Question

A bar magnet of magnetic moment ${10}^4\text{J/T}$ is free to rotate in a horizontal plane. The work done in rotating the magnet slowly from a direction parallel to a horizontal magnetic field of $4\times {10}^{-5}\text{T}$ to a direction ${60}^{\circ }$ from the field will be

• A. $0.2\text{J}$
• B. $2.0\text{J}$
• C. $4.18\text{J}$
• D. $4.2\times {10}^2\text{J}$

Answer 1

The correct option is A $0.2\text{J}$

$M={10}^4\text{J/T},B=4\times {10}^{-5}\text{T}$

Work done in rotating the magnet from

${\theta }_i=0^{\circ }$ to ${\theta }_f=\theta$

Work done is given by,

$W=MB(1-\cos \theta )$

So, $W={10}^4\times 4\times {10}^{-5}(1-\cos {60}^{\circ })$

$W=2\times {10}^{-1}=0.2\text{J}$

Hence, option $\left(A\right)$ is the correct answer.