A square loop having 100 turns, an area of $2.5\times {10}^{-3} m^2$ and a resistance of $100\Omega$. The magnetic field has a magnitude $B=0.40T$. Find the work done in pulling the loop out of the field, slowly and uniformly in 1.0 s.

The side of the square is $l=\sqrt{2.5\times {10}^{-3}{m}^2}=0.05m$

As it is uniformly out in 1.0 s, the speed of the loop is $\upsilon =0.05m{s}^{-1}.$

The emf induced in the left arm of the loop is $E=N\upsilon Bl$

$=100\times \left(0.05m{s}^{-1}\right)\times \left(0.40T\right)\times \left(0.05m\right)=0.1V.$

The current in the loop is $i=\frac{0.1V}{100\Omega }=1.0\times {10}^{-3}A$

The force on the left arm due to the magnetic field is  

$F=NilB=100\times (1.0\times {10}^{-3}A)\times \left(0.05m\right)t\left(0.40T\right)=2.0\times {10}^{-3}N$

This force is towards left in the figure. To pull the loop uniformly, an external force of $2.0\times {10}^{-3}N$ towards right must be applied. The work done by the force is $W=(2.0\times {10}^{-3}N)\times \left(0.05m\right)=1.0\times {10}^{-4}J$