A uniform magnetic field $$B$$ of 0.3 T is along the positive Z-direction. A rectangular loop (abcd) of sides 10 cm $$\times$$ 5 cm carries a current $$I$$ of 12 A. Out of the following different orientations which one corresponds to stable equilibrium?

The magnetic field is given to be uniform and directed along the positive $$Z$$-axis, so we write

$$\vec B = B\,\hat k, \qquad B = 0.3\ \text{T}.$$

For a current-carrying loop the magnetic moment is defined first. We state the formula:

$$\vec \mu = I\,\vec A = I\,A\,\hat n,$$

where

$$I = 12\ \text{A},\qquad A = (\text{length})\times(\text{breadth}) = 0.10\ \text{m}\times0.05\ \text{m}=0.005\ \text{m}^2,$$

and $$\hat n$$ is the unit vector perpendicular to the plane of the loop, obtained from the right-hand rule (curl the fingers in the sense of the current; the extended thumb gives the direction of $$\hat n$$).

Hence the magnitude of the magnetic moment is

$$\mu = I\,A = 12 \times 0.005 = 0.06\ \text{A·m}^2.$$

The torque acting on the loop in a magnetic field is obtained from the vector relation

$$\vec \tau = \vec \mu \times \vec B,$$

and the magnetic potential energy is

$$U = -\vec \mu\cdot\vec B = -\mu B\cos\theta,$$

where $$\theta$$ is the angle between $$\vec \mu$$ and $$\vec B$$. Stable equilibrium occurs when the potential energy is minimum, i.e. when $$\theta = 0^\circ$$ so that $$\vec \mu$$ is parallel to $$\vec B$$. Unstable equilibrium corresponds to $$\theta = 180^\circ$$, and any other angle produces a restoring torque that tends to rotate the loop toward $$\theta = 0^\circ$$.

We now examine each orientation given in the options.

Option A. The loop lies in the $$XZ$$ plane. A plane perpendicular to $$\hat n$$ equal to $$\pm\hat j$$ (along $$\pm Y$$). Thus $$\theta = 90^\circ$$ and

$$U = -\mu B \cos 90^\circ = 0,\qquad \tau = \mu B \sin 90^\circ = \mu B \neq 0.$$

With non-zero torque and zero slope in potential, the loop is driven toward alignment with $$\vec B$$; this is not a position of stable equilibrium.

Option B. The geometry is the same as in Option A; only the sense of current (and hence the sign of $$\hat n$$) is reversed. The moment still points along $$\pm Y$$, so $$\theta = 90^\circ$$, giving the same $$U=0$$ and a non-zero torque. This too is not a stable configuration.

Option C. The loop lies in the $$XY$$ plane, and the current circulates a-b-c-d. By the right-hand rule the thumb points along $$+\hat k$$, i.e. $$\vec \mu = +\mu\,\hat k$$. Therefore

$$\theta = 0^\circ,\qquad U = -\mu B \cos 0^\circ = -\mu B = -0.06 \times 0.3 = -0.018\ \text{J}.$$

This is the minimum attainable potential energy, so any small disturbance will be opposed by a restoring torque. Hence this orientation represents stable equilibrium.

Option D. The loop is again in the $$XY$$ plane but the current is reversed (a-d-c-b), giving $$\vec \mu = -\mu\,\hat k$$. Now

$$\theta = 180^\circ,\qquad U = -\mu B \cos 180^\circ = +\mu B = +0.018\ \text{J}.$$

This is the maximum potential energy, corresponding to an unstable equilibrium.

Comparing all four possibilities, only Option C provides $$\vec \mu$$ parallel to $$\vec B$$ and therefore minimum potential energy.

Hence, the correct answer is Option C.