Griffiths\' treatment of EM propagation in waveguides assumes the walls are perf
ID: 1289773 • Letter: G
Question
Griffiths' treatment of EM propagation in waveguides assumes the walls are perfect conduc tors. A rigorous calculation of the fields for the case where losses (such as Joule heating of the walls) are included is complicated, but is usually unnecessary. Waveguides are normally constructed from good conductors. The electric fields inside the conductor are thus small and their influence can be treated as as a perturbation to the ideal (perfect conductor) so lution. Note that the geometry of the problem outlined below is identical to that shown in Fig. 9.24 of your textbook and/or your notes from class. The dimensions a and b correspond to the widths of the broad and narrow walls of the guide, along and respectively. Wave propagation is in the direction 1. Starting from the TE10 wave solutions to Maxwell's equations in a rectangular waveg uide kz-cut 10e and 10 where Tr Sin 10 X COS sin 10 10 and Bo characterizes the amplitude of the magnetic field, write down expressions for the tangential components of the field Hw just inside each wall (i.e in the metal) Note that H Re Hwy expli(kz wt) 2. Next, think of the field w as part of a plane wave solution to Maxwell's equations, propagating into the wall parallel to the surface normal. Use the fact that? for a TEM wave in a good conductor (cf. Griffiths Egs. 9.137 and 9.138) to obtain expressions for Ew just inside each wall (i.e. in the metal). You will have to think about the implied direction of the resulting complex Poynting vector Sw Ew x Hw in order to set the right orientation (and phasing) for Ew. In a perfect conductor or/we oo, and so Ewl 0; in a good conductor (o/we 1) and so Ewl 10 As long as this is true, the equations describing E10 and B10 remain validExplanation / Answer
Introduction to Electrodynamics, 3
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