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Conceptually, I understand that coaxial cables operate in a TEM mode due to the central conductor, and hence the electric field is directed radially outward from the voltage $V_0$ at the centre of the cable, to the supposedly grounded outer cable (the derivation for this involving Poisson's equation and such). Similarly, as the current flows down the cable, the magnetic field around this conductor has direction only in terms of $\phi$ and does not change with $z$. However, I've been asked:

How are the electric and magnetic fields inside a coaxial cable related to the V and I in the wave of the equation:

$V_+(t-z/v) = Z_0I_+(t-z/v)$

where the subscript $+$ signifies the voltage and current in the $+z$ direction (Note that there is a similar formula for the -ve reflected direction).

I've been searching all night, but I have yet to find a solid explanation for this that doesn't involve derivations from first principles. How can I turn the above equation into something involving B and E fields?

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A unit length of a cable will have some resistance $R$; some conductance $G$ by which charge can leak through the dielectric; some inductance $L$; and some capacitance $C$:

cable model

These quantities combine to give the cable's impedance at a given signal frequency $\omega$, $$ Z_0 = \sqrt\frac{R+j\omega L}{G+j\omega C} \approx \sqrt\frac LC $$ Most common coaxial cables have impedances of 50 Ω or 75 Ω. Other common transmission lines have comparable impedances; 100 Ω is common.

Note that if you're using the cable to carry DC or very low frequency currents, the impedance becomes $Z \approx \sqrt{R/G}$ and the relationship between current and voltage in the cable will (hopefully) be dominated by the impedance of the load.

You can find the fields in the dielectric from the capacitance and inductance. A cylindrical capacitor with inner,outer radius $a,b$, permittivity $\epsilon$, and voltage difference $V$ has charge per unit length $\lambda$, where $$ \lambda = \frac{ 2\pi \epsilon V}{\ln(b/a)} $$ and field strength $$E = \frac{\lambda}{2\pi\epsilon r} = \frac{VC}{2\pi\epsilon r}.$$ Remember that $C$ is the capacitance per unit length, for dimensional consistency. The magnetic field will simply be the Biot-Savart field due to the current on the central wire (in the limit where the bend radius for the cable is long compared with the cable's radius).

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  • $\begingroup$ How does that relate to E and B fields? $\endgroup$
    – Yoshi
    Jun 15, 2014 at 4:00
  • $\begingroup$ Whoops! Better? $\endgroup$
    – rob
    Jun 15, 2014 at 4:44
  • $\begingroup$ Thanks! This is very similar to where my head was going, but as I am working on a report for a third-year physics lab, I figured they may want something slightly more in-depth in terms of reflection coefficients and forwards and backwards travelling voltages. Suffice to say I've skipped this particular question, and will likely fill it in with this knowledge after I've completed another report. $\endgroup$
    – Yoshi
    Jun 15, 2014 at 4:56
  • $\begingroup$ For the fields this is the entire answer; it's the instantaneous voltage and current that matter. (There'd be some phase delay if the signal speed were so slow that the length of an oscillation were comparable to the radius of the cable, but real cables conveniently stop working before those frequencies.) $\endgroup$
    – rob
    Jun 15, 2014 at 5:06
  • $\begingroup$ The convenient way to think about reflection coefficients is by impedance matching. An open-ended cable gives you a same-sign reflected pulse, because the charge on the pulse can't leave the center conductor; a shorted cable gives you an opposite-sign reflected pulse, because the voltage difference at the short is exactly zero. The midpoint with zero reflected pulse is when the termination matches the cable impedance. You can work out the relation from there. $\endgroup$
    – rob
    Jun 15, 2014 at 5:08

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