Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
Paper format. National Examinations, May 2014 — 07-Elec-A7 Electromagnetics. Three hours, closed book, one approved Casio or Sharp calculator. Eight questions of equal value (20 marks each); the paper states that any five constitute a complete paper and that only the first five in the answer book are marked, but all eight are solved here so the set works as a complete study resource. Aids printed on the paper: $\epsilon_0 = 8.85\times10^{-12}$ F/m and $\mu_0 = 4\pi\times10^{-7}$ H/m, from which $\eta_0 = \sqrt{\mu_0/\epsilon_0} = 376.7\ \Omega$ and $c = 1/\sqrt{\mu_0\epsilon_0} = 3.00\times10^{8}$ m/s.
Reference texts. Sadiku, Elements of Electromagnetics, 7th ed.; Hayt & Buck, Engineering Electromagnetics, 9th ed.; Ulaby & Ravaioli, Fundamentals of Applied Electromagnetics, 8th ed.; Pozar, Microwave Engineering, 4th ed.; Balanis, Antenna Theory: Analysis and Design, 4th ed.
Given. A coaxial line whose inner conductor is 1 mm in diameter (so $a = 0.5$ mm), filled with a dielectric of relative permittivity $\epsilon_r = 2.25$ (polyethylene), and designed for a characteristic impedance of 50 Ω. The line is lossless and carries the TEM mode.
Find. (i) the inner diameter of the outer conductor, and (ii) the phase velocity of signals on the line.
Figure 3.1 — Cross-section of the coaxial line. In the TEM mode E is purely radial and H purely azimuthal, which is what makes the logarithmic impedance formula exact.
Approach. Invert the standard TEM coaxial impedance formula for the radius ratio, then convert to a diameter; the velocity follows directly from the filling permittivity, and both answers can be checked against the per-metre inductance and capacitance.
Characteristic impedance of a coaxial line. For the TEM mode,$$Z_0 = \frac{\eta_0}{2\pi\sqrt{\epsilon_r}}\ln\frac{b}{a} \approx \frac{60}{\sqrt{\epsilon_r}}\ln\frac{b}{a}\ \ \Omega,$$where $a$ and $b$ are the inner and outer radii and $\eta_0/2\pi = 59.96 \approx 60\ \Omega$.
Solve for the radius ratio. With $\sqrt{2.25} = 1.5$,$$\ln\frac{b}{a} = \frac{Z_0\sqrt{\epsilon_r}}{60} = \frac{50(1.5)}{60} = 1.250,$$$$\frac{b}{a} = e^{1.250} = 3.490.$$This ratio is the reason 50 Ω polyethylene cable always looks roughly three-and-a-half times wider than its centre wire.
Outer conductor diameter. Because the ratio applies equally to radii and to diameters,$$D = 2b = 3.490\,(2a) = 3.490(1.000\ \text{mm}),$$$$\boxed{D = 3.49\ \text{mm}.}$$
Propagation velocity. A TEM line is filled with one homogeneous dielectric, so the wave travels at the plane-wave velocity of that dielectric and the geometry drops out entirely:$$v_p = \frac{c}{\sqrt{\epsilon_r}} = \frac{3.00\times10^{8}}{1.5},$$$$\boxed{v_p = 2.00\times10^{8}\ \text{m/s}\ \ (66.7\ \%\ \text{of } c).}$$
Independent check through L and C. Using the same geometry,$$L = \frac{\mu_0}{2\pi}\ln\frac{b}{a} = 0.250\ \mu\text{H/m},\qquad C = \frac{2\pi\epsilon_0\epsilon_r}{\ln(b/a)} = 100\ \text{pF/m},$$which return $\sqrt{L/C} = 50.0\ \Omega$ and $1/\sqrt{LC} = 2.00\times10^{8}$ m/s, confirming both answers.
The numbers land on a familiar product: a 1 mm centre conductor inside a 3.5 mm dielectric is essentially RG-58-class cable, and its 5 ns/m delay (the reciprocal of $2\times10^{8}$ m/s) is the figure normally quoted on the data sheet as a velocity factor of 0.667.
