$A$ rectangular wire loop of sides $8 \;cm$ and $2 \;cm$ with a small cut is moving out of a region of uniform magnetic field of magnitude $0.3 \;T$ directed normal to the loop. What is the emf developed across the cut if the velocity of the loop is $1 \;cm \,s^{-1}$ in a direction normal to the $(a)$ longer side,$(b)$ shorter side of the loop? For how long does the induced voltage last in each case?

(N/A) Given:
Length of the rectangular wire,$l = 8 \;cm = 0.08 \;m$
Width of the rectangular wire,$b = 2 \;cm = 0.02 \;m$
Magnetic field strength,$B = 0.3 \;T$
Velocity of the loop,$v = 1 \;cm \,s^{-1} = 0.01 \;m \,s^{-1}$
$(a)$ When the velocity is normal to the longer side $(l)$:
The induced emf is $e = B l v = 0.3 \times 0.08 \times 0.01 = 2.4 \times 10^{-4} \;V$.
The time taken to move out of the field is $t = \frac{b}{v} = \frac{0.02}{0.01} = 2 \;s$.
Thus,the induced voltage is $2.4 \times 10^{-4} \;V$ and it lasts for $2 \;s$.
$(b)$ When the velocity is normal to the shorter side $(b)$:
The induced emf is $e = B b v = 0.3 \times 0.02 \times 0.01 = 0.6 \times 10^{-4} \;V$.
The time taken to move out of the field is $t = \frac{l}{v} = \frac{0.08}{0.01} = 8 \;s$.
Thus,the induced voltage is $0.6 \times 10^{-4} \;V$ and it lasts for $8 \;s$.