# Electromagnetic waves question

by Misr
Tags: electromagnetic, waves
 P: 391 Electromagnetic waves consist of electric and magnetic fields having the same phase how?
 HW Helper P: 3,443 For an electromagnetic wave in vacuum, for a polarised plane wave, we have: $$\vec{K} \wedge \vec{E} = \omega \vec{B}$$ (where $\vec{K}$ is the wavenumber, i.e. $\frac{2\pi}{\lambda}$ times by the unit vector of the wave's propagation. And $\omega$ is the angular frequency of the wave). Therefore, the magnetic and electric fields have different direction and magnitude, but they have the same phase. The reason you get this equation is by solving the electric and magnetic fields in a vacuum, for a polarised plane wave. It is probably also true for a non-polarised plane wave, and for a spherical wave, but I'm not sure about them.
 HW Helper Thanks PF Gold P: 4,759 In vacuo, they are in phase, a direct result of solving the Maxwell equations. However: in a conducting medium they are not in phase. Also a result of solving the Maxwell equations.
 P: 391 Electromagnetic waves question Oh that's complicated I don't know anything about maxwell's equations isn't there some simpler ways to explain this?
 HW Helper P: 3,443 I think the best I can do is explain the meaning of Maxwell's equations, since that is where it all comes from. In loose terms, a changing magnetic field causes an electric field and a changing magnetic field causes an electric field. Also, charge causes electric field and moving charges cause magnetic field. In a vacuum, there are no charges, so the electric and magnetic fields depend only on each other. Then when we specify that the wave must be a polarised plane wave, there are even less possibilities for the form of the electromagnetic wave. In fact, the magnetic and electric fields must be perpendicular to each other. And they must both be perpendicular to the direction of wave propagation. And they must be in phase. The reason for these things is because of the way that the changing magnetic and electric fields affect each other. And the precise way they do this is given by Maxwell's equations.
 HW Helper Thanks PF Gold P: 4,759 Misr - afraid not. If you're not willing or able to dig into the Maxwell equations you shouldn't be too worried about the phase relationship between E and B fields either. Is the question really so important to you? And if it is - which is great - why don't you want to go into the subject more deeply? Although I must say this - without calculus including vector calculus you won't get very far. There is however a publication called 'The Radio Amateur's Handbook' published by ARRL, also "The AARL Antenna Book'. Both those pubs address e-m waves in a non-calculus manner, and they might answer your question adequately or they may not.
P: 391
 Misr - afraid not. If you're not willing or able to dig into the Maxwell equations you shouldn't be too worried about the phase relationship between E and B fields either. Is the question really so important to you? And if it is - which is great - why don't you want to go into the subject more deeply? Although I must say this - without calculus including vector calculus you won't get very far. There is however a publication called 'The Radio Amateur's Handbook' published by ARRL, also "The AARL Antenna Book'. Both those pubs address e-m waves in a non-calculus manner, and they might answer your question adequately or they may not.
Well,I'm interested in digging into Maxwell's equations but I'm still at school and I have no time for something like this
I read the statement above in my text book"Electromagnetic waves consist of electric and magnetic fields having the same phase"
and I couldn't understand it but I didn't know that the subject is complicated
Thanks very much
 HW Helper P: 3,443 The energy density of the electromagnetic field is equal to $\frac{B^2}{2\mu_0}+\frac{\varepsilon_0 E^2}{2}$ and for an electromagnetic plane wave in vacuum, we have: $\vec{K}\wedge \vec{E}=\omega \vec{B}$. If we substitute the second equation into the first, we find that the contributions to the energy density from the electric and magnetic fields are exactly equal. So the potential energy stored in the electromagnetic field is half due to the magnetic field and half due to the electric field (in the case of zero charges).