Finding the Magnetic Field from a Time-Varying Electric Potential. In Cylindrical coordinates.
$$\require{physics}$$
The Voltage at the Collector swings between 0V and 2Vcc
$$ V_{c} = V_{cc} (1 + e^{ i \omega t} ) $$
The Electric Field is equal to negative the Gradient of the Voltage.
$$ V = V_{c} $$
$$ \vb{E} = - \grad{ V } $$
$$
\vb{E} = - \pdv{V}{r} \vectorunit{r} - \frac{1}{r} \pdv{V}{\phi} \vectorunit{\phi} - \pdv{V}{z} \vectorunit{z}
$$
$$
E_{r} = - \pdv{V}{r} \quad
E_{\phi} = - \frac{1}{r} \pdv{V}{\phi} \quad
E_{z} = - \pdv{V}{z}
$$
$$ \vb{E} = E_{r}\vectorunit{r} + E_{\phi}\vectorunit{\phi} + E_{z}\vectorunit{z} $$
Let's find the Magnetic Field from the time-varying Electric Field. To do so, consider Maxwell's Equation for the Curl of the Electric Field.
$$ \curl{ \vb{E} } = - \pdv{ \vb{B} }{t} $$
The Curl of E is defined as the Determinant of the following matrix:
$$ \curl{ \vb{E} } = \frac{1}{r}
\begin{vmatrix}
\vectorunit{r} & r \vectorunit{\phi} & \vectorunit{z} \\
\pdv{}{r} & \pdv{}{\phi} & \pdv{}{z} \\
E_{r} & r E_{\phi} & E_{z} \\
\end{vmatrix}
$$
$$ \frac{1}{r}
\begin{vmatrix}
\vectorunit{r} & r \vectorunit{\phi} & \vectorunit{z} \\
\pdv{}{r} & \pdv{}{\phi} & \pdv{}{z} \\
E_{r} & r E_{\phi} & E_{z} \\
\end{vmatrix} =
\left( \frac{1}{r} \pdv{E_{z}}{\phi} - \pdv{E_{\phi}}{z} \right) \vectorunit{r} +
\left( \pdv{E_{r}}{z} - \pdv{E_{z}}{r} \right) \vectorunit{\phi} +
\frac{1}{r} \left( \pdv{ (r E_{\phi} ) }{r} - \pdv{E_{r}}{\phi} \right) \vectorunit{z}
$$
Therefore, the Determinant of this matrix is equal to negative the partial derivative of the Magnetic Field, with respect to time.
$$
\frac{1}{r}
\begin{vmatrix}
\vectorunit{r} & r \vectorunit{\phi} & \vectorunit{z} \\
\pdv{}{r} & \pdv{}{\phi} & \pdv{}{z} \\
E_{r} & r E_{\phi} & E_{z} \\
\end{vmatrix}
= - \pdv{ \vb{B} }{t}
$$
Let's integrate both sides of the equation with respect to time.
$$
\int{
\frac{1}{r}
\begin{vmatrix}
\vectorunit{r} & r \vectorunit{\phi} & \vectorunit{z} \\
\pdv{}{r} & \pdv{}{\phi} & \pdv{}{z} \\
E_{r} & r E_{\phi} & E_{z} \\
\end{vmatrix}
dt}
= - \int{ \pdv{\vb{B}}{t} dt}
$$
$$
\int{
\frac{1}{r}
\begin{vmatrix}
\vectorunit{r} & r \vectorunit{\phi} & \vectorunit{z} \\
\pdv{}{r} & \pdv{}{\phi} & \pdv{}{z} \\
E_{r} & r E_{\phi} & E_{z} \\
\end{vmatrix}
dt}
= - \vb{B}
$$
$$
- \int{
\frac{1}{r}
\begin{vmatrix}
\vectorunit{r} & r \vectorunit{\phi} & \vectorunit{z} \\
\pdv{}{r} & \pdv{}{\phi} & \pdv{}{z} \\
E_{r} & r E_{\phi} & E_{z} \\
\end{vmatrix}
dt}
= \vb{B}
$$
$$
- \int{
\left(
\left( \frac{1}{r} \pdv{E_{z}}{\phi} - \pdv{E_{\phi}}{z} \right) \vectorunit{r} +
\left( \pdv{E_{r}}{z} - \pdv{E_{z}}{r} \right) \vectorunit{\phi} +
\frac{1}{r} \left( \pdv{ (r E_{\phi} ) }{r} - \pdv{E_{r}}{\phi} \right) \vectorunit{z}
\right)
dt}
= \vb{B}
$$
Now let's try an example. I wrote some Python code to help us. (coming soon...)