A small bar magnet is placed with its axis at 30° with an external magnetic field of 0.06 T experiences a torque of 0.018 Nm. The minimum work required to rotate it from its stable to unstable equilibrium position is:

We are told that a bar magnet placed in a uniform magnetic field experiences a torque. For a magnetic dipole, the relation between the torque $$\tau$$, the magnetic dipole moment $$M$$, the magnetic field $$B$$ and the angle $$\theta$$ between $$\vec M$$ and $$\vec B$$ is stated first:

$$\tau \;=\; M\,B\,\sin\theta$$

The numerical values given are

$$\tau = 0.018\ \text{N m}, \qquad B = 0.06\ \text{T}, \qquad \theta = 30^\circ.$$

We substitute these values into the formula to isolate $$M$$. We begin by writing

$$M = \dfrac{\tau}{B\,\sin\theta}.$$

Now, since $$\sin 30^\circ = \dfrac12,$$ we have

$$B\,\sin\theta \;=\; 0.06 \times \dfrac12 \;=\; 0.03.$$

So

$$M = \dfrac{0.018}{0.03} = 0.6\ \text{A m}^2.$$

Next we calculate the work needed to turn the magnet from its stable to its unstable equilibrium position. The potential energy $$U$$ of a magnetic dipole in a field is

$$U = -M\,B\,\cos\theta.$$

The stable equilibrium corresponds to $$\theta = 0^\circ$$ (magnet aligned with the field), and the unstable equilibrium corresponds to $$\theta = 180^\circ$$ (magnet anti-aligned). We compute the energies at these two orientations.

At $$\theta = 0^\circ$$:

$$U_{\text{stable}} = -M\,B\,\cos 0^\circ = -M\,B \times 1 = -M\,B.$$

At $$\theta = 180^\circ$$:

$$U_{\text{unstable}} = -M\,B\,\cos 180^\circ = -M\,B \times (-1) = +M\,B.$$

The work required is the increase in potential energy while going from the stable to the unstable state:

$$W = U_{\text{unstable}} - U_{\text{stable}} = (M\,B) - (-M\,B) = 2\,M\,B.$$

Substituting the values $$M = 0.6\ \text{A m}^2$$ and $$B = 0.06\ \text{T}$$ we get

$$W = 2 \times 0.6 \times 0.06 = 2 \times 0.036 = 0.072\ \text{J}.$$

Writing this in scientific notation,

$$W = 7.2 \times 10^{-2}\ \text{J}.$$

Hence, the correct answer is Option C.