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Radiation Pressure

  1. Feb 7, 2012 #1
    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?
    Last edited: Feb 7, 2012
  2. jcsd
  3. Feb 7, 2012 #2


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    Staff: Mentor

    When you want an equation to appear "in line", use "itex" in the opening and closing tags, not "tex".
  4. Feb 7, 2012 #3
    Thank you.
    And what about the subject ?
  5. Feb 8, 2012 #4
    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 ?
  6. Feb 9, 2012 #5

    Meir Achuz

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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.
  7. Feb 11, 2012 #6
    Yep I figured this out.
    Thank you for you answer.
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