Question 4 of 10: Rectangular Tunnel as a Waveguide — Cutoff and Evanescent Decay
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
17-Phys-A3, Electromagnetics — National Exam, May 2019. 3-hour closed-book exam (Casio or Sharp approved calculators only, one 8.5″×11″ aid sheet); any FIVE of the printed questions constitute a complete exam paper and only the first five as they appear in a candidate's answer book are marked, each of equal value, with full justification required for marks. All ten printed questions are solved below as a complete study resource.
Reference texts: Sadiku, Elements of Electromagnetics (7th ed.) — plane waves and boundaries, Gauss's law, resistance and current density, magnetostatics; Hayt & Buck, Engineering Electromagnetics (9th ed.) — transmission-line transients and bounce diagrams; Pozar, Microwave Engineering (4th ed.) — transmission-line theory, rectangular waveguide cutoff; Balanis, Antenna Theory: Analysis and Design (4th ed.) — short-dipole radiation resistance and efficiency.
Question 4: Rectangular Tunnel as a Waveguide — Cutoff and Evanescent Decay (20 marks)
Given. Air-filled ($\varepsilon_r=1$) rectangular tunnel/waveguide, $a=7$ m (wide wall), $b=4.5$ m; AM signal at $f=1$ MHz with entrance field $|E_y|=0.025$ V/m at the cross-section centre.
Find. The dominant mode and its cutoff frequency $f_c$; the field pattern of that mode; and a physical explanation (with a supporting number) for the rapid signal decay at 1 MHz.
Approach. The dominant mode of a rectangular guide with $a>b$ is $\text{TE}_{10}$, with $f_c=c/(2a)$. Compare $f=1$ MHz to $f_c$: since $f\ll f_c$, no mode propagates and the field decays exponentially (evanescent), governed by $\alpha=(2\pi f_c/c)\sqrt{1-(f/f_c)^2}$.
Part (a) — Dominant mode and cutoff. For $a>b$, $\text{TE}_{10}$ has the lowest cutoff among all $\text{TE}_{mn}/\text{TM}_{mn}$ modes:
$$f_{c,10}=\frac{c}{2a}=\frac{3\times10^8}{2(7)}=\boxed{21.4\ \text{MHz}\ (\text{TE}_{10})}.$$
(For comparison $f_{c,01}=c/2b=33.3$ MHz, confirming $\text{TE}_{10}$ is lower and hence dominant.)
Part (b) — Field pattern. $\text{TE}_{10}$ has a single transverse electric component, $E_y(x)=E_0\sin(\pi x/a)$, independent of $y$: zero at the side walls ($x=0,a$), maximum at mid-width ($x=a/2$), uniform along the height.
TE₁₀ electric field over the tunnel cross-section: a half-sine in $x$ (the 7 m width), uniform in $y$ (the 4.5 m height), directed along $y$ (vertical, matching the AM signal's polarization).
Part (c) — Why the signal fades quickly. At $f=1$ MHz, $f/f_{c,10}=1/21.4=0.0467\ll1$: the tunnel is driven far BELOW the cutoff of every mode it supports, so no mode can propagate — the propagation constant becomes purely real (attenuative, "evanescent") instead of imaginary (propagating):
$$\alpha=\frac{2\pi f_{c,10}}{c}\sqrt{1-\left(\frac{f}{f_{c,10}}\right)^2}=\boxed{0.448\ \text{Np/m}=3.89\ \text{dB/m}}.$$
Over just 10 m this is $44.8$ Np $=38.9$ dB of loss, i.e. the field is attenuated by $e^{-4.48}=0.0113$, or to about $1.1\%$ of its entrance value: $E_y(10\text{m})\approx(0.025)(0.0113)=2.8\times10^{-4}$ V/m. A rectangular tunnel this size simply cannot carry a 1 MHz wave as a propagating mode; only the near-field of the source leaks a short distance in before dying off exponentially, which is exactly the "quickly reducing" behaviour observed.
