Q. The work done in turning a magnet of magnetic moment ‘$M$’ by an angle of $90{^\circ}$ from the meridian is ‘$n$’ times the corresponding work done to turn it through an angle at $60{^\circ}$, when ‘$n$’ is given by
A) 1/2
B) 2
C) 1/4
D) 1
Explanation
(b) $W_{1} = MB\left( \cos{0{^\circ}} - \cos{90{^\circ}} \right) = MB(1 - 0) = MB$
$W_{2} = MB\left( \cos{0{^\circ}} - \cos{60{^\circ}} \right) = MB\left( 1 - \frac{1}{2} \right) = \frac{MB}{2}$
$\therefore W_{1} = 2W_{2} \Rightarrow n = 2$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.