1. Magnetic dipole moment of a current loop

For a current-carrying loop, the magnetic dipole moment is

$\vec{m} = I\, \vec{A}$

where $\vec{A}$ is the area vector perpendicular to the plane of the loop, given by the right-hand rule.

The magnitude is

$m = IA = I(lb)$

Here,

• $l = 10\text{ cm} = 0.10\text{ m}$
• $b = 5\text{ cm} = 0.05\text{ m}$
• $I = 12\text{ A}$

So,

$A = 0.10 \times 0.05 = 5 \times 10^{-3}\text{ m}^2$

$m = 12 \times 5 \times 10^{-3} = 0.06\text{ A m}^2$

This magnitude is not actually needed to identify equilibrium type, but it confirms the loop behaves like a magnetic dipole.

2. Condition for equilibrium

A magnetic dipole in a uniform magnetic field experiences torque

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

For equilibrium,

$\tau = 0$

which happens when $\vec{m}$ is either:

• parallel to $\vec{B}$, or
• antiparallel to $\vec{B}$.

The potential energy is

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

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

• If $\theta = 0$, then $U = -mB$ which is minimum $\Rightarrow$ stable equilibrium.

• If $\theta = \pi$, then $U = +mB$ which is maximum $\Rightarrow$ unstable equilibrium.

3. Given magnetic field direction

The magnetic field is in the positive $z$-direction.

So:

• Stable equilibrium: loop orientation for which $\vec{m}$ is along $+z$.
• Unstable equilibrium: loop orientation for which $\vec{m}$ is along $-z$.

4. Identify from the given figures

Using the right-hand rule for the current direction in the loop:

• In orientation (B), the area vector (hence magnetic moment) points along $+z$. Therefore (B) is stable equilibrium.

• In orientation (D), the area vector points along $-z$. Therefore (D) is unstable equilibrium.

5. Match with options

Thus,

• Stable equilibrium: (B)
• Unstable equilibrium: (D)

So the correct option is

$\boxed{\text{A}}$

which states: $(B)$ and $(D)$, respectively.