Force on a Current-Carrying Conductor

A magnetic field is produced by moving charges or permanent magnets. A wire carrying current $I$ in a field of flux density $B$ feels a force $F=BIL\sin\theta$ , where $\theta$ is the angle between the wire and the field. The force is greatest when they're perpendicular and zero when parallel. Its direction is given by Fleming's left-hand rule. Magnetic flux density is defined as the force per unit current per unit length on a wire at right angles to the field; $1$ T $=1$ N A⁻¹ m⁻¹.

Cambridge 9702 Paper 4Edexcel YPH11 WPH14AQA 7408 Paper 2≈ 35 minRevision 5 · updated 2026-10-27

Section 1 of 14

Learning objectives

Draw magnetic field lines for magnets, wires and solenoids.
Use $F=BIL\sin\theta$ .
Apply Fleming's left-hand rule.
Define magnetic flux density and the tesla; describe a current-balance measurement.

Pause and think

A loudspeaker has no moving parts except a cone attached to a coil of wire in a magnet. How does a changing current make sound?

Section 2 of 14

Core idea

Start hereWeighing a magnetic field

Put a magnet on a top-pan balance, run a current through a wire between its poles, and the balance reading changes. The wire and magnet push on each other. Measuring that force is how we define the strength of a magnetic field.

The big idea

$F=BIL\sin\theta$ (Fleming's left-hand rule). Defines $B=\frac F{IL}$ in tesla.

1. The force

Force on a wire as its angle to the field changes (B = 0.2 T, I = 3 A, L = 0.5 m)

Maximum (0.3 N) when the wire is perpendicular to the field; zero when parallel.

F=BIL\sin\theta

$B$: flux density (T).
$I$: current (A).
$L$: length in the field (m).
$\theta$: angle between wire and field.

Worked exampleB = 0.2 T, I = 3 A, L = 0.5 m

Perpendicular, $30^\circ$ , parallel.

Goal: Three forces.
Strategy: $\sin\theta$ factor.

$0.2\times3\times0.5=0.30\text{ N}$
$\sin90^\circ=1$ .
$0.30\sin30^\circ=0.15\text{ N}$
$30^\circ$ .
$0$
$\sin0=0$ .

Answer

0.30

0.15

0

✓ CheckOnly the perpendicular component of the current matters ✓.

2. Measuring B

Measuring B with a current balance

The wire is pushed down; by Newton's third law the magnet (on a balance) is pushed up, changing the reading by

\Delta m

. Then

B=\dfrac{\Delta m\,g}{IL}

Worked exampleCurrent balance

Reading changes by 2.4 g with 4.0 A in 5.0 cm of wire.

Goal: $B$ .
Strategy: Force from the mass change.

$F=0.0024\times9.81=0.0235\text{ N}$
Equivalent weight.
$B=\frac{0.0235}{4.0\times0.050}=0.12\text{ T}$
$F/IL$ .

Answer

0.12

✓ CheckTypical permanent magnet strength ✓.

Common misconceptionRight hand Force on a current: left hand.

Summary

$F=BIL\sin\theta$ .
Left-hand rule.

Tesla: N A $^{-1}$ m $^{-1}$ .

HistoryØrsted's lecture surprise. In

1820

Hans Christian Ørsted noticed, during a lecture, that a compass needle swung when he switched on a nearby current. It was the first evidence that electricity and magnetism are linked. Within weeks, Ampère had worked out the forces between currents.

Fun factRailguns. A railgun uses

F=BIL

with currents of millions of amps. The projectile completes a circuit between two rails; the huge current and its own magnetic field drive it forward at over

2

km s⁻¹ without any explosive.

Keep reading for free

Create a free account to read the rest of this lesson: key formulae, worked examples, explore the graph, common mistakes, exam tip, guided practice and more. Your answers are checked and your progress is saved.

Create free account Sign in