Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

EM wave right above and right below the conducting surface.

  1. Jun 30, 2011 #1
    Assume a infinite depth good conductor block with width in y and length in x direction. Boundary is at z=0 and air is at z=-ve and conductor at z=+ve. Let a voltage apply across the length of the block in x direction so a current density established in +ve x direction. We want to look at the EM at the boundary just above and just below the surface.

    1) We know at [itex] z=0^-[/itex](in air):

    [tex]\tilde{E_-}(z) = \hat x E_0 e^{-\alpha z} e^{-j\beta z} \;\hbox { and }\; \tilde B = \frac 1 {\eta_0} \hat z \times \vec E_-(z)\;=\; \hat y \frac {\vec E_-(z)}{\eta_0}[/tex]

    In air, H is perpendicular to E only because [itex]\eta_0[/itex] is real and don't have a phase shift.

    2) We know tangential E continuous cross boundary therefore at [itex] z=0^+[/itex](in conductor):

    [tex]\tilde{E_+}(z) = \hat x E_0 e^{-\alpha z} e^{-j\beta z} \;\hbox { and }\; \tilde B_+ = \frac 1 {\eta_c} \hat z \times \vec E_-(z) [/tex]

    Here [itex]\eta_c[/itex] is complex and therefore has a phase shift, actually the phase angle is [itex]\theta=45^0[/itex]. This mean B is no longer perpendicular to E even though B also propagate in z direction.

    The problem is the book still give:

    [tex] \tilde B_+(z)=\hat y \frac {\tilde E_+(z)}{\eta_c}[/tex]

    This mean B is still perpendicular to E inside the good conductor. I don't understand this. Please help me.

    Last edited: Jun 30, 2011
  2. jcsd
  3. Jun 30, 2011 #2
    I think I understand it now. I just want to verify this:

    Even in good conductor, the EM wave is still TEM where E and B are perpendicular to each other and both perpendicular to the direction of propagation. The difference compare to the TEM in lossless free space is the E and B are out of phase in good conductor because [itex]\eta_c[/itex] of metal is complex. In free space that is lossless, [itex]\eta_0[/itex] is 377 ohm and is real, so E and B are in phase.

    Can anyone verify this
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook