Question

Potential energy of a bar magnet of magnetic moment $M$ placed in a magnetic field of induction $B$ Such that it makes an angle $\theta$ with the direction of $B$ is (take $\theta ={90}^o$ as datum)

• A. $-MB\sin \theta$
• B. $-MB\cos \theta$
• C. $MB(1-\cos \theta )$
• D. $MB(1+\cos \theta )$

Answer 1

The correct option is B $-MB\cos \theta$

$dw=-\tau .d\theta =-MB\sin \theta .d\theta$ where $\theta$ is angle between $\overrightarrow{M}$ and $\overrightarrow{B}$.

Since work done=Potential energy lost

$⟹du=-dw=MB\sin \theta .d\theta$

$u=\int MB\sin \theta .d\theta =-MB\cos \theta$