A small bar magnet placed with its axis at $30^{\circ}$ with an external field of $0.06 \mathrm{~T}$ experiences a torque of $0.018 \mathrm{Nm}$. The minimum work required to rotate it from its stable to unstable equilibrium position is :
Correct Option: , 3
(3) Here, $\theta=30^{\circ}, \tau=0.018 \mathrm{~N}-\mathrm{m}, B=0.06 \mathrm{~T}$
Torque on a bar magnet :
$\tau=M B \sin \theta$
$0.018=M \times 0.06 \times \sin 30^{\circ}$
$\Rightarrow 0.018=M \times 0.06 \times \frac{1}{2} \Rightarrow M=0.6 \mathrm{~A}-\mathrm{m}^{2}$
Position of stable equilibrium $\left(\theta=0^{\circ}\right)$
Position of unstable equilibrium $\left(\theta=180^{\circ}\right)$
Minimum work required to rotate bar magnet from stable
to unstable equilibrium
$\Delta U=U_{f}-U_{i}=-M B \cos 180^{\circ}-\left(-M B \cos 0^{\circ}\right)$
$W=2 M B=2 \times 0.6 \times 0.06$
$\therefore W=7.2 \times 10^{-2} \mathrm{~J}$