Waveguide modes (Homework 1-323)

Consider TE waves in the rectangular waveguide in the $TE_{01}$ mode.

a) What are all the components of the E and B fields?

b) Show that the energy transmitted is along the axis of the waveguide. (Hint: you can use the average Poynting flux).


In a guided wave one just has to know the values of the longitudinal components of E and B ($E z$ and $B z$). The other x,y components are related to the z components by simple equations (equation 9.180).

The values of the longitudinal components are computed knowing the propagation mode (mn) and the dimensions of the guide (x=a and y=b).

$E_z=E_0  \cos⁡ (k_x x)*\cos⁡(k_y y)$  (for TM modes)   and $B_z=B_0 \cos⁡ (k_x x) \cos⁡(k_y y)$   (for TE modes)
where $k_x=mπ/a$      and  $k_y=nπ/b$   (like for quantum wells)
For (transverse electric-meaning that longitudinal electric component is zero)  $TE_01$  one has

$B_z=B_0*\cos⁡((π/b)*y)$     and $E_z=0$

And from (equation 9.180)
$E_x=i/((ω/c)^2-k^2 )  (k*(∂E_z)/∂x+ω*(∂B_z)/∂y)=$

$=-ωi/((ω/c)^2-k^2 ) B_0*π/b *\sin⁡(π/b*y)$

$E_y=i/((ω/c)^2-k^2 )  (k*(∂E_z)/∂x-ω*(∂B_z)/∂y)=$

$=+ωi/((ω/c)^2-k^2 ) B_0*π/b *\sin⁡(π/b*y)$

$B_x=i/((ω/c)^2-k^2 )  (k*(∂B_z)/∂x-ω/c^2 *(∂E_z)/∂y)=$

$=-k i/((ω/c)^2-k^2 ) B_0*π/b* \sin⁡(π/b*y)$

$B_y=i/((ω/c)^2-k^2 )  (k*(∂B_z)/∂x+ω/c^2 *(∂E_z)/∂y)=$

$=-k i/((ω/c)^2-k^2 ) B_0*π/b* \sin⁡(π/b*y)$


The Poynting vector is the vector that has the direction of energy propagation and the module equal to the energy per unit area per unit time (is called energy flux density).

$\overrightarrow{S}=(1/μ)(\overrightarrow{E}  ×\overrightarrow{B})$

$\overrightarrow{S}=(1/μ)[(E_x B_y-E_y B_x)\hat z +(E_y B_z-E_z B_y)\hat x +(E_z B_x-E_x B_z)\hat y]$   (circular permutations of x,y,z indexes)

$\overrightarrow{S}=(1/μ)*[(E_x B_y-E_y B_x )\hat z +E_y B_z*\hat x-E_x B_z*y ̂]$

$\overrightarrow{S}=(1/μ)*[Cz ̂*\sin^2⁡((π/b)*y)+Dx ̂*\sin⁡((π/b)*y)*\cos⁡((π/b)*y)-$

$-Fy ̂*\sin⁡((π/b) y)*\cos⁡((π/b) y)]$

Where C, D, F are constants (pure numbers).

When mediating over a wavelength one has

$\overline{sin^2⁡(…) )} ? 0$    and $\overline{sin⁡(…)  cos⁡(…) )}=1/2*\overline{(sin⁡(2*…) ) }=0$
$1/2π ∫_0^{2π} \sin^2⁡ x*dx=1/2$

So that

$\overrightarrow{S}=(1/μ)*C/2*\hat z$   is in the direction of propagation of the wave