A wire of length l in the form of a square loop is in a plane...

For the square loop, the side length is $a = \dfrac{l}{4}$.

Area of the square loop is $A_1 = a^2 = \dfrac{l^2}{16}$.

The induced e.m.f. $e_1$ is given by $e_1 = \left| \dfrac{d\Phi_1}{dt} \right| = B \dfrac{dA_1}{dt} = B \dfrac{d}{dt} \left( \dfrac{l^2}{16} \right) = \dfrac{Bl}{8} \left| \dfrac{dl}{dt} \right|$.

For the circular loop, the circumference is $l$, so the radius is $r = \dfrac{l}{2\pi}$.

Area of the circular loop is $A_2 = \pi r^2 = \pi \left( \dfrac{l}{2\pi} \right)^2 = \dfrac{l^2}{4\pi}$.

The induced e.m.f. $e_2$ is given by $e_2 = \left| \dfrac{d\Phi_2}{dt} \right| = B \dfrac{dA_2}{dt} = B \dfrac{d}{dt} \left( \dfrac{l^2}{4\pi} \right) = \dfrac{Bl}{2\pi} \left| \dfrac{dl}{dt} \right|$.

Taking the ratio of the two induced e.m.f.s:

$\dfrac{e_1}{e_2} = \dfrac{\dfrac{Bl}{8} \left| \dfrac{dl}{dt} \right|}{\dfrac{Bl}{2\pi} \left| \dfrac{dl}{dt} \right|} = \dfrac{2\pi}{8} = \dfrac{\pi}{4}$.

Answer: $\dfrac{\pi}{4}$