Question

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

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

Answer 1

The correct option is D $6J$

Given that:

The initial direction of magnet = parallel to the field $=0^o$

The final direction of the magnet $={60}^o$

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

Magnetic Moment, $M=2\times {10}^{4}J{T}^{-1}$

Work done in rotating a magnet in the magnetic field is given by;

$W=B.T(cos{\theta }_1-cos{\theta }_2)$

$W=6\times {10}^{-4}\times 2\times {10}^4(1-0.5)$

$W=6J$

Thus, It requires a $6J$ of work to rotate the magnet from $0^o$ to ${60}^o$ in the magnetic field.