EM wave right above and right below the conducting surface.

[yungman]

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 $z=0^-$(in air):

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

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





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

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

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





The problem is the book still give:

$$\tilde B_+(z)=\hat y \frac {\tilde E_+(z)}{\eta_c}$$

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

Thanks

 

[yungman]

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 $\eta_c$ of metal is complex. In free space that is lossless, $\eta_0$ is 377 ohm and is real, so E and B are in phase.

Can anyone verify this