A rectangular wire loop of sides 8 cm and 3 cm with a small cut, is...

Given:
Magnetic field, $B = 0.3$ T
Velocity of the loop, $v = 2$ cm s$^{-1} = 2 \times 10^{-2}$ m s$^{-1}$
Length of the shorter side, $l = 3$ cm $= 3 \times 10^{-2}$ m

Since the loop is moving in a direction normal to the shorter side, the velocity vector is perpendicular to the shorter side. The motional emf is induced across the side that is perpendicular to the direction of motion.

The induced emf $e$ is given by the formula:
$e = B l v$

Substituting the given values:
$e = 0.3 \times (3 \times 10^{-2}) \times (2 \times 10^{-2})$
$e = 1.8 \times 10^{-4}$ V

Answer: $1.8 \times 10^{-4}$ volt