
#1
Apr1808, 02:42 PM

P: 1,445

1. The problem statement, all variables and given/known data
Consider a vacuum filled infinitely long metal cylindrical waveguide of radius, a. Suppose the fields in the waveguide are as follows: [tex] \vec{E}=\vec{E_{0}}\left(s,\phi\right)\exp i(kz\omega t)[/tex] [tex] \vec{B}=\vec{B_{0}}\left(s,\phi\right)\exp i(kz\omega t)[/tex] Find [itex]E_{0z}[/itex] 2. The attempt at a solution Usually when i post my questions ill have a clue as to what to do. But this time around i have no clue whatsoever. Since it is a metal waveguide we can assume that hte parallel component of E and the perpedicular component of B is zero. The E0 given to us depends on s and phi. But this doesn't mean that it only has s and phi components (?) [tex] \vec{E_{0}}=E_{s}\hat{s}+E_{\phi}\hat{\phi}+E_{z}\hat{z}[/tex] But what now? How could the Laplacian be useful? Since there is no charge [tex] \nabla \cdot E = 0 [/tex] so does that imply [tex] \nabla^2 E=0 [/tex]? So if we did that we would get [tex] \frac{1}{s}\frac{\partial }{\partial s}\left(s\frac{\partial E_{s}}{\partial s}\right)+\frac{1}{s^2}\frac{\partial^2 E_{\phi}}{\partial \phi^2}+\frac{\partial^2 E_{z}}{\partial z^2} = 0[/tex] Since they are all equal to zero should be use separation of variables to solve this? I think we hav to use Bessel functions? But the Laplacian would solve for the potential, not the electric field? Please help! Thanks in advance! 



#2
Apr1908, 04:24 PM

P: 1,877

This is a coax cable? Couldn't you simply project the radial (s) unit vector onto the x unit vector?




#3
Apr2008, 10:13 PM

P: 1,445

so instead of s i would write xcos phi? but wouldnt that just make things messier? 



#4
Apr2208, 11:55 AM

P: 1,445

Waveguide
can anyone else provide some input?
Thank you in advance! 



#5
Apr2208, 03:10 PM

P: 455

You have to know whether it is a coaxial cable or not.
If it is not a coaxial cable, you must know whether it is TE or TM. If you don't understand this, read your textbook again. 



#6
Apr2208, 11:35 PM

P: 1,445

it is just a hollow pipe with the fields as stated above it is not specified if it is a TE or TM wave as well 



#7
Apr2308, 06:55 AM

P: 455

The solution is a Bessel function times a Legendre polynomial.
If it is TM, the Bessel must vanish at the surface. If it is TE, E_z=0. 



#8
Apr2308, 09:03 AM

P: 1,445

We didnt study Bessel functions in our class... and he put this on our exam...
im looking at the general solution of laplace equation in cylindrical coords but the boundayr conditions imposed are [tex] u(s,\theta,0)=u(s\phi,\pi)=0[/tex] [tex] u(a,\theta,z)=g(\phi,z) [/tex] this is from my PDE book and they go on to solve that but here we are talking about a conductor so [tex] \hat{n}\cdot (\vec{B}\vec{B_{c}}) = 0 [/tex] [tex] \hat{n}\times (\vec{E}\vecE_{c}}) = 0 [/tex] wjhere E is the electric field on the conductor so then our boundary coniditons will turn into (if u=E) [tex] u(a,\theta,z) = 0 [/tex] [tex] u(s,\theta,z) = \vec{E_{0}}(s,\phi) \exp i(kz\omega t)[/tex] is this the right way to go? 



#9
Apr2308, 06:43 PM

P: 455

[tex]E_0=J_m(ks)\cos(m\phi)[/tex], and [tex]J_m(ka)=0[/tex] determines k.
I was wrong about the Legendre polynomials. They are for spherical coords. 



#10
Apr2308, 10:18 PM

P: 1,445

is that the electric field [itex] E_{0z}[/itex]?? 



#11
Apr2408, 10:07 AM

P: 455

It is E_0z in your equation.
If you separate the last Eq,. on your first pulse, you get the cos(m\phi) from the angjular equation. Then the radial equation is Bessel's equation. You need to look at a math physics or EM text. 


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