A small bar magnet placed with its axis at 30° 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 :

Given information,

The bar is in stable position when  and in unstable position when 

The Angle which is made with horizontal axis 

$\theta ={30}^{\circ }$

Applied magnetic Field applied, $B=0.06T$

The torque which is experienced, 

$\tau =0.018\mathrm{Nm}$

Here, the Torque can be given as

$\tau =MB\sin \theta$

Put the values,

$\begin{array}{l}\Rightarrow 0.018=M\times 0.06\times \sin {30}^{\circ }\\ \Rightarrow M=\frac{0.018}{0.06\times \sin {30}^{\circ }}=0.6{\mathrm{Am}}^2\end{array}$

As, the bar is in the stable position when the value of angle $\theta =0^{\circ }$ and it in the unstable position when the value of angle $\theta ={180}^{\circ }$

So, work done to rotate this from the stable to unstable position can be given as,

$\begin{array}{l}W=-MB\cos {180}^{\circ }-\left(-MB\cos 0^{\circ }\right)\\ \Rightarrow W=2MB=2\times 0.6\times 0.06\\ \Rightarrow W==0.072J=7.2\times {10}^{-2}J\end{array}$