Torque on a Current Loop

Lesson

Part 1 — why a coil carrying current turns in a magnetic field, and why a small loop behaves like a tiny magnet.

In this lesson you will

Derive the torque on a rectangular coil in a uniform field, τ = NIAB sin θ
Define the magnetic moment m = NIA and find its direction with the right-hand rule
Treat a current loop as a magnetic dipole and find the orbital magnetic moment of an electron in a Bohr orbit

Diagram

Forces on the four sides of a coil

A rectangular coil PQRS ( $N$ turns, current $I$ ) can turn about an axis through its centre, perpendicular to a uniform field $\vec B$ . Each side feels $\vec F = NI\,\vec l \times \vec B$ .

QR and SP (length $l$ ) stay perpendicular to $\vec B$ : forces $NIlB$ , equal and opposite, on different lines: a couple.
PQ and RS (length $b$ ): equal and opposite forces along the same line (the axis): they cancel without turning.
Net force on the coil in a uniform field is zero.

Diagram

Deriving τ = NIAB sin θ

Seen from above, the two forces $F = NIlB$ act along parallel lines a distance $b\sin\theta$ apart, where $\theta$ is the angle between the normal to the coil and $\vec B$ : $\tau = NIlB \times b\sin\theta = NIAB\sin\theta \qquad (A = lb)$

The result holds for a flat coil of any shape of area $A$ .
$\theta$ is measured from the normal, not from the plane of the coil.

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