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In a conductor, the relation between the phase of the magnetic field $\delta_B$ and that of the electric field $\delta_E$ is given by $$\delta_B-\delta_E=\tan^{-1}(\frac{\beta}{\alpha})\tag{1}$$ where $\alpha$ and $\beta$ are the real and imaginary parts of the complex wavevector $k=\alpha+i\beta$.

EDIT: From equation (1), as I understand, the phase of magnetic field is ahead of the electric field only when $\tan^{-1}(\frac{\beta}{\alpha})$ is positive. However, in Griffith's electrodynamics book, he uses this equation to claim the opposite i.e., the magnetic field lags the electric field (without saying anything about the sign of $\tan^{-1}(\frac{\beta}{\alpha})$).

Can someone explain this assertion? In particular, what is wrong in my reasoning?

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I'm not sure about that particular equation. There's always issues with how quantities are defined, and signs of particular values. To understand what that equation is saying exactly, you have to look at the derivation. So to solve this problem, I'm going back to some basic electromagnetic relations for a plane wave.

$$E = c B$$

$$c = \frac{1}{\sqrt{\epsilon \mu}}$$

We use the generalized $\epsilon$ that includes conductivity. If we use the time dependence of $e^{i\omega t}$, then $\epsilon = \epsilon'-i\sigma/\omega$

$$\frac{B}{E} = \sqrt{\epsilon \mu}$$

$$phase \left(\frac{B}{E}\right) = phase \left(\sqrt{\epsilon'-i\sigma/\omega}\right) $$

For good conductors at low frequencies, $\sigma/\omega >> \epsilon'$, and the phase difference between the magnetic and electric fields is very close to -45 degrees. This implies the magnetic field phase lags the electric field, just like Griffiths says.

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  • $\begingroup$ How does it follow from Eq.(1) in the question? @David $\endgroup$
    – SRS
    Commented Nov 4, 2018 at 12:53
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Too long for a comment so have to provide as an answer.

The relevant em equations to derive the wave equations in a conductor are: $\nabla . {\bf E} = 0, \nabla . {\bf B} = 0$ and $\nabla \times {\bf E} = -\frac{\partial {\bf B}}{\partial t}, \nabla \times {\bf B} = \mu \epsilon \frac{\partial {\bf E}}{\partial t} + \mu {\bf J}_f$. The first two equations are symmetrical between ${\bf E}$ and ${\bf B}$ but there is an important difference between the second set of equations (besides the units and the $-ve$ sign that is needed to get waves), namely the presence of the ${\bf J}_f = \sigma {\bf E}$ term. This term is the origin for the (slightly) different behaviours of the ${\bf E}, {\bf B}$ fields, namely the phase difference. It might be argued that this extra contribution to the ${\bf B}$ means that you first need the electric field to be present that causes the change in the magnetic field? But I'm not sure if this argument would suggest that the magnetic field lags the electric field?

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Think of the electric field in a capacitor which lags the magnetic field. The capacitor uses a dielectric to increase capacitance. The same thing happens with the electric field in light - it lags the magnetic field as it passes through the dielectric.

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Actually, in Griffith's electrodynamics book (I think) he wanted to say the real part of the magnetic field lags behind the real part of the electric field...
You can understand it easily by the figure.
Let $z=t=\delta_E=0$ then the total phase of $E$ will be $0$, and we know $\cos 0 =1$ but for $B$, its phase will be $\tan^{−1}\frac{\beta}{\alpha}$ and $\cos (0 + \tan^{−1}\frac{\beta}{\alpha})$ is always less than $1$ (as $0\le\cos\theta\le1$ and $\tan^{−1}\frac{\beta}{\alpha}$ is a +ve quantity) thus $B$ lags behind the $E$
but for imaginary parts the there cames $\sin\theta$
and we all know $1\le\sin\theta\le 0$ so in that case $E$ lags behind the $B$.

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$$ \tilde{H} = \frac{1}{\eta_c} \hat k \times \tilde{E} \tag{1}$$

$$\eta_c = \sqrt{\frac{\mu}{\epsilon}}\Big(1-j\frac{\sigma}{\omega \epsilon}\Big)^{-\frac{1}{2} } \tag{2}$$

Rewrite (2), $$\eta_c =\sqrt{\frac{(\frac{\mu}{\epsilon})}{1-j\frac{\sigma}{\omega \epsilon}}} \tag{3}$$

From (3) we see that any conductivity will cause the denominator to have complex $\angle \theta$ S.T. $\theta <0$.

Therefore, $\eta_c$ will have $\angle \theta_2$ S.T. $\theta_2 >0$

Finally, (1) tell us that $\eta_c$ will make $\delta_H < 0$.

This gives a general form of the Magnetic field, $H = \hat y H_{y0}\cos(\omega t - \gamma z + \delta_H)$.

Because $\delta_E = 0$, the magnetic field will lag the electric field.

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Actually leading or lagging is always considered w.r.t time . In Griffith book, it is taken as cos(kz-wt+phi) here you are saying B will lead, because your analysis is w.r.t space (i.e. here z), as here it is (kz-wt). To know how it behaves in time only, we have to take a fixed co-ordinate and analyze it, to know which is leading. Let's take z=0,

Now it will be cos(-wt+phi)= cos (wt-phi), hence it is clear B is lagging in time.

You are thinking about phi is +ve, but look (-wt) is here, not (wt), to analyse correctly always try to analyze with +wt. Also in diagram in the Griffith book it is of a picture when time t=const, but to see which is leading, we need to take z=constant and graph should be w.r.t time 't'

Hope, I cleared your doubt.

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