 Solution

Choose the plate polarity so that the gap electric field points along $+z$ when $V>0$. Away from the plate edges, the quasistatic electromagnetic approximation gives $E_z=V_0\sin \omega t/d$. A circular Ampere-Maxwell law contour of radius $r<a$ then givesIt is displacement current, rather than a conduction current across the air, that supplies the enclosed current term. The direction follows the right-hand rule for increasing positive electric flux; it reverses every half cycle. In the ideal extension that sets the gap electric field to zero outside the plate footprint, $B_{\varphi }={\mu }_0{ϵ}_0\omega V_0{a}^2\cos \omega t/(2dr)$ for $r>a$. That extension is an edge approximation, not an exact exterior solution for finite plates. Neglect fringing field of a capacitor and retardation, requiring $d\ll a$ and $\omega a/c\ll 1$.