Explanation

(a) Consider a loop of radius $r( < R)$ between the two circular plates, placed coaxially with them. The area of the loop =$\ \pi r^{2}$
By symmetry magnetic field is equal in magnetic at all points on the loop. If $i_{D}^{'}$ is the displacement current crossing the loop and $i_{D}$ is the total displacement current between plates $i_{D}^{'} = \frac{i_{D}r}{\pi R^{2}}\ \times \pi r^{2}$ . Using Ampere Maxwell’ law we have, $\oint\overrightarrow{\mathbf{B}}\mathbf{\bullet}\overrightarrow{\mathbf{dI}} = \mu_{0}i_{D}^{'}$
or $2\pi r = \mu_{0}i_{D}\frac{\pi r^{2}}{\pi R^{2}}\ \ or\ B = \frac{\mu_{0}i_{D}r}{2\pi R^{2}}$