Question

Work done in rotating a bar magnet from 0 to angle $\text{ }\\!\\!\theta\\!\\!\text{ }$ 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)

$MH(1-cos\theta )$

Answer 2

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