A uniform magnetic field $B$ of $0.3$ T is along the positive $Z$-direction. A rectangular loop $(abcd)$ of sides $10\ \text{cm}\times5\ \text{cm}$ carries a current $I$ of $12$ A. Out of the following different orientations, which one corresponds to stable equilibrium? (In each figure the current flows round the loop in the direction of the arrows.)
Answer: (C) see figure
A current loop is in stable equilibrium when its magnetic moment $\vec m$ is parallel to $\vec B$: then the torque is zero and the potential energy $U = -\vec m\cdot\vec B$ is minimum.
$\vec m$ must therefore point along $+Z$, so the loop must lie in the $XY$ plane with the current circulating anticlockwise when viewed from $+Z$ (from $+X$ towards $+Y$).
- Loops in the $YZ$ or $XZ$ plane have $\vec m$ perpendicular to $\vec B$ (maximum torque): not equilibrium.
- In (3) the current goes $a\to b\to c\to d$, i.e. from the $+X$ side round to the $+Y$ side: $\vec m$ along $+Z$. Stable.
- In (4) the current is reversed, $\vec m$ along $-Z$: equilibrium, but unstable.
Solution by Sreeraj P, M.Sc Physics