A vertical rod of length l is moved with constant velocity v towards east. The vertical component of earth magnetic field is B and angle of dip is $\mathrm{\theta }$. The induced e.m.f. in the rod is

$$\begin{gathered} \mathrm{When} \mathrm{the} \mathrm{Vertical} \mathrm{rod} \mathrm{moves} \mathrm{towards} \mathrm{east} \mathrm{horizontal} \mathrm{component} \mathrm{of} \mathrm{earth} \\ \mathrm{magnetic} \mathrm{field} \mathrm{is} \mathrm{perpendicular} \mathrm{to} \mathrm{both} \mathrm{length} \mathrm{of} \mathrm{the} \mathrm{conductor} \mathrm{and} \mathrm{velocity}. \end{gathered}$$

$$\begin{gathered} \mathrm{angle} \mathrm{of} \mathrm{dip} \mathrm{is} \mathrm{\theta } \mathrm{so} \frac{{\mathrm{B}}_{\mathrm{v}}}{{\mathrm{B}}_{\mathrm{H}}}=\mathrm{tan\theta } \\ {\mathrm{B}}_{\mathrm{H}}= {\mathrm{B}}_{\mathrm{v}}\cot \mathrm{\theta } \mathrm{or} \mathrm{B} \mathrm{cot\theta } (\mathrm{given} {\mathrm{B}}_{\mathrm{v}}=\mathrm{B}) \end{gathered}$$

$\mathrm{induced} \mathrm{emf} \mathrm{e}={\mathrm{B}}_{\mathrm{H}}\mathrm{lv}=\mathrm{Bcot\theta lv}$