GR9277 #34


Problem


\prob{34}
A conudcting cavity is driven as an electromagnetic resonator. If perfect conductivity is assumed, the transverse and normal field components must obey which of the following conditions at the inner cavity walls?
 ,
 ,
 ,
 ,
 None of the above

Electromagnetism}TEM Waves
The full formalism of a conducting cavity can be solved via TEM (transverse electromagnetic) wave guides. However, to solve this problem, one needs only the two boundary conditions from the reflection at a conducting surface, and .
The electric field parallel to the cavity is the transverse field, and thus one has choice (D), exactly the conditions above.


Alternate Solutions 
Daw6 20091106 14:39:07  I just thought about this question in terms of a plain old electromagnetic field in a cavity. If we just imagine some electromagnetic field source inside of an arbitrary enclosure, then only choice (D) makes sense;
E is a curlless field (i.e. it only diverges), so for the Efield lines to be parallel with the inner walls of an enclosure is just silly.
B is a divergenceless field, (i.e. it only curls), so for the Bfield lines to be perpendicular with the inner walls of the enclosure is equally as silly.
So only choice (D) makes sense. When you get down to it, it's really a very basic E+M question.  

Comments 
archard 20100530 18:03:54  Is it valid to solve this by thinking that the charges at the surface of the conductor must be stationary in order for the delicate arrangement of the charges to be such that the field inside remains 0? If that's the case then it's pretty obvious that the E field can't have a transverse component, because that would push charges along the surface. Likewise a normal B component would push the charges along the surface.   Daw6 20091106 14:39:07  I just thought about this question in terms of a plain old electromagnetic field in a cavity. If we just imagine some electromagnetic field source inside of an arbitrary enclosure, then only choice (D) makes sense;
E is a curlless field (i.e. it only diverges), so for the Efield lines to be parallel with the inner walls of an enclosure is just silly.
B is a divergenceless field, (i.e. it only curls), so for the Bfield lines to be perpendicular with the inner walls of the enclosure is equally as silly.
So only choice (D) makes sense. When you get down to it, it's really a very basic E+M question.
heypete 20101105 16:12:03 
That's a rather simple way of looking at it. I like it. Thanks.

  chrisfizzix 20081006 12:16:10  Conducting cavity = waveguide. You can find the first condition without knowing anything about waveguides by remembering that in a perfect conductor there can't be an electric field parallel to the surface.   sirius 20080624 14:27:26  i have never heard of conducting cavities, what topic is this from? is there a text i should look at for this?
Poop Loops 20081005 15:22:16 
Griffiths E&M covers it.

GREmania 20081012 07:30:31 
Looks like most of the problems are related to Griffiths Stuff.

  kevglynn 20060925 12:28:15  Could someone explain what a conducting cavity is (in general terms, an example of one, etc.)? Also, when it comes to the boundary conditions, what are the others?
Thanks
kevglynn 20061015 19:16:01 
nevermind, must have been groggy that morning. feel free to edit this out ;)
p.s. Yosun is the man

 

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