A conducting rod AC of length $4\mathcal{l}$ is rotated about a point O in a uniform magnetic field B directed into the plane of the paper $AO=\mathcal{l}$ and $OC=3\mathcal{l}$. Then

Potential drop across ${\mathrm{V}}_{\mathrm{OA}},{\mathrm{V}}_{\mathrm{O}}-{\mathrm{V}}_{\mathrm{A}}=\frac{\mathrm{B\omega }{\mathcal{l}}^2}{2}\_\_\_\_\_\left(\mathrm{i}\right)$

Potential drop across ${\mathrm{V}}_{\mathrm{OC}},{\mathrm{V}}_{\mathrm{O}}-{\mathrm{V}}_{\mathrm{C}}=\frac{\mathrm{B\omega }{\left(3\mathcal{l}\right)}^2}{2}=\frac{9\mathrm{B\omega }{\mathcal{l}}^2}{2}\_\_\_\_\_\left(\mathrm{ii}\right)$

On subtracting equation (ii)from Equation (i), we get

${\mathrm{V}}_{\mathrm{A}}-V_C=4\mathrm{B\omega }{\mathcal{l}}^2$