As l2=25cm2, then l=5cm=0.05m.
Given: t=1s
Velocity of the square loop, v=10.05=0.05ms−1.
Induced current,
i=RV=RBlv=1040×0.05×0.05=0.01A
Now force acting on the side of the square loop, F=Bil=40×0.01×0.05
⇒F=0.02N
Therefore, work done
W=Fl=0.02×l=0.02×0.05
⇒W=1×10−3J