Force on a current-carrying conductor: $F = BIL \sin \theta$

Magnetic Fields & Forces — AQA A-Level Physics

A currentThe rate of flow of chargeA property of matter that causes it to experience a force in an electromagnetic field. Measured in coulombs (C).. Measured in amperes (A).-carrying conductor has its own magnetic field. Place it inside an external field and the two fields interact — the conductor feels a force.
This is the principle behind every electric motor.

F = BIL \sin \theta

$F$: force on the conductor (N)
$B$: magnetic flux densityMass per unit volume of a material. Measured in kg m⁻³.The strength of a magnetic field. The force per unit length per unit currentThe rate of flow of chargeA property of matter that causes it to experience a force in an electromagnetic field. Measured in coulombs (C).. Measured in amperes (A). on a currentThe rate of flow of chargeA property of matter that causes it to experience a force in an electromagnetic field. Measured in coulombs (C).. Measured in amperes (A).-carrying conductor perpendicular to the field. Measured in teslaThe SI unit of magnetic flux density. One tesla is the flux density when a force of 1 N acts on a 1 m conductor carrying 1 A perpendicular to the field. (T). of the external field (T)
$I$: current in the conductor (A)
$L$: length of conductor within the field (m)
$θ$: angle between the conductor and the magnetic field (°)

L is the length of conductor actually inside the field, not the total wire length.
Maximum force at θ = 90° (perpendicular to field): $F = BIL$.
Zero force at θ = 0° (parallel to field): sin 0° = 0.
The sin θ term isolates the component of current that cuts across the field lines. Wire running along the field lines doesn't cut through any, so no force.

Worked Example

A current of 0.87 A flows in a wire of length 1.4 m placed at 30° to a magnetic field of flux density 80 mT. Calculate the force on the wire.

Show Solution

List known values

Flux densityMass per unit volume of a material. Measured in kg m⁻³.: $B = 80 \text{ mT} = 80 \times 10^{-3} \text{ T}$
Current: $I = 0.87 \text{ A}$
Length: $L = 1.4 \text{ m}$
Angle: $\theta = 30°$

Choose the correct equation

The wire carries a current in a magnetic field, so use:

$$F = BIL \sin \theta$$

Substitute values

$$F = (80 \times 10^{-3}) \times 0.87 \times 1.4 \times \sin 30°$$

Evaluate

Since $\sin 30° = 0.5$:

$$F = (80 \times 10^{-3}) \times 0.87 \times 1.4 \times 0.5$$ $$= 0.049 \text{ N (2 s.f.)}$$

Answer

$F = 0.049$ N

Common Mistake MEDIUM

Students often: Using the angle between the wire and the force, instead of between the wire and the field.
Instead: θ is always the angle between the current direction and the magnetic field direction. Draw both vectors and measure the angle between them.