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Electromagnetic Wave

  1. Sep 14, 2010 #1
    Hi Folks,

    I understand that a changing magnetic field induces an electric field and a changing electric field induces a magnetic field. I also understand that the greater the time rate of change of one, the greater is the other.

    Now in free space, the electric and magnetic field of a wave are in phase. However, as I see it, at the peak of the sign wave of the electric field, the electric field time rate of change is zero. So it seems like this should coincide with a zero magnetic field but since they are in phase, the magnetic field is also at a peak.

    It seems like they should have a phase difference of plus or minus pi/2.

    What am I missing here?

    Thank you,
  2. jcsd
  3. Sep 14, 2010 #2
    Where did you read that?
    Last edited: Sep 14, 2010
  4. Sep 14, 2010 #3
    I've seen it proven in a book on Electricity and Magnetism and I've come across it many times. An illustration shows the plane of the magnetic wave at a ninety degree angle to that of the electric wave but they are in phase meaning one reaches a maximum when the other reaches a maximum and one reaches a minimum when the other reaches a minimum.
  5. Sep 14, 2010 #4
    90 degrees out-of-phase is more appropriate.
  6. Sep 14, 2010 #5
    I've seen it drawn that way, too.
  7. Sep 14, 2010 #6
    It's not simply a drawing error. I just checked three separate E&M books and they are all drawn the same way. Additionally, the mathematics takes you there using Maxwell's equations!
  8. Sep 14, 2010 #7


    User Avatar

    Staff: Mentor

    The E and B field vectors are oriented at 90 degrees with respect to each other in an electromagnetic wave; however, we do not describe this by saying that they are "90 degrees out-of-phase."

    "90 degrees out of phase" means that the waves are described mathematically by something like

    [tex]\vec E = \vec E_0 \sin (kx - \omega t)[/tex]

    [tex]\vec B = \vec B_0 \sin (kx - \omega t + 90^{\circ})[/tex]

    or better,

    [tex]\vec B = \vec B_0 \sin (kx - \omega t + \pi/2)[/tex]

    This is not true for electromagnetic waves in a vacuum.
    Last edited: Sep 14, 2010
  9. Sep 14, 2010 #8


    Staff: Mentor

    Wannabeagenius is correct. They are in-phase, not 90 degrees out of phase.

    If you look at Maxwell's laws in vacuum you will find that it is not quite corect that "a changing magnetic field induces an electric field". It is more correct to say "a changing magnetic field induces curl of an electric field" or in other words "a changing magnetic field (in time) induces a spatially changing electric field". When you express it correctly you immediately see that the electric and magnetic fields should be in phase.
  10. Sep 14, 2010 #9
    I get it. Thank you.

  11. Sep 14, 2010 #10
    So, there is NO angular difference in the propagated electric/magnetic fields?
  12. Sep 14, 2010 #11
    No difference in phase in a vacuum.

    Perhaps someone could address whether or not this is true in general. I'm thinking of material media and waveguides.

  13. Sep 14, 2010 #12
    If you can see maxwell's equations, you will notice that a rate of change in the magnetic field creates a gradient in the electric field in the perpendicular direction and vice versa.
  14. Sep 14, 2010 #13
    Ch4_4-3.gif [PLAIN]http://elementaryteacher.files.wordpress.com/2008/08/maxwells-equations.gif [Broken]
    Last edited by a moderator: May 4, 2017
  15. Sep 14, 2010 #14


    Staff: Mentor

    There is no PHASE difference. I.e. When the E field is at its peak the B field is also at its peak.

    Do not confuse this with the DIRECTION of the fields. If the E field points along the x axis then the B field will point in the y-z plane (90 degrees). That is not at all the same as the phase relationship.
  16. Sep 14, 2010 #15
    OK, thanks...
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