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Does E field and B field has a 90 degree phase difference in EM wave

  1. Feb 25, 2004 #1
    A simple question.

    I though Maxwell said that :

    1. The change of electric field generates magnetic field.
    2. The change of Magnetic field generates Electric field.

    So, simple algorithm tells me there shall be a 90 degree of phase difference between the peaks of E field and B field,because the change amount of a sinuous wave is zero at its high and amximum at its zero.

    I am seeing a chart and equation showing that there is no phase difference between E field and B field.

    Why is that?
  2. jcsd
  3. Feb 25, 2004 #2


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    When an electro Magnetic wave is traveling in free space, yes the B field is normal to and 90deg out of phase with the E field, both are normal to the direction of propagation.

    This may not be always true in a medium.
  4. Feb 25, 2004 #3
    Whaa!? Errhumm! Let's try again fellas. I know you didn't mean this, Integral. , although I'm not sure about Sammy.

    Standard physics from Maxwell's eqns.say: For EM radiation in free space (or in a waveguide) the E and B fields are in phase, both reaching their maximums at the same time.

  5. Feb 25, 2004 #4


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    Humm.. am I thinking of how the energy is transfered rather then the B and E fields? Did this off the top of my head that is always dangerous!
  6. Feb 25, 2004 #5
    You were probably thinking of an electromagnetic cavity oscillator which is somewhat like an LC circuit. In it, due to the boundary conditions, the electric field is at a maximum when the magnetic field is zero, and vice versa.

  7. Feb 26, 2004 #6
    For instance, I though @B/@t=E. Now, this formula says E(x,t)=Emax*sin(kx+wT). B=Bmax*(sin(kx+wt). But @B/@t= Bmax*w*cos(kx+wt). If you see Emax=Bmax*w, then E and B shall have 90 degree out of phase.
  8. Feb 26, 2004 #7
    OK. I was probably wrong. @Bz/@t=@Ey/@x-@Ex/@y. This looks a better chance that Ex and Ey have the same phase as Bz.
  9. Feb 26, 2004 #8


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  10. Feb 26, 2004 #9


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    Good. And don't forget your factor of c2 (propagation constant that takes care of the k's and ω's that pop out).
  11. Feb 27, 2004 #10
    EZ+, Thanks. The article further clarified some questions.
  12. Feb 28, 2004 #11
    hey... if i got the question rite, E and B are not out of phase... they only propagate in different planes.... its the plane of propagation that are perpendicular to each other, there's no phase difference... u said u noticed that e is max where b is least.... hmmm where did u see that?
  13. Mar 1, 2004 #12

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    Perhaps the confusion arises from the relationship of E-field and B-field ofelectromagnetic induction in a transformer? i.e., the E-field (and thus current) set up in the secodary coil is 90 degrees out of phase with the B-field in the iron core, etc.

    The reason this shift does not apply to light propagation is that the collapsing B-field at any point is space is not creating the E-field at the same point; rather, the collapsing B-field at one point in space is inducing the B-field(s) at subsequent points in space.
  14. Mar 1, 2004 #13
    I'm glad to see this question. I distinctly remember being told that the enrgy is a max. in the Efield, then max in the B field and the wave carries energy in a sort of bucket brigade in lower division physics. The idea being that changing B creates a changing E and vice versa.

    Then I took an engineering course and saw the derivation of the wave equation where the E & B fields of a plane wave are clearly in phase.

    I've been trying to track down the source of the idea that EM energy travels as a bucket brigade implying falsely that E &B are 90 deg out of phase. I don't see it in my undergraduate physics book, so maybe it is just a common misconception? Has anyone read such a statement in a text book?
  15. Mar 1, 2004 #14


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    No, but, now that you mention it, I too remember being taught thusly. Hmm. TGFPF.
  16. Mar 3, 2004 #15

    Your mentioning of the induced current or E feild has a 90 degree phase difference apparently shows there is some foundamental differences between EM wave propagation and induced current even though both seem to come from the same principle.

    Isn't that interesting?
  17. Mar 3, 2004 #16


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    Not exactly. An induced current is a phenomenon that can occur during propagation. There is a parameter called "intrinsic impedance" that will tell you how "in phase" the fields are. The more reactive the intrinsic impedance, the more out of phase the fields will be. If the intrinsic impedance is completely reactive, then the fields will be 90o out of phase. In other words:

    E = ηH

    where E is the electric field phasor, H is the magnetic field phasor, and η is the intrinsic impedance.

    If it seems counterintuitive that the phase difference should increase with the relative reactance, then don't worry. It should seem counterintuitive (based on the idea of circuit impedance, that is). The intrinsic impedance isn't that kind of impedance. It is basically, pretty much just a convenient way to characterize the relative magnitude and phase of a propagating EM wave. A more reactive intrinsic impedance actually corresponds to a more conductive (lossy) propagation medium.

    The intrinsic impedance contains the propagation constant of the medium (and therefore the conductance), γ:

    η = jωμ/γ

    where γ = σ + jωε

    and σ = the conductivity of the medium.

    Note that all of these parameters are composite parameters. That is why in a transformer, for instance, the intrinsic impedance (as much sense as you can make of it in a transformer) is extremely reactive, and therefore the fields are far from in phase.
  18. Mar 4, 2004 #17



    I have a problem with your document for EM field.

    The first formula introduced makes the unit of mue-zero as Newton*meter/Ampere^2. Isn't it supposed to be Newton/Ampere^2?

    Anything wrong with my interpretation of this formula?

    F = 2*(Mue-zero/4Pie )*I1*I2/r,
  19. Mar 6, 2004 #18
    I think it says force per meter of current. Soo this probably shall be:

    F/dL = 2*(Mue-zero/4Pie )*I1*I2/r,

    That will balance the units on the two sides of this equation.
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