Consider a rectangular current loop ABCD placed in a uniform magnetic field $\vec{B}=0.5\text{T}\hat{k}$ (out of the page). The sides $AB$ and $CD$ are each $6\text{cm}$ long, and the sides $BC$ and $DA$ are each $4\text{cm}$ long. A current $I=3\text{A}$ flows clockwise when viewed from above. The plane of the loop makes an angle $\theta =30^{\circ }$ with the direction of $\vec{B}$ (the angle is between the normal to the loop and $\vec{B}$).

Step-by-step solution

1. Compute the area of the rectangle: $A=(6\text{cm})(4\text{cm})=24{\text{cm}}^2=24\times 10^{-4}{\text{m}}^2=2.4\times 10^{-3}{\text{m}}^2$.
2. Find the magnetic dipole moment: $\mu =IA=3\text{A}\times 2.4\times 10^{-3}{\text{m}}^2=7.2\times 10^{-3}{\text{A m}}^2$.
3. Torque magnitude is $\tau =\mu B\sin \theta$. Substitute $\mu =7.2\times 10^{-3}$, $B=0.5\text{T}$, $\sin 30^{\circ }=0.5$: $\tau =7.2\times 10^{-3}\times 0.5\times 0.5=1.8\times 10^{-3}\text{N m}$.
4. Direction of torque is given by $\vec{\tau }=\vec{\mu }\times \vec{B}$. The current is clockwise, so the area vector $\vec{\mu }$ points into the page ($-\hat{k}$). $\vec{B}$ points out of the page ($+\hat{k}$). Using the right‑hand rule, $\vec{\mu }\times \vec{B}$ points out of the page (+$\hat{k}$).
5. Potential energy is $U=-\vec{\mu }\cdot \vec{B}=-\mu B\cos \theta$. Use $\cos 30^{\circ }=0.866$: $U=-7.2\times 10^{-3}\times 0.5\times 0.866=-3.12\times 10^{-3}\text{J}$.
6. Round to two significant figures: $U\approx -3.1\times 10^{-3}\text{J}$.
7. Thus (a) $\tau =1.8\times 10^{-3}\text{N m}$ out of the page; (b) $U=-3.1\times 10^{-3}\text{J}$.

Concept tested: Torque on a current loop

Published 6 September 2026 · How this solution was prepared and checked