MYP Physics · topic MYP-14.03
Force on a current-carrying conductor
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.
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.