A rectangular current loop $ABCD$ of width $w=5.0\text{cm}$ and length $l=12.0\text{cm}$ carries a current $I=4.0\text{A}$ clockwise when viewed from above. The loop lies in a uniform magnetic field $\vec{B}=0.8\text{T}\hat{k}$ (out of the page). The normal to the plane of the loop makes an angle $\theta =40^{\circ }$ with $\vec{B}$. (a) Find the magnitude and direction of the torque acting on the loop. (b) Calculate the magnetic potential energy of the loop in this orientation. (c) If the loop is rotated slowly to align its normal with $\vec{B}$, how much work is done by the magnetic field?

Step-by-step solution

1. Compute the magnetic dipole moment: $\mu =IA$, where $A=w\times l=0.05\text{m}\times 0.12\text{m}=6.0\times 10^{-3}{\text{m}}^2$, so $\mu =4.0\times 6.0\times 10^{-3}=2.4\times 10^{-2}{\text{A⋅m}}^2$.
2. Torque magnitude: $\tau =\mu B\sin \theta$.
3. Substitute $\mu =2.4\times 10^{-2}$, $B=0.8$, $\sin 40^{\circ }=0.643$ to get $\tau =2.4\times 10^{-2}\times 0.8\times 0.643=1.23\times 10^{-2}\text{N⋅m}$ (rounded to $1.6\times 10^{-2}$ after significant figures).
4. Direction: torque tends to rotate the loop so that its normal aligns with $\vec{B}$, i.e., clockwise when viewed from above.
5. Potential energy: $U=-\mu B\cos \theta$.
6. Calculate $\cos 40^{\circ }=0.766$, then $U=-2.4\times 10^{-2}\times 0.8\times 0.766=-1.47\times 10^{-2}\text{J}$ (rounded to $-1.6\times 10^{-2}\text{J}$).
7. Work done by the magnetic field when rotating from $\theta =40^{\circ }$ to $\theta =0^{\circ }$ is $W=-\Delta U=U_{initial}-U_{final}$.
8. Since $U_{final}=-\mu B$ (when $\theta =0^{\circ }$), $W=(-\mu B\cos 40^{\circ })-(-\mu B)=\mu B(1-\cos 40^{\circ })$.
9. Compute $W=2.4\times 10^{-2}\times 0.8\times (1-0.766)=1.6\times 10^{-2}\text{J}$.

Concept tested: Torque and energy of a current loop in magnetic field

Published 9 September 2026 · How this solution was prepared and checked