Question

A bar magnet having a magnetic moment of $2\times {10}^4{JT}^{-1}$ is free to rotate in a horizontal plane. A horizontal magnetic field
$B=6\times {10}^{-4}T$ exits in the space. the work done in taking the magnet slowly from a direction parallel to the field to a direction ${60}^0$ from the field is

• A. $2J$
• B. $0.6J$
• C. $12J$
• D. $6J$

Answer 1

The correct option is D $6J$

$\text{Given}$: $M=2\times {10}^{4}J{T}^{-1}$

$B=6\times {10}^{-4}T$

${\theta }_1=0^{\circ }$ and ${\theta }_2={60}^{\circ }$

$\text{Solution}$;

We know,

$W=MB(cos{\theta }_1-cos{\theta }_2)=MB(1-cos{60}^{\circ })$

$\therefore W=2\times {10}^4\times 6\times {10}^{-4}(1-\frac{1}{2})=6J$

$\text{Hence D is the correct option}$