Source: High school physics (Chinese)
Problem Sets:
Problem
A rectangular coil is wound from $N = 50$ turns of insulated fine wire; its sides are $10$ cm and $5$ cm long, and it carries a current $I = 0.10$ A. The coil can rotate about one of its sides, $OO'$ (see Figure 14.34). A uniform external magnetic field of magnitude $B = 0.050$ T is applied so that $\vec{B}$ makes a $30^{\circ}$ angle with the plane of the coil. Find the torque on the coil.
The torque on a planar coil is
$$\tau = NIAB\sin\theta,$$where $\theta$ is the angle between $\vec{B}$ and the coil's normal. Since $\vec{B}$ makes $30^{\circ}$ with the coil plane, it makes $90^{\circ}-30^{\circ} = 60^{\circ}$ with the normal; equivalently $\tau = NIAB\cos 30^{\circ}$.
With $A = 0.10\times 0.05 = 5.0\times 10^{-3}\ \text{m}^2$:
$$\tau = 50\times 0.10 \times 5.0\times10^{-3}\times 0.050 \times \cos 30^{\circ}.$$Multiplying:
$$\tau = 1.25\times 10^{-3} \times \frac{\sqrt{3}}{2} \approx 1.08\times 10^{-3}\ \text{N}\cdot\text{m}.$$