20.2 Force on a current-carrying conductor
- Syllabus
- 9702–2028–2029
- Topic
- 20.2
- Level
- A2
A current-carrying conductor experiences magnetic force when its current has a component perpendicular to an external magnetic field.
Reverse current or field direction to reverse force. Parallel current and field produce no force in the ideal straight-wire model.
A wire between magnet poles deflects when current flows, forming the basis of a simple motor.
The force is not caused by current alone; it requires interaction with an external magnetic field.
forcemagnitudeF=BILsinθ
| Symbol | Meaning |
|---|---|
| B | magnetic flux density in T |
| I | conventional current in A |
| L | wire length within the field in m |
| θ | angle between conventional current and magnetic field |
| Fleming's left-hand digit | Direction |
|---|---|
| First finger | magnetic Field, N to S |
| seCond finger | conventional Current, + to - |
| thuMb | Motion / force on conductor |
At θ=90°, F=BIL is maximum. At θ=0° or 180°, F=0. Reversing either I or B reverses force; reversing both leaves force direction unchanged.
A 0.20 m wire carrying 3.0 A at 30° to a 0.50 T field experiences F=(0.50)(3.0)(0.20)sin30°=0.150 N. Use the left hand separately to give direction.
Use conventional current, not electron flow, with Fleming's rule. θ is the angle between I and B, not between force and field; the force is perpendicular to both I and B.
For a wire perpendicular to a field, B=F/(IL), measured in tesla, so 1 T=1 N A⁻¹ m⁻¹.
The definition assumes the conductor is perpendicular; otherwise divide by IL sinθ or resolve the perpendicular component.
A 0.40 N force on a 0.20 m wire carrying 2.0 A gives B=1.0 T when perpendicular.
Tesla is not force per charge; that relates to electric field, while B describes magnetic force on current or moving charge.