= Solution
\b[The integral field equation.] The <Ampere-Maxwell law> is $\nabla\times\mathbf H=\mathbf J+\partial_t\mathbf D$. Apply <Stokes theorem> to any fixed oriented surface bounded by the chosen contour:
$$
\boxed{\oint\mathbf H\cdot d\boldsymbol\ell
=\int\mathbf J\cdot d\mathbf S
+\int\partial_t\mathbf D\cdot d\mathbf S.}
$$
The added <displacement current> is required for compatibility with <electric charge conservation>. Taking a divergence and using $\nabla\cdot\mathbf D=\rho_f$ gives $\nabla\cdot\mathbf J+\partial_t\rho_f=0$.
\b[Field geometry.] Choose the upper plate positive and let $z$ increase upward. In the long-plate, slow-discharge approximation, $\mathbf E=-(V/d)\widehat{\mathbf z}$. Conventional wire current runs downward from the positive plate. The <magnetic field intensity> circulates clockwise as viewed from above. The <Poynting vector> $\mathbf S=\mathbf E\times\mathbf H$ points radially toward the resistive wire.
\Image[/past-exam-of-the-physics-b-course-of-the-university-of-cambridge/2019/paper-1-b7-capacitor-energy-flow.png]
{title=Original meridional section of the discharging capacitor; electric field and wire current point down, magnetic field circles the wire, and electromagnetic energy flows inward}
\b[Wire and displacement currents.] The <capacitance> and discharge law are
$$
C=\frac{\varepsilon_0\pi a^2}{d},\qquad
I=\frac VR=-C\dot V.
$$
Because $V$ decreases, $\partial_t\mathbf D$ points upward, opposing the downward wire current. Through a disk of radius $r$ between the plates its magnitude is
$$
I_D(r)=-\frac{\varepsilon_0\pi r^2}{d}\dot V
=I\frac{r^2}{a^2}.
$$
The effective downward current enclosed by the circular contour is therefore $I-I_D(r)$. The <Ampere-Maxwell law> gives
$$
2\pi rH=I\left(1-\frac{r^2}{a^2}\right),\qquad
\boxed{H(r)=\frac{V}{2\pi rR}\left(1-\frac{r^2}{a^2}\right).}
$$
This is the magnitude; with the stated coordinates, $\mathbf H=-H\widehat{\boldsymbol\phi}$. The factor $1-r^2/a^2$ expresses cancellation between conduction and displacement currents. The approximation requires distances outside the thin wire and away from detailed electrode-edge fields.
\b[Energy flux and power.] Since $\widehat{\mathbf z}\times\widehat{\boldsymbol\phi}=-\widehat{\mathbf r}$,
$$
\boxed{\mathbf S(r)=-\frac{V^2}{2\pi rdR}
\left(1-\frac{r^2}{a^2}\right)\widehat{\mathbf r}.}
$$
Its units are watts per square metre. Integrating over the cylinder's side area $2\pi rd$, the signed outward power is negative:
$$
P_{\rm out}(r)=-\frac{V^2}{R}\left(1-\frac{r^2}{a^2}\right),
\qquad
\boxed{P_{\rm in}(r)=\frac{V^2}{R}\left(1-\frac{r^2}{a^2}\right).}
$$
There is no leading axial flux through the cylinder's ends in this ideal field geometry.
\b[Where the energy goes.] The <Poynting flux in a centrally discharged parallel-plate capacitor> carries field energy through the dielectric space into the resistor, where it becomes <Joule heating> at rate $V^2/R$. Inside radius $r$ the stored electric <energy> is
$$
U_{<r}=\frac{\varepsilon_0\pi r^2V^2}{2d},\qquad
-\frac{dU_{<r}}{dt}=\frac{V^2}{R}\frac{r^2}{a^2}.
$$
Hence
$$
\boxed{P_{\rm in}(r)-\frac{dU_{<r}}{dt}=\frac{V^2}{R}.}
$$
The resistor receives the inward flux from outside that cylinder plus the field energy released inside it. As $r$ approaches the wire radius, essentially all its power arrives as radial electromagnetic flux. Over the whole capacitor, $-d(CV^2/2)/dt=V^2/R$. The <Poynting theorem> thus accounts for the complete discharge without treating energy as transported only along the conducting wire. High resistance makes the quasistatic electric-field approximation and neglect of changing magnetic energy consistent.
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