I was reading EM Theory from a book when I came across the topic of Wave Impedance. In the book, it says that the wave impedance of a medium is given by the ratio of the magnitudes of E and H of a wave in that medium.

$Z=\frac{|\vec E|}{|\vec H|}$

But, for a conductor we have

$\frac{|\vec H|}{|\vec E|}=\sqrt{\frac{\sigma}{\omega \mu}}$

So, for a good conductor we should have

$|\vec H|>>|\vec E|$

This implies that a conductor should have very low wave impedance.

But, an EM wave is attenuated in a conductor due to the imaginary part of the wave vector. How can both of these be correct?

  • $\begingroup$ What makes you think that a low (real part of) characteristic impedance is not compatible with a high attenuation? $\endgroup$
    – The Photon
    Aug 27, 2020 at 18:44
  • $\begingroup$ @ThePhoton Doesn't a high impedance cause a higher attenuation? $\endgroup$
    – SaptarshiS
    Aug 27, 2020 at 19:33
  • $\begingroup$ @SaptarshiSarkar Absolutely not. $\endgroup$
    – DanielSank
    Aug 27, 2020 at 21:39

1 Answer 1


You are mixing up the usual usage of the term impedance (the ratio of voltage to current in a circuit element) with characteristic impedance (the ratio of voltage to current signals in a transmission line, or of E to H fields in an EM wave in space).

When talking about characteristic impedance, any value that is purely real (has no imaginary component) indicates lossless propagation. Only if there is an imaginary component to the characteristic impedance will the wave be attenuated as it propagates.


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