Q. A bar magnet of magnetic moment $10^{4}J/T$ is free to rotate in a horizontal plane. The work done in rotating the magnet slowly from a direction parallel to a horizontal magnetic field of $4 \times 10^{- 5}T$ to a direction $60{^\circ}$ from the field will be
A) 0.2 $J$
B) 2.0 $J$
C) 4.18 $J$
D) $2 \times 10^{2}J$
Explanation
(a) Magnetic moment of bar $M = 10^{4}\ J/T$
$B = 4 \times 10^{- 5}T$
Hence work done $W = \overrightarrow{M}.\overrightarrow{B}$
$= 10^{4} \times 4 \times 10^{- 5} \times \cos{60{^\circ}} = 0.2\ J$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.