Question

The work done in turning a magnet of magnetic moment $M$ by an angle of ${90}^o$ from the, is $n$ times the corresponding work done to turn it through an angle of ${60}^o$. The value of $n$ is given by:

• A. $2$
• B. $1$
• C. $0.5$
• D. $0.25$

Answer 1

The correct option is A $2$

$w_1=MB(\cos 0-\cos 90)=MB.......\left(1\right)$

$w_2=MB(\cos 0-\cos 60)=\frac{MB}{2}.......\left(2\right)$

It is given that $w_1=n{w}_2.....\left(3\right)$

Comparing equation 1, 2 and 3

$n=2$