= 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 gives
$$
2\pi r B_\phi=\mu_0\epsilon_0\frac{d}{dt}\int_0^r E_z\,2\pi s\,ds,
\qquad
\boxed{\mathbf B(r,t)=\frac{\mu_0\epsilon_0\omega V_0}{2d}\,r\cos\omega t\,\widehat{\boldsymbol\phi}\quad(0\leq r<a).}
$$
It 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, $B_\phi=\mu_0\epsilon_0\omega V_0a^2\cos\omega t/(2dr)$ 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 $\omega a/c\ll1$.
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