 to obtain 0.040 N m. Torque is greatest when the normal is perpendicular to the field.
- Instantaneous mechanical power is τω = 0.060(50) = 3.0 W. Opposite forces on opposite sides form a couple, even if the net force on the whole coil is zero.
- Without reversal, the magnetic couple would reverse after half a turn and oppose continued rotation. Inertia carries the coil through positions of zero torque.
- Electrical input is divided between mechanical output and losses such as I²R heating, bearing friction and magnetic losses; power output cannot exceed input.
- Increasing current increases torque but raises resistive heating quadratically. A lifting motor needs sufficient torque while meeting temperature and efficiency constraints.
Answer (a) Maximum torque = 0.080 N m; at 30° = 0.040 N m; mechanical power = 3.0 W. (b) A split-ring commutator reverses coil current each half-turn, maintaining torque in the rotation direction.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.
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.
Reducing grid losses while respecting transformer limits
Optional enrichment: this challenge may extend beyond the usual syllabus depth and assumes confidence with algebra and graph interpretation. 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.