# Waveguide problem help

1. Homework Statement
Consider a vacuum filled infinitely long metal cylindrical waveguide of radius, a. Suppose the fields in the waveguide are as follows:
$$\vec{E}=\vec{E_{0}}\left(s,\phi\right)\exp i(kz-\omega t)$$
$$\vec{B}=\vec{B_{0}}\left(s,\phi\right)\exp i(kz-\omega t)$$

Find $E_{0z}$

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 (?)
$$\vec{E_{0}}=E_{s}\hat{s}+E_{\phi}\hat{\phi}+E_{z}\hat{z}$$

But what now? How could the Laplacian be useful? Since there is no charge
$$\nabla \cdot E = 0$$ so does that imply
$$\nabla^2 E=0$$?
So if we did that we would get
$$\frac{1}{s}\frac{\partial }{\partial s}\left(s\frac{\partial E_{s}}{\partial s}\right)+\frac{1}{s^2}\frac{\partial^2 E_{\phi}}{\partial \phi^2}+\frac{\partial^2 E_{z}}{\partial z^2} = 0$$

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?

Related Advanced Physics Homework Help News on Phys.org
This is a coax cable? Couldn't you simply project the radial (s) unit vector onto the x unit vector?

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

so instead of s i would write xcos phi?

but wouldnt that just make things messier?

can anyone else provide some input?

pam
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.

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.
it is not a coaxial cable

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

pam
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.

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

$$u(s,\theta,0)=u(s\phi,\pi)=0$$
$$u(a,\theta,z)=g(\phi,z)$$
this is from my PDE book and they go on to solve that

but here we are talking about a conductor so
$$\hat{n}\cdot (\vec{B}-\vec{B_{c}}) = 0$$
$$\hat{n}\times (\vec{E}-\vecE_{c}}) = 0$$

wjhere E is the electric field on the conductor

so then our boundary coniditons will turn into (if u=E)
$$u(a,\theta,z) = 0$$
$$u(s,\theta,z) = \vec{E_{0}}(s,\phi) \exp i(kz-\omega t)$$

is this the right way to go?

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

$$E_0=J_m(ks)\cos(m\phi)$$, and $$J_m(ka)=0$$ determines k.
I was wrong about the Legendre polynomials. They are for spherical coords.
how did u get that?

is that the electric field $E_{0z}$??

pam
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.