Question

Work done in rotating a bar magnet from $0$ to angle $\theta$ is:

• A. $MH(1-\cos \theta )$
• B. $\frac{M}{H}(1-\cos \theta )$
• C. $\frac{M}{H}(\cos \theta -1)$
• D. $MH(\cos \theta -1)$

Answer 1

Answer: (A)

Work done in rotating a magnet is given by
$W=\underset{0}{\overset{\theta }{\int }}\tau d\theta$
where $\tau$ is torque and $d\theta$ angular charge
Also, $\tau =MH\sin \theta$
$\therefore W=\underset{0}{\overset{\theta }{\int }}MH\sin \theta d\theta$
$\Rightarrow W=MH\underset{0}{\overset{\theta }{\int }}\sin \theta d\theta$
$\Rightarrow W=MH[-\cos \theta {]}_0^{\theta }$
$\Rightarrow W=MH[-\cos \theta +\cos 0]$
$\Rightarrow W=MH[1-\cos \theta ]$