A horizontal wire $0.8 m$ long is falling at a speed of $5 m/s$ perpendicular to a uniform magnetic field of $1.1 T$, which is directed from east to west. Calculate the magnitude of the induced emit in the rod .

$$\begin{gathered} The rod is aligned along north -south . \\ e=Bvl=1.1\times 5\times 0.8 \\ =4.4 V \\ Magnetic force on each free electron is given by ,\overrightarrow{F_e}=\left(-e\right)\overrightarrow{v}\times \overrightarrow{B} and so this force is directed towards south .Therefore free electrons will be accumulated at the south end \\ and the north end of the rod will acquire positive polarity . \end{gathered}$$