Question

A small circular coil with a mass of m consists of N turns of fine wire and carries a current of I. The coil is located in a uniform magnetic field of B with the coil axis originally parallel to the field direction.
If the coil is rotated through an angle of $\theta$ from its equilibrium position and then released, what will be its angular speed when it passes through equilibrium position?

Answer 1

Apply conservation of energy: $(KE+PE{)}_i=(KE+PE{)}_f$
$\Rightarrow 0-MBcos{10}^o=\frac{1}{2}I{\omega }^2-MBcos{)}^o$
$\Rightarrow \frac{1}{2}\left(\frac{1}{2}m{R}^2\right)omeg{a}^2=Mg(1-cos\theta )$
$\Rightarrow \frac{1}{4}m{R}^2{\omega }^2=NI\pi R^{2}B(1-cos\theta )$
$\Rightarrow \omega =\sqrt{\frac{4NI\pi B(1-cos\theta )}{m}}$