The figure shows a bent rod rotating about its end O in a plane perpendicular to the magnetic field $\overrightarrow{B}.$ The part OA of the rod is non-conducting while the part AB is conducting. Find the induced emf between the ends A and B.

 

Suppose straight length of the rod is $l$. The distance of the end B from O

$O{B}^2=O{A}^2+{\left(2R\right)}^2=l^2+{\left(2R\right)}^2$

The potential difference between A and B can be calculated as:

$V_A-V_B=\left(V_O-V_B\right)-\left(V_O-V_A\right)\mathrm{......}\left(i\right)$

Here $V_0-V_B=\frac{B\omega {\left(OB\right)}^2}{2}=\frac{B\omega }{2}\left[l^2+4R^2\right]$ And $V_0-V_B=\frac{B\omega }{2}{\left(OA\right)}^2=\frac{B\omega l^2}{2}$

Substituting these values in equation (i), we get

$V_A-V_B=\frac{B\omega }{2}\left[\left(l^2+4R^2\right)-l^2\right]=\frac{3\omega }{2}\times 4R^2=2B\omega R^2.$