Question

If the work done in turning a magnet of magnetic moment $M$ by an angle of 90$^{\circ}$ from the magnetic meridian is $n$ times the corresponding work done to turn it through an angle of 60$^{\circ}$, then the value of $n$ is

• A. 1
• B. 2
• C. $\frac{1}{2}$
• D. $\frac{1}{4}$

Answer 1

Answer: (B)

2

Answer 2

We have, $W=-M B\left(\cos \theta_{2}-\cos \theta_{1}\right)$
So, $W_{1}=-M B\left(\cos 90^{\circ}-\cos 0^{\circ}\right)=M B$
and $W_{2}=-M B\left(\cos 60^{\circ}-\cos 0^{\circ}\right)=\frac{1}{2} M B$
As $W_{1}=n W_{2}$
$\therefore n=\frac{W_{1}}{W_{2}}=\frac{M B}{\frac{1}{2} M B}=2$