Current passes through the circuit while a capacitor is charging (or draining). However, no actual charge transfer occurs in the insulated region between the capacitors, which is incompatible with current flow. As a result, displacement current is the current flowing through the insulated zone as the electric flux changes.
For a charged capacitor, electric field between the plates is given by:
$$
\begin{array}{l}
\mathrm{E}=\frac{\mathrm{Q}}{\varepsilon_{0} \mathrm{~A}} \\
\mathrm{Q}=\varepsilon_{0} \phi_{\mathrm{E}}
\end{array}
$$
Displacement current is given by:
$$
\mathrm{i}_{\mathrm{d}}=\frac{\mathrm{dQ}}{\mathrm{dt}}=\varepsilon_{\mathrm{0}} \frac{\mathrm{d} \phi \mathrm{E}}{\mathrm{dt}}
$$
According to Maxwell’s Ampere circuital Law, the line integral of magnetic field along a closed path is equal to $\mu_{0}$ times the total current.
$$
\oint \overrightarrow{\mathrm{B}} \cdot \overrightarrow{\mathrm{dl}}=\mu_{\mathrm{o}}\left(\mathrm{i}_{\mathrm{c}}+\mathrm{i}_{\mathrm{d}}\right)
$$
where $\mathrm{i}_{\mathrm{c}}$ : Conduction current
$\mathrm{i}_{\mathrm{d}}$ : Displacement current
$$
\oint \overrightarrow{\mathrm{B}} \cdot \overrightarrow{\mathrm{dl}}=\mu_{0}\left(\mathrm{i}_{\mathrm{c}}+\varepsilon_{0} \frac{\mathrm{d} \phi}{\mathrm{dt}}\right)
$$