A copper rod of length l is rotated about one end perpendicular to the magnetic field B with constant angular velocity $\omega .$ The induced e.m.f. between the two ends is

If in time the rod turns by an angle $\theta ,$ the area generated by the rotation of rod will be

$=\frac{1}{2}\mathrm{l}\times \mathrm{l\theta }=\frac{1}{2}{\mathrm{l}}^2\mathrm{\theta }$

So the flux linked with the area generated by the rotation of rod

$\mathrm{\varphi }=\mathrm{B}\left(\frac{1}{2}{\mathrm{l}}^2\mathrm{\theta }\right)\cos 0°=\frac{1}{2}{\mathrm{Bl}}^2\mathrm{\theta }=\frac{1}{2}{\mathrm{Bl}}^2\mathrm{\omega t}$

$\text{ So, }\mathrm{e}=\frac{\mathrm{d\varphi }}{\mathrm{dt}}=\frac{\mathrm{d}}{\mathrm{dt}}\left(\frac{1}{2}{\mathrm{Bl}}^2\mathrm{\omega t}\right)=\frac{1}{2}{\mathrm{Bl}}^2\mathrm{\omega }$