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When an EM wave travels inside a conductor , we find that there is a phase difference between the Electric and magnetic fields within the conductor. The magnetic field lags behind the E field and this lag is roughly equal to 45 degrees within a good conductor (all from Griffiths chp 9.4). What is the cause of this phase difference though? If an EM wave is incident on a lossless dielectric and then travels within that dielectric, the transmitted electric and magnetic waves are in phase with each other as they travel through/inside the dielectric medium. But if an EM wave is incident on a conductor, the transmitted wave travelling within that conductor ends up having a phase difference between its electric and magnetic fields.

I know that in a driven LR circuit there is a phase difference ($\phi_o $) between the driving voltage and the current through the inductor. Since the magnetic field produced by an inductor is proportional to the current, we know that there is an equal phase ($\phi_o$) difference between the driving voltage and the magnetic field produced by the circuit. Is this phase difference in any way related to the phase difference we get when an EM wave travels in a conductor?

I can account for the phase difference in a driven LR circuit easily: The non-zero inductance of the inductor creates a back emf which "fights" against a change in current which ultimately leads to the current lagging behind the voltage. For large inductances and/or high frequencies and/or low resistance (ie good conductance), the phase lag approaches 90 degrees though. Not the 45 degrees we get for a good conductor when a EM wave travels within it. I can't seem to account for these seemingly conflicting behaviours. Is the phase delay between the electric and magnetic components of an EM wave in a conductor even caused by a back emf at all or does it arise because of a different reason entirely?

Any help on this issue would be most appreciated as it has been driving me mad recently.

SalahTheGoat
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1 Answers1

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The inductance is essentially zero, so not that same mechanism.

But since it’s not an inductor, that removes both aspects of your statement - i.e. does not just eliminate the 90-deg phase difference, but also the fact that magnetic field is proportional to current.

So what causes the field? We can go to Maxwell’s equations for electric and magnetic fields, $E$ and $H$. First note that for a good conductor that follows Ohm’s law: $$\vec{I}=\sigma \vec{E}\text{ , }\sigma :=1/r\text{ , } \sigma>>\epsilon_0 \omega$$

Assume axes with the conductor along $z$, and assume $E$ is polarized along $x$. For $\frac{\partial E_x}{\partial t}, ~ \frac{\partial H_y}{\partial t} $, Maxwell’s equations are: $$\epsilon_0 \frac{\partial E_x}{\partial t} = \frac{\partial H_z}{\partial y} - \frac{\partial H_y}{\partial z} -I_x $$ $$ \epsilon_0 \frac{\partial E_x}{\partial t} + \sigma E_x = \frac{\partial H_z}{\partial y} - \frac{\partial H_y}{\partial z}$$ $$\implies \epsilon_0 \frac{\partial E_x}{\partial t} + \sigma E_x = - \frac{\partial H_y}{\partial z}$$ And: $$ \mu_0 \frac{\partial H_y}{\partial t} = \frac{\partial E_z}{\partial x}- \frac{\partial E_x}{\partial z}$$ $$\implies \mu_0 \frac{\partial H_y}{\partial t}= - \frac{\partial E_x}{\partial z}$$

With $\sigma >>\epsilon_0 \omega $, this is governed by: $$E_x = -\frac{1}{\sigma} \frac{\partial H_y}{\partial z}$$

$$ \frac{\partial H_y}{\partial t} = -\frac{1}{\mu_0} \frac{\partial E_x}{\partial z} $$

Which is solved by, $k:= -\tfrac{\sqrt{2}}{2}\mu_0 \omega$ $$ E_x= E_0 e^k\text{ } cos(\omega t+k)$$ $$ H_y= E_0 e^k\sqrt{\frac{\sigma }{\mu_0 \omega}} \text{ } cos(\omega t+ k -\frac{\pi}{4})$$

Re-summarized and strengthening of intuition:

Remember that because of high conductance ($\sigma=\frac{1}{r}$ so $\sigma / \omega >>\epsilon_0$), we ignored the effect of $\frac{\partial E_x}{\partial t}$.

To make clearer, rewrite the system (listed above, just above “Which is solved”). To get rid of extraneous characters for a second and focus, use: $E’:= -\frac{1}{\mu_0} E_x ~,~ H’:= H_z ~,~ k:= \mu_0 \sigma$

$$\frac{\partial H’}{\partial z}= k E’ $$ $$ \frac{\partial H’}{\partial t} = \frac{\partial E’}{\partial z} $$

Al Brown
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