Explanation:
Let l is the length of the wire. such that, side of square is l/4. Area of square loop is, [tex]A_s=\dfrac{l^2}{16}[/tex]
Radius of the circular loop,
[tex]l=2\pi r\\\\r=\dfrac{l}{2\pi}[/tex]
Area of the circular loop,
[tex]A_a=\pi r^2\\\\A_a=\pi \times (\dfrac{l}{2\pi})^2\\\\A=\dfrac{l^2}{4\pi}[/tex]
Torque in magnetic field is given by :
[tex]\tau=NIAB\sin\theta[/tex]
It is clear that, [tex]\tau\propto A[/tex]
So,
[tex]\dfrac{\tau_s}{\tau_a}=\dfrac{A_s}{A_a}\\\\\dfrac{\tau_s}{\tau_a}=\dfrac{\dfrac{l^2}{16}}{\dfrac{l^2}{4\pi}}\\\\\dfrac{\tau_s}{\tau_a}=\dfrac{\pi}{4}[/tex]
So, the ratio of maximum torque on the square loop to the maximum torque on the circular loop is [tex]\pi :4[/tex].