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Electromagnetic waves question 
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#1
Sep311, 10:48 PM

P: 391

Electromagnetic waves consist of electric and magnetic fields having the same phase
how? 


#2
Sep411, 07:40 AM

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P: 3,443

For an electromagnetic wave in vacuum, for a polarised plane wave, we have:
[tex] \vec{K} \wedge \vec{E} = \omega \vec{B} [/tex] (where [itex]\vec{K}[/itex] is the wavenumber, i.e. [itex]\frac{2\pi}{\lambda}[/itex] times by the unit vector of the wave's propagation. And [itex]\omega[/itex] 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 nonpolarised plane wave, and for a spherical wave, but I'm not sure about them. 


#3
Sep511, 05:48 AM

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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. 


#4
Sep511, 10:00 AM

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? 


#5
Sep511, 10:50 AM

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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. 


#6
Sep511, 03:52 PM

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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 em waves in a noncalculus manner, and they might answer your question adequately or they may not. 


#7
Sep511, 05:40 PM

P: 391

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 


#8
Sep611, 05:04 PM

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PF Gold
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#9
Sep711, 08:38 AM

P: 391

Yep,you are right



#10
Sep811, 08:34 AM

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Maxwell's equations are very much empirical. That is, we believe them because they agree with experiment so profoundly; they are not derived from a more fundamental source.
In other words, E and B fields in an EM wave have the same phase, because they just do, that's how we observe them to be. Claude. 


#11
Sep811, 09:57 AM

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It has to be said that em waves are odd compared with other waves. Mechanical waves have a uniform flow of energy because the kinetic energy variations (movement) are out of phase with the potential energy variations (displacement). In em waves, the E field (regarded as potential energy) is in phase with the B field (associated with kinetic energy). The energy comes in 'bursts'. A kind of anomaly.



#12
Sep811, 07:35 PM

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P: 3,443

The energy density of the electromagnetic field is equal to [itex]\frac{B^2}{2\mu_0}+\frac{\varepsilon_0 E^2}{2}[/itex] and for an electromagnetic plane wave in vacuum, we have: [itex]\vec{K}\wedge \vec{E}=\omega \vec{B}[/itex]. 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). 


#13
Sep911, 12:43 PM

P: 617

In regular traveling electromagnetic waves, the electric field causes the magnetic field and the magnetic field causes the electric field. They are in phase, because this what allows the wave to keep traveling, like waves down a slinky. There are situations in free space where the electric field is out of phase with the magnetic field, but this just does not create sustained traveling waves. It's like how jerking sporadically the end of a slinky does create slinky movement, but does not send traveling waves down it.



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