Question

A bar magnet having a magnetic movement of $2\times 10^{4}J{T}^{-1}$ is free to rotate in a horizontal plane. A horizontal magnetic field $B=6\times 10^{-4}T$ exists in the space. The work done in taking the magnet slowly from a direction parallel to the field to a direction $60^{\circ }$ from the field is :

• A. 12 J
• B. 6 J
• C. 2 J
• D. 0.6 J

Answer 1

Answer: (B)

The work done in rotating a magnetic dipole against the torque acting on it, when placed in magnetic field is stored inside it in the form of potential energy.
When magnetic dipole is rotated from intial position $\theta ={\theta }_1$ to final position $\theta ={\theta }_2$,
then work done $=MB\left(\cos {\theta }_1-\cos {\theta }_2\right)$
$=MB\left(1-\frac{1}{2}\right)=\frac{2\times 10^4\times 6\times 10^{-4}}{2}=6J$