Answer
$7.39\times 10^{-4}V $
Work Step by Step
We know that the circumference of a square is given as
$4a=L $
$\implies a=\frac{L}{4}$ where $ a $ is a side of a square
$\implies a=\frac{1.22m}{4}=0.305m $
Now $ A_{square}=a^2$
$ A_{square}=(0.305m)^2=0.093025m^2$
Similarly, for a circular loop $ r=\frac{L}{2\pi}$
$\implies r=\frac{1.22m}{2\pi}=0.194m $
Now $ A_{circle}=\pi r^2$
$ A_{circle}=\pi(0.194m)^2=0.11811704m^2$
The average magnitude of induced emf of the coil when its shape is changed from square to circular is given as
$\epsilon=NB|\frac{A_{circle}cos\theta-A_{square}cos\theta}{\Delta t}|$
We plug in the known values to obtain:
$\epsilon=(1)(0.125T)|\frac{(0.11817704m^2)cos0^{\circ}-0.093025m^2cos0^{\circ}}{4.25s}|$
$\epsilon=7.39\times 10^{-4}V $
