Electromagnetic Fields and Waves: Notes, Solved Problems and Interactive Lab

The time-average Poynting vector $\frac{1}{2} Re {E \times H^{*}} = (∣ E_{ρ} ∣^{2} /2 η) a_{z}$ points along the line. Integrating it over the annulus $a < ρ < b$ ,

P = \int_{0}^{2 π} \int_{a}^{b} \frac{∣ E _{ρ} ∣ ^{2}}{2 η} ρ d ρ d ϕ = \frac{2 π ∣ V _{0} ∣ ^{2}}{2 η ln ^{2} ( b / a )} \int_{a}^{b} \frac{d ρ}{ρ} = \frac{π ∣ V _{0} ∣ ^{2}}{η ln ( b / a )} = \frac{∣ V _{0} ∣ ^{2}}{2 Z _{0}} . (7.5.9)

The circuit formula $\frac{1}{2} Re {V I^{*}} = ∣ V_{0} ∣^{2} / (2 Z_{0})$ gives the same number. In the TEM mode the power really flows in the dielectric, guided by the conductors, and Exercise 7.5.3 shows that it crowds towards the inner conductor. The field is largest on the inner conductor,

E_{a} = \frac{∣ V _{0} ∣}{a ln ( b / a )} ⟹ P = \frac{π a ^{2} E _{a}^{2}}{η} ln \frac{b}{a} . (7.5.10)

If the dielectric breaks down at the field $E_{br}$ , the line can carry at most

Peak power of a coaxial line

P_{m a x} = \frac{π a ^{2} E _{br}^{2}}{η} ln \frac{b}{a} (breakdown at ρ = a), V_{m a x} = E_{br} a ln \frac{b}{a} . (7.5.11)

$E_{br} \approx 3$ MV/m for dry air at sea level (peak), much more for solid PTFE or polyethylene. In practice the heating of the inner conductor limits the average power long before breakdown, and a standing wave on a mismatched line raises the peak voltage above $V_{0}$ .

Where the field is strongest

Since $E \propto 1/ ρ$ , a thin inner conductor raises $E_{a}$ for a given voltage, while a thick one leaves little room for the field and little $ln (b / a)$ . Somewhere in between lies an optimum, found in Section 7.5.5. Exercise 3.3.10 of Section 3.3 found it already for the highest voltage, $a = b / e$ .

7.5.3 Power flow and the strongest field