Answer
(a) less than
(b) $\frac{\pi}{4}$
Work Step by Step
(a) We know that $\tau=NAIBsin\theta$
It is given that the coil has one loop and the magnetic field is parallel to the coil, so
$\tau=IAB$
The above equation shows that torque $\propto$ Area. As we know that the circular loop has a larger area than the square loop, thus, the maximum torque of the square is less than the maximum torque of the circular loop.
(b) We can calculate the required ratio as follows:
$\frac{t_s}{t_c}=\frac{IA_sB}{IA_cB}$
$\frac{t_s}{t_c}=\frac{A_s}{A_c}$
$\frac{t_s}{t_c}=\frac{(\frac{L}{4})^2}{\pi(\frac{L}{2\pi})^2}$
$\frac{t_s}{t_c}=\frac{\pi}{4}$
