Question

A short bar magnet placed with its axis at ${30}^{\circ }$ with a uniform external magnetic field of $0.16\text{T}$ , experiences a torque of $0.032\text{Nm}.$ If the bar magnet is free to rotate, its potential energy when it is in stable and unstable equilibrium are-

• A. $0.032\text{J};-0.032\text{J}$
• B. $0.064\text{J};-0.128\text{J}$
• C. $-0.032\text{J};0.032\text{J}$
• D. $-0.064\text{J};0.064\text{J}$

Answer 1

The correct option is D $-0.064\text{J};0.064\text{J}$

Given,

$B=0.16\text{T};\tau =0.032\text{Nm};\theta ={30}^{\circ }$

As , $\tau =MB\sin \theta =0.032$

$\Rightarrow M=\frac{\tau }{B\sin \theta }=\frac{2\times 0.032}{0.16}=0.4$

For stable equilibrium, $\theta =0^{\circ }$

$\text{P.E}=-MB\cos \theta =-(0.4\times 0.16\times 1)=-0.064\text{J}$

For unstable equilibrium, $\theta ={180}^{\circ }$

$\text{P.E}=-MB\cos \theta =-(0.4\times 0.16\times (-1\left)\right)=0.064\text{J}$

<!--td {border: 1px solid #ccc;}br {mso-data-placement:same-cell;}--> Hence, $\left(D\right)$ is the correct answer.