Question

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 :

• A. $9.2 \times 10^{-3} J$
• B. $6.4 \times 10^{-2} J$
• C. $11.7 \times 10^{-3} J$
• D. $7.2 \times 10^{-2} J$

Answer 1

Answer: (D)

Torque on a bar magnet : $I = MB \sin \theta$
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 $\left(\theta=0^{\circ}\right)$ $u_{i}=-M B$
Position of unstable equilibrium $\left(\theta=180^{\circ}\right)$ :
$u _{ f }= MB$
$\Rightarrow$ work done $: \Delta U$
$\Rightarrow W =2 MB$
$\Rightarrow W =2 \times 0.6 \times 0.06$
$\Rightarrow W =7.2 \times 10^{-2} J$
option (4) is correct