Skip to main content
TASPhysicsSyllabus dot point

How does a current in a magnetic field produce motion?

Describe the magnetic force on currents and charges and explain the operation of a DC motor.

The force on a current-carrying conductor and a moving charge in a magnetic field, the right-hand rule, and how torque on a coil drives a direct current motor.

Generated by Claude Opus 4.77 min answer

Reviewed by: AI editorial process; not yet individually human-reviewed

Have a quick question? Jump to the Q&A page

What this dot point is asking

This dot point is about the motor effect: the force that a magnetic field exerts on moving charge, whether that charge flows as a current in a wire or moves as a free particle.

Force on a current-carrying conductor

When a straight conductor of length LL carrying current II sits in a uniform magnetic field of strength BB, it experiences a force:

F=BILsinθF = BIL\sin\theta

where θ\theta is the angle between the current direction and the field. The force is maximum when the wire is perpendicular to the field (θ=90\theta = 90^\circ) and zero when it is parallel to the field. The magnetic field strength BB is measured in tesla (T).

The direction of the force is perpendicular to both the current and the field. Use the right-hand rule: point your fingers in the direction of conventional current, curl them toward the field BB, and your thumb gives the force. Many texts use the right-hand slap or palm rule with the same result.

Force on a moving charge

A single charge qq moving at speed vv through a field BB feels:

F=qvBsinθF = qvB\sin\theta

If the charge moves perpendicular to a uniform field, this force is constant in magnitude and always perpendicular to the velocity, so the charge follows a circular path, exactly like centripetal motion. Setting qvB=mv2rqvB = \dfrac{mv^2}{r} gives the radius r=mvqBr = \dfrac{mv}{qB}.

The DC motor

A direct current motor converts electrical energy into rotational kinetic energy using the motor effect. A rectangular coil carrying current sits in a magnetic field. The two sides of the coil that lie across the field experience forces in opposite directions (because the current flows in opposite directions in them), producing a turning effect, or torque, on the coil.

The torque on a coil of NN turns and area AA is:

τ=NBIAcosϕ\tau = N B I A \cos\phi

where ϕ\phi is the angle between the coil plane and the field. Torque is maximum when the coil is parallel to the field and zero when the coil is perpendicular to it.

The key engineering problem is that as the coil passes through the vertical position the torque would reverse and stop the rotation. A split-ring commutator solves this by reversing the current direction in the coil every half turn, so the torque always acts to keep the coil rotating the same way.

When answering motor questions in the exam, describe the forces on each side of the coil, explain how they create a torque, and clearly state the role of the commutator in maintaining continuous rotation. A labelled diagram strongly supports your explanation.

Exam-style practice questions

Practice questions written in the style of TASC exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

2024 TASC4 marksA rectangular coil is made of 30 turns of wire, 5.00 cm by 2.00 cm, inside a uniform magnetic field of flux density 0.2 T. A DC current of 5 A flows in the coil. Calculate the magnitude of the force on the side AB (the 5.00 cm side), and explain why the commutator strips are wrapped only part way around the axle.
Show worked answer →

The force on a current-carrying conductor in a field is F = n B I L (L perpendicular to B), with n turns.

F = 30 x 0.2 x 5 x 0.05 = 1.5 N.

So each long side experiences 1.5 N, and the two forces on opposite sides act in opposite directions to produce a torque that turns the coil.

Commutator: the split commutator strips are wrapped only part way around the axle so that they reverse the direction of current in the coil every half turn. As the coil passes the vertical (plane parallel to the field), the gaps between the strips break contact, and when contact resumes the current direction has flipped. This keeps the torque acting in the same rotational sense, so the motor keeps spinning one way instead of oscillating. Markers want F = nBIL and that the commutator reverses current each half-turn to maintain rotation.

2022 TASC5 marksAn old moving coil ammeter has a rectangular coil 1.00 x 1.00 cm with 20 turns rotating in a radial magnetic field. The restoring springs require 2.67 x 10^-3 N on each active side per degree of rotation. When 1.00 A flows, the pointer moves through 9 degrees. Calculate the force on each side of the coil, and hence the magnetic flux density of the field.
Show worked answer →

At the 9 degree reading the magnetic force on each side balances the spring restoring force.

Spring force on each side = 2.67 x 10^-3 N per degree x 9 degrees = 2.40 x 10^-2 N.
So the magnetic force on each active side is F = 2.40 x 10^-2 N.

The magnetic force on each side with 20 turns is F = n B I L:
2.40 x 10^-2 = 20 x B x 1.00 x 0.01
2.40 x 10^-2 = 0.20 B
B = 0.12 T.

The flux density between the magnets is about 0.12 T. Markers want the spring force at 9 degrees, then B = F / (n I L) using the 1.00 cm side length.

2023 TASC1 marksStudents carry out a magnetic-force-on-a-current against angle experiment, passing 10 A through a 1.0 cm wire in a field and reading the force from a balance as the wire is rotated through an angle. Which force equation applies to the wire?
Show worked answer →

The force on a straight current-carrying wire in a magnetic field depends on the angle between the wire (current direction) and the field:

F = B I L sin(theta)

where B is the flux density, I the current, L the length of wire in the field, and theta the angle between the current and the field.

This explains the data: the force is zero when the wire is parallel to the field (theta = 0, sin0 = 0) and maximum when perpendicular (theta = 90, sin90 = 1). Plotting force against sin(theta) gives a straight line whose gradient is B I L, allowing B to be found. Markers want F = B I L sin(theta).