The magnetic field between the plates of parallel plate capacitor is given by $B=\frac{{\mu }_{0}Ir}{2{\mathrm{\pi R}}^2}$, when 

Consider a closed loop of radius $r$ in a region between the plates around the line joining center of two parallel plates of capacitor, where $r<R$.

If $I_D$ is the total displacement current, then current threading the loop is $I_D=\frac{I_D}{{\mathrm{\pi R}}^2}\times {\mathrm{\pi r}}^2$

Using Ampere's circuital law, we have $\oint \overrightarrow{B}.\overrightarrow{dl}={\mu }_0{I}_D$

 $$\begin{gathered} \mathrm{or} B\times 2\mathrm{\pi r}={\mathrm{\mu }}_0\frac{I_D{r}^{\mathit{2}}}{R^{\mathit{2}}} \\ \Rightarrow B\mathit{=}\frac{{\mu }_{\mathit{0}}{I}_{D}r}{\mathit{2}\pi R^{\mathit{2}}} \end{gathered}$$