1. Torque on a magnetic dipole

For a bar magnet of magnetic moment $M$ placed in a uniform magnetic field $B$ at angle $\theta$, the torque is

$\tau = MB\sin\theta$

Given:

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

So,

$0.018 = M(0.06)\sin 30^\circ$

Since $\sin 30^\circ = \tfrac{1}{2}$,

$0.018 = M(0.06)(0.5) = 0.03M$

Hence,

$M = \frac{0.018}{0.03} = 0.6\,\text{A m}^2$

2. Potential energy of a magnetic dipole

The potential energy is

$U = -MB\cos\theta$

• In stable equilibrium, the dipole is parallel to the field: \theta = 0^

$U_{\text{stable}} = -MB$

• In unstable equilibrium, the dipole is antiparallel to the field: $\theta = 180^\circ$

$U_{\text{unstable}} = +MB$

3. Minimum work required

The minimum external work required to rotate the magnet slowly from stable to unstable equilibrium equals the increase in potential energy:

$W_{\min} = U_{\text{unstable}} - U_{\text{stable}} = MB - (-MB) = 2MB$

Substitute the values:

$W_{\min} = 2(0.6)(0.06)$

$W_{\min} = 0.072\,\text{J}$

$W_{\min} = 7.2 \times 10^{-2}\,\text{J}$

4. Option check

• A: $6.4 \times 10^{-2}\,\text{J}$
• B: $9.2 \times 10^{-3}\,\text{J}$
• C: $7.2 \times 10^{-2}\,\text{J}$
• D: $11.7 \times 10^{-3}\,\text{J}$

Therefore, the correct option is C.