Question

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 of ${60}^{\circ }$. The value of $n$ is-

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

Answer 1

The correct option is D $2$

Work done in turning a bar magnet is,

$W=MB[1-\cos \theta ]$

$\Rightarrow W_1=MB[1-\cos {90}^{\circ }]=MB.....\left(1\right)$

$\Rightarrow W_2=MB[1-\cos {60}^{\circ }]=\frac{MB}{2}......\left(2\right)$

$\Rightarrow W_2=\frac{W_1}{2}$

$\Rightarrow W_1=2W_2$

$\therefore n=2$

<!--td {border: 1px solid #ccc;}br {mso-data-placement:same-cell;}--> Hence, $\left(D\right)$ is the correct answer.