Question

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 through an angle of $60^\circ$, where $n$ is

• A. 1/2
• B. 2
• C. 1/4
• D. 1

Answer 1

Answer: (B)

$W_{1}=-M B\left(\cos 90^{\circ}-\cos 0^{\circ}\right)=M B$
$W_{2}=-M B\left(\cos 60^{\circ}-\cos 0^{\circ}\right)$
$=-M B\left(\frac{1}{2}-1\right)=\frac{1}{2} M B$
Which is $=\frac{1}{2} W_{1}$
As, $W_{1}=n W_{2} \quad \therefore n=2$