Q. A bar magnet having a magnetic moment of $2 \times 10^{4}\ JT^{- 1}$ 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) 0.6 $J$
B) 12 $J$
C) 6 $J$
D) 2 $J$
Explanation
(c) $W = MB\left( 1 - \cos\theta \right) = 2 \times 10^{4} \times 6 \times 10^{- 4}\left( 1 - \cos 60 \right) = 6J$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.