Question 9.28: how that the mode TE00 cannot occur in a rectangular wave gu...

how that the mode TE00 cannot occur in a rectangular wave guide. [Hint: In this case ω/c = k, so Eqs. 9.180 are indeterminate, and you must go back to Eq. 9.179. Show that B_{z} is a constant, and hence—applying Faraday’s law in integral form to a cross section—that B_{z}=0 , so this would be a TEM mode.]

  \left. \begin{matrix} \text { (i) } E_{x}=\frac{i}{(\omega / c)^{2}-k^{2}}\left(k \frac{\partial E_{z}}{\partial x}+\omega \frac{\partial B_{z}}{\partial y}\right) \text {, } \\ \text { (ii) } E_{y}=\frac{i}{(\omega / c)^{2}-k^{2}}\left(k \frac{\partial E_{z}}{\partial y}-\omega \frac{\partial B_{z}}{\partial x}\right), \\ \text { (iii) } B_{x}=\frac{i}{(\omega / c)^{2}-k^{2}}\left(k \frac{\partial B_{z}}{\partial x}-\frac{\omega}{c^{2}} \frac{\partial E_{z}}{\partial y}\right), \\ \text { (iv) } B_{y}=\frac{i}{(\omega / c)^{2}-k^{2}}\left(k \frac{\partial B_{z}}{\partial y}+\frac{\omega}{c^{2}} \frac{\partial E_{z}}{\partial x}\right) \text {. } \end{matrix} \right\}                          (9.180)

\left. \begin{matrix} \text { (i) } \frac{\partial E_{y}}{\partial x}-\frac{\partial E_{x}}{\partial y}=i \omega B_{z}, & \text { (iv) } \frac{\partial B_{y}}{\partial x}-\frac{\partial B_{x}}{\partial y}=-\frac{i \omega}{c^{2}} E_{z}, \\ \text { (ii) } \frac{\partial E_{z}}{\partial y}-i k E_{y}=i \omega B_{x} \text {, } & \text { (v) } \frac{\partial B_{z}}{\partial y}-i k B_{y}=-\frac{i \omega}{c^{2}} E_{x}, \\ \text { (iii) } i k E_{x}-\frac{\partial E_{z}}{\partial x}=i \omega B_{y}, & \text { (vi) } i k B_{x}-\frac{\partial B_{z}}{\partial x}=-\frac{i \omega}{c^{2}} E_{y}. \end{matrix} \right\}                           (9.179) 

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\text { Here } E_{z}=0( TE ) \text { and } \omega / c=k(n=m=0) \text {, so Eq. } 9.179( ii ) \Rightarrow E_{y}=-c B_{x} \text {, Eq. 9.179(iii) } \Rightarrow E_{x}=c B_{y},

\text { Eq. 9.179(v) } \Rightarrow \frac{\partial B_{z}}{\partial y}=i\left(k B_{y}-\frac{\omega}{c^{2}} E_{x}\right)=i\left(k B_{y}-\frac{\omega}{c} B_{y}\right)=0 \text {, Eq. 9.179(vi) } \Rightarrow \frac{\partial B_{z}}{\partial x}=i\left(k B_{x}+\frac{\omega}{c^{2}} E_{y}\right)=

i\left(k B_{x}-\frac{\omega}{c} B_{x}\right)=0 \text {. So } \frac{\partial B_{z}}{\partial x}=\frac{\partial B_{z}}{\partial y}=0 \text {, and since } B_{z} \text { is a function only of } x \text { and } y \text {, this says } B_{z} \text { is in fact } a constant (as Eq. 9.186 also suggests). Now Faraday’s law (in integral form) says \oint E \cdot d l =-\int \frac{\partial B }{\partial t} .

B_{z}=B_{0} \cos (m \pi x / a) \cos (n \pi y / b)                    (9.186)

da, and Eq. 9.176 \Rightarrow \frac{\partial B }{\partial t}=-i \omega B , \text { so } \oint E \cdot d l =i \omega \int B \cdot d a . Applied to a cross-section of the waveguide this gives \oint E \cdot d l =i \omega e^{i(k z-\omega t)} \int B_{z} d a=i \omega B_{z} e^{i(k z-\omega t)}(a b)\left(\text { since } B_{z}\right. is constant, it comes outside the integral). But if the boundary is just inside the metal, where E = 0, it follows that B_{z}=0. So this would be a TEM mode, which we already know cannot exist for this guide. 

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