Magnetism
Beginner
Current in Magnetic Field
Source: High school physics (Chinese)
Problem Sets:
magnetic field
Problem
A small circular coil of $N = 20$ turns and radius $r = 4$ cm is placed in a uniform magnetic field of magnitude $B = 0.050$ T. The normal to the coil plane makes a $60^{\circ}$ angle with $\vec{B}$. The coil carries a current $I = 3$ A. Find the torque on the coil.
$\tau \approx 1.3\times 10^{-2}$ N$\cdot$m.
For a planar coil in a uniform field, the torque is
$$\tau = NIAB\sin\theta,$$where $A$ is the area of one turn and $\theta$ is the angle between $\vec{B}$ and the coil's normal.
Here $A = \pi r^2 = \pi(0.04)^2 = 1.6\pi\times10^{-3}\ \text{m}^2$, $\theta = 60^{\circ}$, so
$$\tau = 20\times 3 \times (1.6\pi\times10^{-3})\times 0.050 \times \sin 60^{\circ}.$$Multiplying:
$$\tau = 4.8\pi\times10^{-3} \times \frac{\sqrt{3}}{2} = 2.4\sqrt{3}\,\pi\times10^{-3} \approx 1.3\times10^{-2}\ \text{N}\cdot\text{m}.$$