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: , 4
Torque on a bar magnet : I = MB $\sin \theta$
Here, $\theta=30^{\circ}, \mathrm{I}=0.018 \mathrm{~N}-\mathrm{m}, \mathrm{B}=0.06 \mathrm{~T}$
$\Rightarrow 0.018=\mathrm{M} \times 0.06 \times \sin 30^{\circ}$
$\Rightarrow 0.018=\mathrm{M} \times 0.06 \times \frac{1}{2}$
$\Rightarrow \mathrm{M}=0.6 \mathrm{~A}-\mathrm{m}^{2}$
Now $\mathrm{v}=-\mathrm{MB} \cos \theta$
Position of stable equilibrium $\left(\theta=0^{\circ}\right)$ :
$\mathrm{u}_{\mathrm{i}}=-\mathrm{MB}$
Position of unstable equilibrium $\left(\theta=180^{\circ}\right)$ :
$\mathrm{u}_{\mathrm{f}}=\mathrm{MB}$
$\Rightarrow$ work done $: \Delta \mathrm{U}$
$\Rightarrow \mathrm{W}=2 \mathrm{MB}$
$\Rightarrow \mathrm{W}=2 \times 0.6 \times 0.06$
$\Rightarrow \mathrm{W}=7.2 \times 10^{-2} \mathrm{~J}$
option (4) is correct