Figure shows a square loop of side 1m and resistance $1\Omega$ the magnetic field on left side of line PQ has a magnitude B = 1T. The work done in pulling the loop out of the field uniformly in 1s  is

Given side of square loop  $l=1\text{\hspace{0.17em}m}$
Time $t=1\text{\hspace{0.17em}s}$ 
Magnetic field  $B=1\text{\hspace{0.17em}T}$
Electro motive force induced in loop due to its velocity
$\epsilon =Blv$ 
Current  $i=\frac{\epsilon }{R}=\frac{Blv}{R}$
$\therefore i=\frac{1\times 1\times 1}{1}=1A$
 Work done in pulling the loop is
$\begin{array}{l}W=M\left[B_1-B_2\right]\\ \Rightarrow W=niA\left[B_1-B_2\right]\\ \Rightarrow W=1\times 1\times 1^2\left[1\right]\\ \therefore W=1\mathrm{J}\end{array}$