Question

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

• A. $1/2$
• B. $2$
• C. $1/4$
• D. $1$

Answer 1

Answer: (B)

$2$

The work done in turning a magnet of magnetic moment M by an angle of 90⁰ from the meridian is

$w_1=MB(\cos {\theta }_1-\cos {\theta }_2)$

$w_1=MB(\cos 0-\cos 90)$

$w_1=MB.......\left(1\right)$

Similarly, The work done in turning a magnet of magnetic moment M by an angle of 60⁰ from the meridian is

$w_2=MB(\cos 0-\cos 60)$

$w_2=\frac{MB}{2}........\left(2\right)$

It is given that

$w_1=n{w}_2$

$MB=n\frac{MB}{2}$

$n=2$