1. Magnetic dipole moment of a current loop

A current-carrying rectangular loop behaves like a magnetic dipole with magnetic moment

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

where:

• $I = 12\,\text{A}$
• $A = (0.10)(0.05) = 5\times 10^{-3}\,\text{m}^2$
• $\hat n$ is the unit vector normal to the plane of the loop, given by the right-hand rule.

So the magnitude is

$m = IA = 12 \times 5\times 10^{-3} = 6\times 10^{-2}\,\text{A m}^2$

2. Condition for equilibrium

In a uniform magnetic field $\vec B$, the torque on the loop is

$\vec \tau = \vec m \times \vec B$

For equilibrium,

$\vec \tau = 0$

which happens when $\vec m$ is either parallel or antiparallel to $\vec B$.

3. Stable vs unstable equilibrium

Potential energy of a magnetic dipole in a magnetic field is

$U = -\vec m \cdot \vec B = -mB\cos\theta$

where $\theta$ is the angle between $\vec m$ and $\vec B$.

• For stable equilibrium, potential energy must be minimum.
• $U$ is minimum when $\theta = 0^\circ$.

Hence stable equilibrium occurs when

$\vec m \parallel \vec B$

Since the magnetic field is along the positive $Z$-direction, the stable orientation is the one in which the loop’s magnetic moment points along $+Z$.

4. Interpretation for the loop

The magnetic moment is perpendicular to the plane of the loop. Therefore, the loop must orient itself so that its area vector (given by current direction through right-hand rule) is along $+Z$.

Among the given figures, this corresponds to Option C.

5. Comparison with stored answer

Derived answer: C

Stored correct answer: C

They match.