Torque on a bar magnet in a uniform magnetic field (qualitative)
In a uniform field a bar magnet feels no net force, only a torque τ = m × B. The torque turns it to line up with the field.
What the field does to a magnet
Picture the N pole being pushed along B and the S pole being pushed against B, with equal forces. The two forces cancel, so a uniform field gives no net force. But they act at different points, so they form a couple that turns the magnet. The torque is τ = m × B, of size mB sin θ, where θ is the angle between m and B. The torque is zero when m is along or opposite to B. It is largest (mB) when m is at right angles to B. This is why a compass needle swings and finally settles along the field. It usually swings to and fro a few times about the field direction before coming to rest. This topic is qualitative only: learn the results, not a derivation.
Torque and energy at key angles
| Angle θ between m and B | Torque τ | Potential energy U | State |
|---|---|---|---|
| 0° | 0 | −mB (lowest) | Stable equilibrium |
| 90° | mB (largest) | 0 | Not in equilibrium |
| 180° | 0 | +mB (highest) | Unstable equilibrium |
Stable vs unstable equilibrium
Worked example: torque and work
A bar magnet of moment 0.40 A m² is in a uniform field of 0.25 T. (a) Find the torque when m makes 30° with B. (b) How much work is needed to turn it from θ = 0° to θ = 90°, and from θ = 0° to θ = 180°?
- (a) Formula: τ = mB sin θ.
- τ = 0.40 × 0.25 × sin 30° N m = 0.10 × 0.5 N m = 0.050 N m.
- (b) Work = U(final) − U(initial) = mB(cos θ₁ − cos θ₂), with mB = 0.10 J.
- From 0° to 90°: W = 0.10 × (1 − 0) J = 0.10 J.
- From 0° to 180°: W = 0.10 × (1 − (−1)) J = 0.20 J.
(a) 0.050 N m; (b) 0.10 J and 0.20 J
How to draw a bar magnet in a uniform field
- Draw parallel, equally spaced arrows for the uniform field B.
- Draw the magnet at an angle θ to the field lines. Mark N and S, and draw m from S to N.
- At N draw a force arrow along B; at S draw an equal force arrow against B.
- Mark θ between m and B. Show the turning direction with a curved arrow.
- Write τ = mB sin θ beside the figure.
A compass needle swings and settles along the field because of this torque. Trekking compasses are filled with liquid, so the needle stops swinging quickly and you can read it sooner.
1 marks: “When is the torque on a bar magnet in a uniform magnetic field largest?” When m is perpendicular to B (θ = 90°); then τ = mB.
2 marks: “Write the torque on, and the potential energy of, a magnetic dipole in a uniform field. Name the positions of stable and unstable equilibrium.” τ = m × B (size mB sin θ); U = −m·B (−mB cos θ). Stable at θ = 0°, unstable at θ = 180°.
2 marks: “A bar magnet of moment 0.5 A m² lies along a uniform field of 0.4 T. Find the work needed to turn it through 90° and through 180°.” mB = 0.2 J. W(90°) = mB(1 − 0) = 0.2 J; W(180°) = mB(1 − (−1)) = 0.4 J.
Sine for torque, minus cos for energy: τ = mB sin θ, U = −mB cos θ.