GR9277 #36
|
|
Problem
|
|
\prob{36}
A plane-polarized electromagnetic wave is incident normally on a flat, perfectly conducting surface. Upon reflection at the surface, which of the following is true?
- Both the electric vector and magnetic vector are reversed.
- Neither the electric vector nor the magnetic vector is reversed.
- The electric vector is reversed; the magnetic vector is not.
- The magnetic vector is reversed; the electric vector is not.
- The directions of the electric and magnetic vectors are interchanged.
|
Electromagnetism}Boundary Conditions
The conductor perfectly reflects the incoming wave, and none is transmitted. The electric field is thus reversed. However, since E and B are perpendicular (related to each other by the Poyting Vector where the direction of propagation is given by the direction of ), the magnitude of B is increased by 2, but its direction stays the same.
Search on the GRE Physics Solutions homepage with keyword conductors for more on this.
|
|
Alternate Solutions |
Herminso 2009-09-13 17:12:50 | By BC's the parallel components of the electric field at the conducting boundary need to be continuous. Since the wave is incident normally on a flat, we know that the electric and magnetic field of the electromagnetic wave are parallel to the surface (Remember ). So the BC for the electric field is:
or
where because the field inside to a conductor must be zero. Thus , the electric field is reversed.
Using now the relation for the reflected wave, we can see the reflected magnetic field is not reversed.
| |
|
Comments |
asdfasdfasdf 2016-08-09 16:43:11 | It\'s not too difficult for me to see that the electric field is reversed, but knowing that only rules out B and D, leaving 3 possibilities.\r\n\r\nCould A and E be ruled out together? Perhaps I\'m misunderstanding what E is saying, but it sounds like the same thing as A, and thus, they can\'t both be correct. If so, then C is left as the only option. | | Herminso 2009-09-13 17:12:50 | By BC's the parallel components of the electric field at the conducting boundary need to be continuous. Since the wave is incident normally on a flat, we know that the electric and magnetic field of the electromagnetic wave are parallel to the surface (Remember ). So the BC for the electric field is:
or
where because the field inside to a conductor must be zero. Thus , the electric field is reversed.
Using now the relation for the reflected wave, we can see the reflected magnetic field is not reversed.
Allenji 2012-09-29 18:30:48 |
concise and intuitive
|
Jovensky 2013-03-24 21:25:25 |
Excellent solution
|
gear3 2013-10-09 06:07:49 |
very good!
|
| | flux 2008-11-07 18:16:24 | Because the surface is a conductor, the E field must go to zero. This is similar to a rope with a fixed end. The E field will thus flip over and head the other direction. This leaves us with choices A and C. Using the right-hand-rule brings us to the conclusion that the B field holds its direction. Choice C it is!
Almno10 2010-11-12 14:40:25 |
Likewise, the B is not necessarily zero at the surface, analogous to a rope that is ties but not fixed to a pole, so it does not change phase.
|
| | Lukav 2008-10-03 12:42:32 | Kinda Silly, but you can just use the right hand rule to do this quickly.
Poop Loops 2008-10-05 15:38:38 |
Yes, but how do you know whether it's the E or B field that reversed?
|
| | Furious 2007-10-29 15:41:36 | This probably is stupid, but why is it that the transverse portion of the E field must be reversed, is this the same reason why two problems prior to this the transverse E field had to be zero? I'm not quite sure which physical concept is responsible for this.
nick1234 2007-10-29 17:41:14 |
There can't be an electric field along the surface of a conductor, otherwise the free electrons would move along the surface until the E-field was canceled. So, any E-field that exists must be perpendicular to the surface. (Griffiths p.98)
|
Poop Loops 2008-10-05 15:37:29 |
In short, yes, exactly that reason. E|| has to go to zero at the edges, whereas B|| can stay the same, so it makes sense that it would be E to flip directions.
|
his dudeness 2010-09-04 13:02:20 |
Basically, the total transverse E-field must be zero, like Furious said (see two problems ago). How do we get that to happen? Well, given that the incoming wave has some transverse E-field E_t, the reflected wave must therefore have transverse E-field -E_t. That way, E_t + (-E_t) = 0 and the boundary condition is met.
|
| |
|
Post A Comment! |
|
Bare Basic LaTeX Rosetta Stone
|
LaTeX syntax supported through dollar sign wrappers $, ex., $\alpha^2_0$ produces .
|
type this... |
to get... |
$\int_0^\infty$ |
|
$\partial$ |
|
$\Rightarrow$ |
|
$\ddot{x},\dot{x}$ |
|
$\sqrt{z}$ |
|
$\langle my \rangle$ |
|
$\left( abacadabra \right)_{me}$ |
|
$\vec{E}$ |
|
$\frac{a}{b}$ |
|
|
|
|
|
The Sidebar Chatbox...
Scroll to see it, or resize your browser to ignore it... |
|
|