The expression of motional emf is $e=BLv\sin \theta$, where, $\theta$ is angle between magnetic field and direction of motion.

Here, $\theta =90°$, 

So, induced emf across the ends of wire is $$\begin{gathered} e= BLv\sin 90° \\ =BLv \end{gathered}$$

$$\begin{gathered} = 2 \times 1 \times 8 \\ = 16 \mathrm{V} \end{gathered}$$