If time-varying electric and magnetic fields create each other, how come they are in phase?

Question

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

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

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

Answer 1

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.

Answer 2

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.

Answer 3

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.