Figure shows a square loop of side 0.5 m and resistance 10Ω. The magnetic field has a magnitude B = 1.0 T. The work done in pulling the loop out of the field slowly and uniformly in 2.0 s is

Speed of the loop should bev=ℓt=0.52=0.25ms−1Induced emf, ε=Bvℓ=(1.0)(0.25)(0.5)=0.125V∴  Current in the loop, i=εR=0.12510=1.25×10−2AThe magnetic force on the left arm due to the magnetic field isFm=iℓB=1.25×10−2(0.5)(1.0)=6.25×10−3NTo pull the loop uniformly an external force of 6.25 x 10-3N towards right must be applied.∴ W=6.25×10−3N(0.5m)=3.125×10−3J