3
$\begingroup$

As I have read that "time varying electric field is source of changing magnetic field and time varying magnetic field is source of changing electric field "

Hence, I have the following doubt : If time varying nature of one field originates other changing field. Then, why are they in phase?

One is changing and producing the second thing hence second must depend upon first's derivative, I. e. If one is sine then second will be cosine Hence a phase difference of pi/2 and not in phase. Where am I wrong?

$\endgroup$
5
$\begingroup$

It is not correct to say that a changing E-field "generates" a changing B-field, and vice-versa.

Maxwell's equations imply that they co-exist. The presence of a changing E-field means that there must be an accompanying B-field.

How and why they are in phase can be seen by solving Maxwell's equations in vacuum. The solutions have the form $$\vec{E} = \vec{E_0} f(\vec{k}\cdot \vec{r} - \omega t),$$ where $\omega/k=c$ and $\vec{E_0}\cdot \vec{k}=0$.

If you take the curl of this field (the right hand side of Faraday's law), you take spatial first derivatives. The corresponding B-field is then found by integrating this with respect to time. $$\vec{B}= -\int \nabla \times \vec{E}\ dt$$

The process of differentiating with respect to spatial coordinates and then integrating with respect to time, ensures that the function $f$ and it's argument remain unchanged and therefore the E- and B-field must be in phase.

$\endgroup$
1
$\begingroup$

It is a bit more complicated than one being derivative of other. Maxwell equations relate time derivative of one field to the curl of another field, which is a difference of several space derivatives.

Thus in terms of equations you'll have the same phase for sine or cosine on both sides.

$\endgroup$
  • $\begingroup$ Well thank you FEDOR, but why is that so? I mean why single derivative cannot explain this. If we take "several times" derivative the value must be very very small. Isn't it? Also "several times" is variable term. Secondly, What can be physical reasons for this? $\endgroup$ – Pranjal Rana Sep 24 '17 at 8:24
  • $\begingroup$ Derivative of the function does not have to be smaller than the function itself. The reason for having derivatives in the first place is phenomenological and based on experiments. It would be hard to explain all of it without studying the subject, so I'd certainly recommend to learn more about the subject from the textbook. $\endgroup$ – Darkseid Sep 24 '17 at 8:35
  • $\begingroup$ Thanks FEDOR, with smaller i mean that its degree of the variable on which it is dependent, but now i think that doesn't matter. Can you please recommended me some excellent books over this as i am in high school and the common course books mostly ignore them. That would be very nice of you. $\endgroup$ – Pranjal Rana Sep 24 '17 at 10:53
  • $\begingroup$ Perhaps that very vague phrase should be "Time varying electric fields is a source of a gradient of the magnetic field." $\endgroup$ – garyp Sep 24 '17 at 13:16
  • $\begingroup$ What phrase appears to be vague here? Is it in my comment or in the original answer. Should I add explanation of it? $\endgroup$ – Darkseid Sep 24 '17 at 18:28
0
$\begingroup$

As Fedor pointed out, the Maxwell equations describe the dependence between E and B fields. However, in order to answer your question, we can simplify them. So consider the simple relationship \begin{align} \textrm{Relation I: }\quad \frac{dB}{dz} &= a \frac{dE}{dt} \\ \textrm{Relation II: }\quad \frac{dE}{dz} &= a \frac{dB}{dt} \end{align} where $a$ is a constant. Now let's start with $E_0 = \sin{(\omega t - kz)}$. We get \begin{align} \frac{dB_1}{dz} &= a \frac{dE_0}{dt} = +a\omega \cos{(\omega t - kz)} \\ \textrm{integrate both sides}\quad&\Rightarrow B_1 = -\frac{a\omega}{k} \sin{(\omega t - kz)} + const\\ \frac{dE_2}{dt} &= a \frac{dB_1}{dt} = -k\left(\frac{a\omega}{k}\right)^2 \cos{(\omega t - kz)} \\ \textrm{integrate both sides}\quad&\Rightarrow E_2 = \left(\frac{a\omega}{k}\right)^2 \sin{(\omega t - kz)} + const\\ \frac{dB_3}{dt} &= a \frac{dE_2}{dt} = k\left(\frac{a\omega}{k}\right)^3 \cos{(\omega t - kz)} \\ \textrm{integrate both sides}\quad&\Rightarrow B_3 = -\left(\frac{a\omega}{k}\right)^3 \sin{(\omega t - kz)} + const\\ \ldots \end{align} The two "waves" $E_0$ and $E_2$ have a phase difference of $0°$. Hence, they are in phase. However, this is not a problem.

$\endgroup$
  • $\begingroup$ Semoi, Phase difference like in this sketch i.stack.imgur.com/nGltr.png ? $\endgroup$ – HolgerFiedler Sep 24 '17 at 18:04
  • $\begingroup$ I don't think that this example has any relationship with Electrodynamics. The magnetic and electric fields have no phase difference in electromagnetic wave. $\endgroup$ – Darkseid Sep 24 '17 at 18:27
  • $\begingroup$ As FEDOR says, em waves are in phase sir, and your explanation makes them under a phase of 90 degrees $\endgroup$ – Pranjal Rana Sep 25 '17 at 1:46
  • $\begingroup$ You are perfectly right. I corrected that. $\endgroup$ – Semoi Sep 25 '17 at 6:10

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.