# A small bar magnet placed with its axis

Question:

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 :

1. (1) $6.4 \times 10^{-2} \mathrm{~J}$

2. (2) $9.2 \times 10^{-3} \mathrm{~J}$

3. (3) $7.2 \times 10^{-2} \mathrm{~J}$

4. (4) $11.7 \times 10^{-3} \mathrm{~J}$

Correct Option: , 3

Solution:

(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

$\Delta U=U_{f}-U_{i}=-M B \cos 180^{\circ}-\left(-M B \cos 0^{\circ}\right)$

$W=2 M B=2 \times 0.6 \times 0.06$

$\therefore W=7.2 \times 10^{-2} \mathrm{~J}$