Past exam of the physics course of the year 4 of the natural sciences course of the University of Cambridge / 2013 / Paper 1 / Section B / B6

For charge per length ${\lambda }_q$, Gauss's law gives $E_r={\lambda }_q/(2\pi {ϵ}_{0}r)$ between the conductors. Integration gives $V={\lambda }_q\ln (b/a)/(2\pi {ϵ}_0)$. Similarly Ampère's law gives $B_{\varphi }={\mu }_{0}I/(2\pi r)$; integrating the magnetic energy per length and setting it equal to $L{I}^2/2$ givesHere $C,L$ are per unit length. The displayed $L$ is the external TEM inductance, appropriate to the ideal surface-current/high-frequency model. Internal magnetic energy in conductors with appreciable current penetration adds internal inductance; it cannot be determined fully from an outer conductor with unspecified wall geometry.

The distributed series resistance and inductance, with shunt capacitance and negligible leakage, give the telegrapher's equationsDifferentiate the first with respect to $x$ and substitute the second to obtainFor $V\propto e^{ikx-i\omega t}$ the forward attenuating branch has $k^2={\omega }^{2}LC+i\omega RC$. When $R\ll \omega L$,The positive imaginary part gives amplitude $e^{-x/{\ell }_{\mathrm{amp}}}$. Power decays twice as fast in its exponent, so its attenuation length is ${\ell }_{\mathrm{amp}}/2$. Finite conductivity changes the speed only at higher order in this constant-parameter weak-loss expansion, apart from any separately retained internal inductance or dielectric response.