Radiation Pressure


by megaflop
Tags: pressure, radiation
megaflop
megaflop is offline
#1
Feb7-12, 03:08 PM
P: 10
Hi everyone ! This is my first post!

How can we prove that during the reflection of an electromagnetic wave on the surface of a perfect conductor, the magnetic field [itex] \vec{b} [/itex] acting on a surface element [itex] ds [/itex] is worth half the total magnetic field [itex] \vec{B} [/itex] using Ampere's Law. That is [itex] \vec{b}=\frac{1}{2}\vec{B} [/itex].
This in order to justify the one half factor in the expression of the force acting on [itex] ds [/itex] which is [itex] d\vec{f}=\frac{1}{2}\vec{j_{s}}\times \vec{B} \cdot ds[/itex], where [itex]\vec{j_{s}}[/itex] is the surface current density on the conductor.
A drawing would be welcome.

Thanks in advance for your answers


PS: Why does it automatically go to a new line when i insert a Latex equation?
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jtbell
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#2
Feb7-12, 03:34 PM
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When you want an equation to appear "in line", use "itex" in the opening and closing tags, not "tex".
megaflop
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#3
Feb7-12, 03:51 PM
P: 10
Quote Quote by jtbell View Post
When you want an equation to appear "in line", use "itex" in the opening and closing tags, not "tex".
Thank you.
And what about the subject ?

megaflop
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#4
Feb8-12, 11:24 AM
P: 10

Radiation Pressure


There's maybe another way to justify the [itex]\frac{1}{2}[/itex] factor without using Ampère's law ? It's how it was justified in my textbook but I couldn't understand it.
Anyone ?
Meir Achuz
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#5
Feb9-12, 10:42 AM
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Consider a sheet of surface current K: KKKKKKKKKKKKKKKKKKKKKKKKKK
with the EM wave incident from above. The B field comes from two sources
1) the wave B_W, and 2)the current, B_K. Below the surface (in the metal) B is zero.
That mean the two components of B cancel so B_W=B_K in magnitude and opposite in sign.
Above the surface, B_K changes sign, but B_W doesn't, so the total B field outside is twice the field B_W which acts on the surface current.
megaflop
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#6
Feb11-12, 11:56 AM
P: 10
Quote Quote by Meir Achuz View Post
Consider a sheet of surface current K: KKKKKKKKKKKKKKKKKKKKKKKKKK
with the EM wave incident from above. The B field comes from two sources
1) the wave B_W, and 2)the current, B_K. Below the surface (in the metal) B is zero.
That mean the two components of B cancel so B_W=B_K in magnitude and opposite in sign.
Above the surface, B_K changes sign, but B_W doesn't, so the total B field outside is twice the field B_W which acts on the surface current.
Yep I figured this out.
Thank you for you answer.


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