E and H ratio (Homework 2-323)

3. In the last homework you found the free-space ratio of magnitude of the transverse components of E and H. Repeat this for the rectangular waveguide TE10 example Griffiths 9.5.2.

For the $TE_10$ mode one has $m=1$ and $n=0$ so that (equation 9.186 with $m=1$ and $n=0$)
$B_z=B_0*\cos⁡ (π/a*x)$ and $E_z=0$ (TE mode!)

We can compute now the other components of the E and B fields (from equation 9.180).
$E_x=i/((ω/c)^2-k^2 ) (k*(∂E_z)/∂x+ω*(∂B_z)/∂y)=-ωi/((ω/c)^2-k^2 ) B_0*π/b sin⁡(π/b*y)$
$E_y=i/((ω/c)^2-k^2 ) (k*(∂E_z)/∂x-ω*(∂B_z)/∂y)=+ωi/((ω/c)^2-k^2 ) B_0*π/b sin⁡(π/b*y)$
$B_x=i/((ω/c)^2-k^2 ) (k*(∂B_z)/∂x-ω/c^2 *(∂E_z)/∂y)=-k i/((ω/c)^2-k^2 ) B_0*π/b sin⁡(π/b*y)$
$B_y=i/((ω/c)^2-k^2 ) (k*(∂B_z)/∂x+ω/c^2 *(∂E_z)/∂y)=-k i/((ω/c)^2-k^2 ) B_0*π/b sin⁡(π/b*y)$

Which means
$E_x=0$
$E_y=i/((ω/c)^2-k^2 )*[π/a*ωB_0 sin⁡(πx/a) ]$
$B_x=i/((ω/c)^2-k^2 )*[-π/a*k B_0*sin⁡(πx/a) ]$
$B_y=0$

So that
$E_y/B_x =-ω/k=c/\sqrt{(1-(ω_{m n}/ω)^2)} >c$ (equation 9.191)
Which comes from the dispersion relation for a waveguide
$ω^2=(ck)^2+ω_{m n}^2$ (see last Homework)
Since $B=μH$ and $c=1/\sqrt{μϵ}$ one has
$E_y/H_x =(μ*c)/√(1-(ω_{m n}/ω)^2 )=\sqrt{(μ/ϵ)}*1/\sqrt{(1-(ω_{m n}/ω)^2}$


valentin68

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