A small bar magnet placed with its axis at \(30^{\circ}\) 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, \(\theta=30^{\circ}, I =0.018\, N-m , B =0.06 T\)

\(\Rightarrow 0.018= M \times 0.06 \times \sin 30^{\circ}\)

\(\Rightarrow 0.018=M \times 0.06 \times \frac{1}{2}\)

\(\Rightarrow M =0.6 A - m ^{2}\)

Now \(v=-M B \cos \theta\)

Position of stable equilibrium \(\left(\theta=0^{\circ}\right)\)

\(u _{ i }=- MB\)

Position of unstable equilibrium \(\left(\theta=180^{\circ}\right)\) :

\(u _{ f }= MB\)

\(\Rightarrow\) work done : \(\Delta U\)

\(\Rightarrow W =2 MB\)

\(\Rightarrow W =2 \times 0.6 \times 0.06\)

\(\Rightarrow W =7.2 \times 10^{-2} J\)