Question

Work done in rotating a bar magnet from $0$ to angle $120^{\circ }$ is

• A. $\frac{1}{2}MB$
• B. $\frac{3}{2}MB$
• C. $MB$
• D. $\frac{2}{3}MB$

Answer 1

Answer: (B)

Work done in rotating a magnet (from angle 0 to $\theta$ ) is given by
$W=\underset{0}{\overset{\theta }{\int }}\tau d\theta$
where, $\tau =$ torque and $d\theta =$ angular change.
Also, $\tau =MB\sin \theta =\underset{0}{\overset{\theta }{\int }}MB\sin \theta d\theta$
$=MB\underset{0}{\overset{\theta }{\int }}\sin \theta d\theta =MB(-\cos \theta {)}_0^{\theta }$
$=MB\left(-\cos 120^{\circ }+\cos 0^{\circ }\right)$
$\Rightarrow MB\left(1+\frac{1}{2}\right)=\frac{3}{2}MB$