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GR0177 #61
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Problem
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This problem is still being typed. |
Electromagnetism }Boundary Conditions
For a conductor, the electric field boundary condition at the interface are  ( is the free charge density) and  .
The plane wave traveling in the x-direction is polarized (say) in the +z-direction. Thus, ) .
There is no component perpendicular to the conductor at the boundary, and thus the first boundary condition implies that the free charge density is 0.
The second boundary condition requires that  . The parallel component of E is polarized in the z-direction, and thus the requirement is  , i.e., the incident plus the reflected is equal to the transmitted wave. However, for a perfect conductor, the transmitted wave is 0. Thus, one has  . This implies that the electric field to the left of the conductor cancels.
Recall the Poynting vector,  , which conveniently points in the direction of the electromagnetic wave propagation. Since the electric field (by the convention used above) is polarized in the z-direction, the magnetic field of the incident wave points in the -y direction. However, since the electric field of the reflecting wave points in the -z-direction, its magnetic field also points in the -y direction. The magnetic field magnitude is thus  . Hence, one has choice (C).
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Alternate Solutions |
ramparts
2009-10-08 12:17:53 | Even more simply put - a conductor sets the E field to 0, and multiplies the B field by 2. That's C.
That's how I solved it, but as mentioned in the comments, knowing that E is set to 0 and B isn't (and B isn't 0 to begin with, as an EM wave can't be all E) is all you need to pick C. |  | SonOfOle
2006-10-31 21:22:48 | More simply put, from a conceptual level.
1) E fields don't propagate through conductors. So E=0.
2) The E field incident on the conductor moves the charges back and forth of the conductor, which in turn set up an oscillating B-field to the right of the conductor. So B is oscillating.
3) Answer (C) is the only one that satisfies both these conditions. | |
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Comments |
Grandpsykick
2010-03-11 15:20:18 | Even more simply put:
1) E fields don't propagate through conductors (if they are normal to the surface).
2) The energy in the field MUST be transmitted through the surface.
3) E is zero (0), thus B cannot be zero (0); thus, the only answer is C.
NOTE: you didn't need any math, or "wave dynamics", or anything but simple knowledge of EM and conservation of energy. |  | ramparts
2009-10-08 12:17:53 | Even more simply put - a conductor sets the E field to 0, and multiplies the B field by 2. That's C.
That's how I solved it, but as mentioned in the comments, knowing that E is set to 0 and B isn't (and B isn't 0 to begin with, as an EM wave can't be all E) is all you need to pick C. | | Gaffer
2007-10-23 13:54:30 | Do be careful about the question. It asks for the fields to the left of the conductor - that is the side of incidence, not transmittance.
To be accurate, you need the original approach.
Gaffer
2007-10-23 13:55:44 |
TO clairify, I was agreeing with the author, not SonOfOle. What he said was correct, but his comment indicates he was looking at the improper side.
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Jeremy
2007-10-25 11:58:59 |
I think SonOfOle's conceptual approach works if you replace his (1) with:
(1) Conductors only allow perpendicular electric fields at their boundaries, so  on both sides.
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| | SonOfOle
2006-10-31 21:22:48 | More simply put, from a conceptual level.
1) E fields don't propagate through conductors. So E=0.
2) The E field incident on the conductor moves the charges back and forth of the conductor, which in turn set up an oscillating B-field to the right of the conductor. So B is oscillating.
3) Answer (C) is the only one that satisfies both these conditions.
anmuhich
2009-03-15 12:54:45 |
Right! And you don't even need to know the B field is oscillating, just that there is a B field and no E field! The electrons in the conductor arrange themselves in such a way with the E field as to cancel its affect. However, you know a B field must be induced anytime you have moving charge.
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anmuhich
2009-03-15 13:02:06 |
Sorry it's the oscillating E field that causes the induced B field.
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