A small bar magnet placed with its axis at 30⁰ with an external 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:

Here, $\mathrm{\theta }={30}^{\circ },\mathrm{\tau }=0.018\mathrm{N}-\mathrm{m},\mathrm{B}=0.06\mathrm{T}$

Torque on a bar magnet:

$\begin{array}{l}\mathrm{\tau }=\mathrm{MBsin}\mathrm{\theta }\\ 0.018=\mathrm{M}\times 0.06\times \sin {30}^{\circ }\\ \Rightarrow 0.018=\mathrm{M}\times 0.06\times \frac{1}{2}\Rightarrow \mathrm{M}=0.6\mathrm{A}-{\mathrm{m}}^2\end{array}$

Minimum work required to rotate bar magnet from stable to unstable equilibrium

$\begin{array}{l}\mathrm{\Delta U}={\mathrm{U}}_{\mathrm{f}}-{\mathrm{U}}_{\mathrm{i}}=-\mathrm{MBcos}{180}^{\circ }-\left(-\mathrm{MBcos}{0}^{\circ }\right)\\ \mathrm{W}=2\mathrm{MB}=2\times 0.6\times 0.06\\ \therefore \mathrm{W}=7.2\times {10}^{-2}\mathrm{J}\end{array}$