Magnetic force on current-carrying conductors (SACE Stage 2 Physics)

Syllabus dot point

“Calculate the force on a current-carrying conductor in a magnetic field and explain its role in the operation of a motor.”

9 min readUpdated 2 Jun 2026

Quick answer

A current is moving charge, so a current-carrying wire in a magnetic field feels a force: $F = BIL\sin\theta$ , where $I$ is the current, $L$ the length of wire in the field, $B$ the field, and $\theta$ the angle between current and field. The direction is given by the right-hand rule. This "motor effect" is what makes electric motors turn.

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What this dot point is asking
The motor effect
Direction of the force
How a DC motor works
How SACE assesses this
Exam-style questions

What this dot point is asking

You need to calculate the force on a current-carrying conductor in a magnetic field and explain how this force drives an electric motor.

Force on a wire in a field

A 0.15 m wire carries 4.0 A perpendicular to a 0.30 T field

Find the force on the wire.

Force (perpendicular, $\sin 90^\circ = 1$ ):

F = BIL = (0.30)(4.0)(0.15) = 0.18\ \text{N}.

If the wire were at $30^\circ$ to the field instead, the force would be $0.18\times\sin 30^\circ = 0.090\ \text{N}$ .

The motor effect

A current is just charges in motion, so the same magnetic force that acts on moving charges acts on a current-carrying wire. Summing the force over all the charges in a length $L$ of wire gives:

Force on a current-carrying conductor

F = BIL\sin\theta

F

is the force (N),

B

the magnetic flux density (T),

I

the current (A),

L

the length of conductor in the field (m), and

\theta

the angle between the current direction and the field. The force is maximum when the wire is perpendicular to the field (

\theta = 90^\circ

) and zero when parallel.

Direction of the force

Use the right-hand rule (treating conventional current as the direction of motion of positive charge): point fingers along the current $I$ , curl toward the field $B$ ; the thumb/palm gives the force. Equivalently the flat-hand rule: fingers point along $B$ , thumb along $I$ , palm pushes in the force direction. The force is perpendicular to both the current and the field, so reversing either the current or the field reverses the force.

How a DC motor works

A simple motor is a current-carrying coil free to rotate in a magnetic field:

Current flows through the coil. The two sides of the coil carry current in opposite directions.
By the motor effect, the two sides feel forces in opposite directions (one up, one down), producing a turning effect (torque) on the coil.
The coil rotates. As it passes the vertical, the forces would now oppose the rotation.
A split-ring commutator reverses the current direction in the coil every half-turn, so the torque keeps driving the coil the same way and rotation continues.

Why the commutator matters

Without the split-ring commutator, the coil would only swing to the vertical and stop (or oscillate). The commutator reverses the current at the right moment each half-turn so the force always pushes the coil in the same rotational sense, giving continuous rotation.

The turning effect is largest when the coil is parallel to the field (forces have the longest lever arm) and zero when the coil is perpendicular to the field, but the coil's momentum carries it through this point.

How SACE assesses this

SACE Stage 2 questions on the motor effect give the field, current and length of conductor and ask for the magnitude and direction of the force using $F = BIL\sin\theta$ (usually with the wire perpendicular to the field, so $\sin\theta = 1$ ). The direction part expects a clear hand rule applied to conventional current, and you should state the resulting direction explicitly (for example "upward, out of the plane toward the top of the page"). A rearrangement to find the current from a measured force is also common. The most reliable approach is to write the formula, note the angle, substitute in SI units, and then determine the direction separately. A conceptual part may ask why a motor needs a commutator: state that it reverses the current each half-turn so the torque keeps driving the coil the same way, giving continuous rotation.

Common errors

Using the wrong length: $L$ is the length of conductor actually inside the field, not the total wire length.
Forgetting $\sin\theta$: When the wire is at an angle to the field, only the perpendicular component matters; the force is $BIL\sin\theta$ .
Confusing current direction: Hand rules use conventional current (positive to negative outside the source), which is opposite to electron flow.
Saying the motor stops at the vertical without a commutator: It is precisely the commutator's job to reverse the current each half-turn so rotation continues.

Exam-style questions

Questions in the style of SACE Board exam questions on this dot point, each with a worked answer. They are written by ExamExplained unless tagged "Past paper"; the year shows the paper a question is modelled on.

SACE 2024-style3 marks

A straight conductor carries a current of

0.120\ \text{A}

in a uniform magnetic field directed into the plane of the page. The field is

0.350\ \text{T}

and the length of conductor within the field is

2.50\times10^{-2}\ \text{m}

. With the current to the right, determine the magnitude and direction of the force on the conductor.

Show worked answer →

The force on a current-carrying conductor is $F = BIL\sin\theta$ . The field is perpendicular to the current, so $\theta = 90^\circ$ and $\sin\theta = 1$ .

F = BIL = (0.350)(0.120)(2.50\times10^{-2}) = 1.05\times10^{-3}\ \text{N}.

Direction: use the right-hand rule with conventional current to the right and field into the page. The force is directed upward (toward the top of the page).
1 mark for $F = BIL$ , 1 mark for the magnitude of about $1.05\times10^{-3}\ \text{N}$ , 1 mark for the correct direction. This sideways force on wires produces the turning effect in a motor.

SACE 2022-style3 marks

A horizontal wire of length

0.080\ \text{m}

in the field of a magnet carries a current

I

. The wire experiences an upward force of

4.8\times10^{-2}\ \text{N}

in a uniform field of

0.30\ \text{T}

perpendicular to the wire. (a) Calculate the current in the wire. (b) State what happens to the force if the current is reversed.

Show worked answer →

(a) The wire is perpendicular to the field, so $F = BIL$ . Rearrange for the current:

I = \dfrac{F}{BL} = \dfrac{4.8\times10^{-2}}{(0.30)(0.080)} = \dfrac{4.8\times10^{-2}}{2.4\times10^{-2}} = 2.0\ \text{A}.

(2 marks)

(b) Reversing the current reverses the force direction (by the right-hand rule), so the force becomes downward with the same magnitude $4.8\times10^{-2}\ \text{N}$ . (1 mark) Markers reward rearranging $F = BIL$ and recognising the force direction depends on the current direction.

Practise this

Exam-style questions 2 on this page
Flashcards Physics deck

Sources & how we know this

Stage 2 Physics subject outline (2025 onwards) — SACE Board
Stage 2 Physics: external assessment — SACE Board