A conducting rod AB of mass M length L is hinged at its end A. It can rotate freely in the vertical plane (in the plane of the figure). A long straight wire is vertical and carrying a current I. The wires passes very close to A. The rod is released from its vertical position of unstable equilibrium. Calculate the emf between the ends of the rod when it has rotated through an angle $\mathrm{\theta }$ (see figure). 

Solution:

We will apply energy conservation to find the angular speed $\left(\mathrm{\omega }\right)$ of the rod.
$\frac{1}{2}{\mathrm{I}}_{\mathrm{A}}{\mathrm{\omega }}^2=$loss in gravitational PE
$$\begin{gathered} \frac{1}{2}\left(\frac{{\mathrm{ML}}^2}{3}\right){\mathrm{\omega }}^2=\mathrm{Mg}\frac{\mathrm{L}}{2}(1-\mathrm{cos\theta }) \\ \Rightarrow \mathrm{\hspace{0.17em} \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{\omega }=\sqrt{\frac{3\mathrm{g}(1-\mathrm{cos\theta })}{\mathrm{L}}}........\left(\mathrm{i}\right) \end{gathered}$$
Now consider an element of length  on the rod.
Speed of the element is $\mathrm{v}=\mathrm{\omega x}$
Magnetic field at the location of the element is $\mathrm{B}=\frac{{\mathrm{\mu }}_0\mathrm{I}}{2\mathrm{\pi d}}=\frac{{\mathrm{\mu }}_0\mathrm{I}}{2\mathrm{\pi xsin\theta }}$
$\therefore \mathrm{Emf}$ induced in the element is $\mathrm{d\epsilon }=\mathrm{Bvdx}=\frac{{\mathrm{\mu }}_0\mathrm{I\omega }}{2\mathrm{\pi sin\theta }}\mathrm{dx}$
$\therefore \mathrm{Emf}$ in the rod is $\mathrm{\epsilon }=\int \mathrm{d\epsilon }=\frac{{\mathrm{\mu }}_0\mathrm{I\omega }}{2\mathrm{\pi sin\theta }}\underset{\mathrm{O}}{\overset{\mathrm{L}}{\int }}\mathrm{dx}=\frac{{\mathrm{\mu }}_0\mathrm{I\omega L}}{2\mathrm{\pi sin\theta }}...........\left(\mathrm{ii}\right)$
$\mathrm{\epsilon }=\frac{{\mathrm{\mu }}_0\mathrm{I}}{2\mathrm{\pi sin\theta }}\sqrt{3\mathrm{gL}(1-\mathrm{cos\theta })}$