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EM Wave: Phase of the electric and magnetic waves?

  1. Feb 29, 2016 #1
    In a vacuum, the plane wave solutions to Maxwell's Equations are...
    ie they are in phase. (See for example
    http://hyperphysics.phy-astr.gsu.edu/hbase/waves/emwv.html )

    I don't understand how they can be in phase. I expected them to be 90 degrees out of phase because a changing electric field causes a magnetic field. So when the electric field is changing the most the magnetic field should be greatest. Since the differential of cos is sin, I'd expected the magnetic field to be 90 degrees out of phase. Where am I going wrong?
  2. jcsd
  3. Feb 29, 2016 #2


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    Staff: Mentor

    Have you tried substituting them into Maxwell's equations to verify that they are indeed a solution? Make sure to use a suitable vector form for the solution. For a plane wave propagating in the z-direction, one such form is $$\vec E = \hat x E_0 \cos (\omega t - kz) \\ \vec B = \hat y B_0 \cos (\omega t - kz)$$ More explicitly in terms of components: $$E_x = E_0 \cos (\omega t - kz) \\ E_y = 0 \\ E_z = 0 \\ B_x = 0 \\ B_y = B_0 \cos (\omega t - kz) \\ B_z = 0$$ Consider for example the equation $$\nabla \times \vec E = - \frac {\partial \vec B}{\partial t}$$ On the left side you have first derivatives with respect to x, y, z, of components of ##\vec E##, which give you (for my example) either zeroes or sines. On the right side you have the first derivatives with respect to t, of components of ##\vec B##, which again give you either zeroes or sines.
  4. Feb 29, 2016 #3


    Staff: Mentor

    Maxwells equations does not say that. It says that when the electric field is changing the most the curl of the Magnetic field should be the greatest.
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