Question 10.8: How many component plane-waves does the TE10 mode contain in......

For the TE_{10} mode, let us rewrite Eq. (10-74) as

\boxed{H_{z}(x,y) = H_{o} \cos \left\lgroup\frac{m\pi }{a} x \right\rgroup \cos \left\lgroup\frac{n\pi }{b}y \right\rgroup }                ( m,n = 0,1,2,…)                     (10-74)

H_{z} = H_{o} \cos \left\lgroup\frac{\pi }{a}x \right\rgroup = H_{o} \frac{1}{2} (e^{j\pi x/a} +e^{- j\pi x/a})                                          (10-89)

As can be seen from Eq. (10-75), the TE_{10} mode has no E_{x}– and no H_{y}–component. Rewriting Eq. (10-75c) for the TE_{10} mode, we have

Combining Eq. (10-89) and Eq. (10-90), assuming a lossless waveguide( \gamma = j\beta _{10}  ), the magnetic field phasor is

From the above equation, we see that the mode TE_{10} comprises two plane waves of the form e^{j \pi x/a -j\beta _{10}z}  and e^{- j \pi x/a -j\beta _{10}z} .