= Solution
The <continuity equation for electric charge> is $\nabla\cdot J=-\partial_t\rho$. Taking the <divergence> of the magnetostatic <Ampère's law> would instead require $\nabla\cdot J=0$. The <displacement current> repairs this conflict: using <Gauss's law>, $\nabla\cdot(\epsilon_0\partial_tE)=\partial_t\rho$, so the <Ampère-Maxwell equation> is
$$
\nabla\times B=\mu_0(J+\epsilon_0\partial_tE).
$$
In the <parallel-plate capacitor>, neglect <fringing fields> and take $E_z=V_0e^{-i\omega t}/d$. A circular <Ampère loop> of radius $s<a$ encloses <displacement current> $-i\omega\epsilon_0\pi s^2V_0e^{-i\omega t}/d$. Therefore the <magnetic field between circular capacitor plates> has phasor and amplitude
$$
\widetilde B_\phi(s)=-\frac{i\omega\mu_0\epsilon_0V_0s}{2d},\qquad
\boxed{B_{\mathrm{amp}}(s)=\frac{\omega\mu_0\epsilon_0|V_0|}{2d}\,s,\quad0\leq s<a.}
$$
It is zero on the axis, grows linearly with radius, and is azimuthal by the <right-hand rule>. This is a <quasistatic approximation>: in addition to $d\ll a$, the uniform-field model requires $\omega a/c\ll1$ and excludes the edge region. The answer gives the oscillation amplitude, rather than a time-independent instantaneous magnitude.
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