A rectangular loop of sides 10 cm and 5 cm, carrying a current $$I$$ of 12 A, is placed in different orientations as shown in the figures below.

If there is a uniform magnetic field of 0.3 T in the positive z direction, in which orientations the loop would be in (i) stable equilibrium and (ii) unstable equilibrium?

First we recall that a current-carrying planar loop behaves like a magnetic dipole. Its magnetic moment $$\vec m$$ is defined by the right-hand rule and is given in magnitude by

$$m = I\,A,$$

where $$I$$ is the current and $$A$$ is the geometrical area of the loop. The torque acting on the loop when it is kept in a uniform magnetic field $$\vec B$$ is obtained from the vector product

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

If $$\theta$$ is the angle between $$\vec m$$ and $$\vec B$$, the magnitude of the torque becomes

$$\tau = mB\sin\theta.$$

We also need the expression for the potential energy of the magnetic dipole in the field; the formula is

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

Equilibrium positions occur when the torque is zero, i.e. when $$\sin\theta = 0$$, or equivalently when $$\theta = 0^{\circ}$$ (parallel) or $$\theta = 180^{\circ}$$ (antiparallel). The nature of the equilibrium follows from the potential energy:

• For $$\theta = 0^{\circ}$$ we get $$U_{\min} = -mB$$. This is the minimum possible energy, so any small rotation raises the energy and the dipole experiences a restoring torque. Thus the equilibrium is stable.
• For $$\theta = 180^{\circ}$$ we obtain $$U_{\max} = +mB$$, the maximum possible energy. A small deflection lowers the energy, hence the dipole is driven further away from that orientation. The equilibrium is therefore unstable.

Let us calculate the magnetic moment of the given loop so that we can identify its direction in each drawing. The sides are 10 cm and 5 cm, so

$$A = 0.10\;\text{m}\times0.05\;\text{m}=0.005\;\text{m}^2.$$

The current is 12 A, hence

$$m = IA = 12\times0.005 = 0.06\;\text{A m}^2.$$

The external magnetic field is $$\vec B = 0.3\;\text{T}\,\hat k$$, i.e. in the positive $$z$$ direction. The direction of $$\vec m$$ for each sketch (a), (b), (c) and (d) follows from curling the right hand around the sense of current shown in that sketch. Making that inspection we find:

• (b) The normal to the plane of the loop points along $$+\hat k$$, i.e. $$\vec m$$ is parallel to $$\vec B$$. Therefore $$\theta = 0^{\circ}$$ and this is a stable equilibrium.
• (d) The normal to the plane of the loop points along $$-\hat k$$, i.e. $$\vec m$$ is antiparallel to $$\vec B$$. Therefore $$\theta = 180^{\circ}$$ and this is an unstable equilibrium.
• (a) and (c) give normals that are neither parallel nor antiparallel to $$\vec B$$, so $$\theta$$ is different from $$0^{\circ}$$ or $$180^{\circ}$$. A non-zero torque acts, the loop tends to rotate, and these positions are not equilibria at all.

Thus the loop is in

(i) stable equilibrium in orientation (b), and

(ii) unstable equilibrium in orientation (d).

Among the given choices only Option D lists “(b) and (d), respectively.” Hence, the correct answer is Option D.