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I want to determine the phase delay between H and E fields in a medium with losses and not electrically charged. In this medium we also have $\sigma_c\approx \varepsilon \omega$. The enunciation of the problem gives the following hint:

$$\tan(2\phi)=\frac{2}{\text{cotan}{\phi}-\tan{\phi}}$$

The solution of this problem is:

$$\phi=\frac{1}{2}\arctan{\left(\frac{\sigma_c}{\varepsilon \omega}\right)}$$

I will show you my first steps on the resolution:


We have:

$$\vec{\bar{E}}=\bar{E_0}e^{\left(\bar{k}z-wt\right)} ,\hspace{15pt}\vec{\bar{H}}=\bar{H_0}e^{\left(\bar{k}z-wt\right)}$$

where the bar symbol is due to complex notation.

From Maxwell's equation on the reciprocal space:

$$\left[\vec{k}\times \vec{E}\right]=w\mu \vec{H}$$

So, by knowing that $\vec{E}, \vec{H}, \vec{k}$ are perpendicular to each other then we can say (after some computing):

$$\bar{H_0}=\frac{\bar{k}\bar{E_0}}{w\mu}$$

By complex notation we can define:

$$\bar{E_0}=E_0e^{i\delta_E}\hspace{2pt} \text{, so}\hspace{3pt}\bar{H_0}=H_0e^{\left(i\delta_E+\phi\right)}$$

where $\phi$ is due to $\bar{k}$ phase ($\bar{k}=ke^{i\phi}$).

The phase delay between $H$ and $E$ will then be $\phi$. And we can calculate $\phi$ by the following way:

$$\phi=\arctan\left(\frac{k_i}{k_r}\right)$$

where $k_i=\omega\sqrt{\frac{\varepsilon \mu}{2} \left(-1+\sqrt{1+\left(\frac{\sigma_c}{\varepsilon \omega} \right)^2}\right)}$ and $k_r=\omega\sqrt{\frac{\varepsilon \mu}{2} \left(1+\sqrt{1+\left(\frac{\sigma_c}{\varepsilon \omega} \right)^2}\right)}$.


My question here is where should I use the approximation $\sigma_c\approx \varepsilon \omega$? Because if I use it directly I can make $\frac{\sigma_c}{\varepsilon \omega}=1$ which will not give me the solution that I enunciated above.

I know that I need to use the hint. The only problem is what to do after the last step of my resolution.

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  • $\begingroup$ I thought that the phase difference was 90 degrees by definition. How could it vary from that? $\endgroup$ – user95006 Oct 8 '15 at 22:00
  • $\begingroup$ @nocomprende What definition? The phase difference cannot be 90 degrees for any conventional medium. It reaches 45 degrees in an extremely good conductor and is zero in an insulator. As for how, well just use Maxwell's equations in a conductor. $\endgroup$ – Rob Jeffries Oct 8 '15 at 23:07
  • $\begingroup$ @ElioPereira - You state in your question "So, by knowing that $\vec{E}$, $\vec{H}$, $\vec{k}$ are perpendicular to each other..." Does this not directly imply that the phase difference between $\vec{E}$ and $\vec{H}$ be $\pi/2$? More importantly, the equation immediately above this statement results in as much, by definition. I guess I am not sure what you mean by phase delay then... $\endgroup$ – honeste_vivere Oct 8 '15 at 23:09
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    $\begingroup$ @honeste_vivere the phase delay has nothing to do with the E- and H-fields being perpendicular. e.g $\vec{E} = E_0 \sin(kz-\omega t) \hat{i}$ and $\vec{H} = H_0 \sin(kz - \omega t - \phi) \hat{j}$ would be an example. $\endgroup$ – Rob Jeffries Oct 8 '15 at 23:16
  • $\begingroup$ @RobJeffries - Ah, I see... my mistake. Thanks for the clarification. $\endgroup$ – honeste_vivere Oct 8 '15 at 23:21
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I just did it!

We have that $$\tan(2\phi)=\frac{2}{\text{cotan}{\phi}-\tan\phi}\Leftrightarrow$$

$$\Leftrightarrow\arctan\left[\tan(2\phi)\right]=\arctan\left\{\frac{2}{\text{cotan}\left[\arctan\left(\frac{ki}{kr}\right)\right]-\tan\left[\arctan\left(\frac{ki}{kr}\right)\right]}\right\}\Leftrightarrow$$

$$\Leftrightarrow \phi=\frac{1}{2}\arctan\left(\frac{2}{\frac{kr}{ki}-\frac{ki}{kr}}\right)$$

And,

$$\frac{2}{\frac{kr}{ki}-\frac{ki}{kr}}=2\frac{\sqrt{-1+\sqrt{1+\left(\frac{\sigma_c}{\varepsilon \omega}\right)^2}}. \sqrt{1+\sqrt{1+\left(\frac{\sigma_c}{\varepsilon \omega}\right)^2}}}{1+\sqrt{1+\left(\frac{\sigma_c}{\varepsilon \omega}\right)^2}-\left(-1+\sqrt{1+\left(\frac{\sigma_c}{\varepsilon \omega}\right)^2}\right)}=2\frac{\frac{\sigma_c}{\varepsilon\omega}}{2}=\frac{\sigma_c}{\varepsilon\omega}$$

So, $$\phi=\frac{1}{2}\arctan\left(\frac{\sigma_c}{\varepsilon\omega}\right)$$

which is the solution of the problem. We didn't even need the approximation $\sigma_c \approx \varepsilon \omega$.

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