Figure shows a square loop of side 0.5 m and resistance 10 Ω. 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=0.25m/sInduced emf, e=Bvℓ=(1.0)(1.0)(0.25)(0.5)= 0.125 V∴      Current in the loop, i=eR=0.12510=1.25×10−2 AThe magnetic force on the left arm due to the magnetic field isFm=iℓ B=1.25×10−2(0.5)(1.0)=6.25×10−3 NTo pull the loop uniformly, an external force of 6.25×10−3 N towards right must be applied.∴W=6.25×10−3 N(0.5m)=3.125×10−3 J