Question 3 of 8: Coaxial line: inner conductor diameter and dielectric constant
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
Paper format. National Exams, December 2015 — 07-Elec-A7 Electromagnetics. Three hours, closed book; one of two approved calculators permitted. Eight questions, all of equal value (20 marks each). The rubric states that any five questions constitute a complete paper and that only the first five presented are marked — all eight are worked here, because this set is a study resource rather than a sat examination. Aids printed on the paper: $\varepsilon_0 = 8.85\times10^{-12}\ \text{F/m}$ and $\mu_0 = 4\pi\times10^{-7}\ \text{H/m}$, from which $\eta_0=\sqrt{\mu_0/\varepsilon_0} = 376.8\ \Omega$ and $c = 1/\sqrt{\mu_0\varepsilon_0} = 3.00\times10^{8}\ \text{m/s}$.
Reference texts. D. M. Pozar, Microwave Engineering, 4th ed. (transmission-line transients, matching stubs, waveguide modes); M. N. O. Sadiku, Elements of Electromagnetics, 7th ed.; W. H. Hayt and J. A. Buck, Engineering Electromagnetics, 9th ed. (plane waves, normal incidence); F. T. Ulaby, Fundamentals of Applied Electromagnetics, 8th ed. (polarisation, interference, radiation); C. A. Balanis, Antenna Theory: Analysis and Design, 4th ed. (short current elements).
Given. A uniformly filled coaxial line whose electrical behaviour is specified and whose outer conductor dimension is known.
Given data
Quantity
Symbol
Value
Characteristic impedance
$Z_0$
$50\ \Omega$
Propagation velocity
$v_p$
$2\times10^{8}\ \text{m/s}$
Inner diameter of the outer conductor
$D$
$1\ \text{cm} = 10\ \text{mm}$
Dielectric
—
uniform, non-magnetic ($\mu_r = 1$)
Find. The diameter $d$ of the inner conductor and the relative permittivity $\varepsilon_r$ of the filling.
Figure Q3. Coaxial cross-section. The velocity fixes the dielectric; the impedance then fixes the conductor diameter ratio.
Approach. The two unknowns separate cleanly: the propagation velocity depends only on the filling, and the characteristic impedance then depends only on the geometry once the filling is known.
Get the permittivity from the velocity. In a TEM line filled with a uniform non-magnetic dielectric the wave travels at the medium's own light speed, $$v_p = \frac{c}{\sqrt{\varepsilon_r}} \;\Longrightarrow\; \sqrt{\varepsilon_r} = \frac{c}{v_p} = \frac{3.00\times10^{8}}{2\times10^{8}} = 1.50$$ so that $$\boxed{\varepsilon_r = 2.25}$$ which is a thoroughly ordinary value — solid polyethylene sits at $\varepsilon_r \approx 2.25$–$2.3$, and a velocity factor of 0.667 is the familiar figure quoted for RG-58 style cable.
Write the impedance of a coaxial TEM line. Solving Laplace's equation between two concentric cylinders and forming $Z_0=\sqrt{L'/C'}$ gives $$Z_0 = \frac{\eta_0}{2\pi\sqrt{\varepsilon_r}}\,\ln\frac{D}{d} = \frac{60}{\sqrt{\varepsilon_r}}\,\ln\frac{D}{d}\ \ \Omega$$ Note that only the ratio of the diameters enters, which is why the question can specify one dimension and one impedance and still fix the geometry uniquely.
Solve for the diameter ratio. Rearranging and substituting, $$\ln\frac{D}{d} = \frac{2\pi\sqrt{\varepsilon_r}\,Z_0}{\eta_0} = \frac{2\pi(1.50)(50)}{376.8} = 1.250$$ so $$\frac{D}{d} = e^{1.250} = 3.490$$
Recover the inner diameter. With $D = 10\ \text{mm}$, $$d = \frac{D}{3.490} = \frac{10}{3.490} = \boxed{2.865\ \text{mm}}$$ an inner conductor of about 2.87 mm diameter, i.e. roughly a 10 AWG wire — entirely realistic for a 10 mm outer shield.
Cross-check through the per-unit-length parameters. The same geometry gives $$L' = \frac{\mu_0}{2\pi}\ln\frac{D}{d} = 250\ \text{nH/m}, \qquad C' = \frac{2\pi\varepsilon_0\varepsilon_r}{\ln(D/d)} = 100.1\ \text{pF/m}$$ from which $\sqrt{L'/C'} = 50.0\ \Omega$ and $1/\sqrt{L'C'} = 2.00\times10^{8}\ \text{m/s}$, reproducing both given quantities and confirming the pair $(\varepsilon_r, d)$ is self-consistent rather than merely arithmetically correct.
Had the question given a magnetic filling the two steps would no longer separate, because $v_p = c/\sqrt{\mu_r\varepsilon_r}$ and $Z_0 \propto \sqrt{\mu_r/\varepsilon_r}$ would then have to be solved simultaneously. The phrase “a dielectric” is the licence to take $\mu_r = 1$.
