I was wondering if in the case of a very thin enameled (i.e. sub-millimeter thick wire insulation dielectric usually made from polyimide) magnet wire the generated electric field of a signal transient (i.e a short pulse) is propagating exclusively inside the insulation dielectric of the wire or also through the air (dielectric) surrounding the wire?
I want to find out the signal velocity in the magnet wire which is:
$$ \mathbf{v}_{\mathrm{s}}=\frac{c}{\sqrt{\varepsilon_{r} \mu_{r}}} \approx \frac{c}{\sqrt{\varepsilon_{r}}} $$ where $\varepsilon_{r}$ is the relative permittivity of the medium, $\mu_{r}$ is the relative permeability of the medium, and $c$ is the speed of light in vacuum. The approximation shown is used in many practical context because for most common materials $\mu_{r} \approx 1$.
I know that for polyimide insulated magnet wire the relative permittivity of the insulation is specified around $k=3$. However, if a large part of the field is traveling via the air and not exclusively through the insulation, which is possible since the insulation of the magnet wire is very thin only a small fraction of a mm, then the calculation of the signal velocity is more complicated. The signal will propagate faster via the air around the wire than via the insulation dielectric therefore arriving faster than expected to the circuit load. The relative permittivity of dry air which is also a dielectric is $k=1.0006$.
How can I calculate accurately the signal velocity of the magnet wire?