The magnetic field between the plate of a capacitor when r < R is given by

$\frac{\mu_0 i_D r}{2\pi R^2}$

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 $i'_D =\frac{i_Dr}{\pi R^2}×\pi r^2$. using Ampere Maxwell's law we have,

$\oint \vec{B}.\vec{dl}= \mu_0 i'_D$

$ or \, 2 \pi r = \mu_0i_D\frac{\pi r^2}{\pi R^2} or B = \frac{\mu_0i_Dr}{2\pi R^2}$