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They say that electric and magnetic fields are perpendicular to each other in electromagnetic waves but i am not getting how will they look together?

Two waves perpendicular to each other!

Can anyone give me an example of two waves perpendicular to each other from a daily life?

All i get from pictures is two sine waves perpendicular to each other but they are confusing. Can anyone provide me a 3D view of electromagnetic wave?

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The fields are perpendicular, not the waves. Looking at a typical web search you get:

enter image description here

Let's say ${\bf \hat k} = {\bf \hat x}$ so the plane wave is:

$$ {\bf \vec E}(x, y, z, t) = E\sin{(kx-\omega t)} {\bf \hat z} $$

and similarly for ${\bf \vec B}$.

The picture only shows the field at $y=0$, $z=0$, and $t=0$. Of course, the plane wave does not depend on $y$ or $z$, so to construct a 3D "view" of the wave, just translate the picture (at fixed $x$) to anywhere, and that's what the field looks like.

Of course, translation in $t$ is just propagation.

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Bare with me, I don't have the capability to give a better picture at the moment. Antenna Field

The rod is a simple antenna. By moving charge to one end, you create a dipole (red lines.) By just having just moved the charge, you create a magnetic field (blue line) perpendicular to the rod (cross product.) these fields propagate outward. Do the in an oscillatory fashion and you get waves. This I find easier to get than most other situations like plane waves. This also being the representation of an accelerating charged particle means it's easily generalized into other set ups through superposition.

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  • $\begingroup$ Does it bother you that the maximum charge separation (electric field) is out of phase with the maximum current (magnetic field)? $\endgroup$ – JEB May 24 '18 at 16:08
  • $\begingroup$ Yes. Posting a proper image is on my to do list $\endgroup$ – Captain Morgan May 25 '18 at 0:01
  • $\begingroup$ I don't mean your drawing--I mean the explanation of dipole radiation. I have seen it many times, but it has the E and B fields out of phase--but in an EM wave, they are in phase. That has always bugged me. $\endgroup$ – JEB May 25 '18 at 1:34
  • $\begingroup$ @CaptainMorgan . do you mean that the red lines are electric (field) waves and blues lines are the magnetic(field) waves ? $\endgroup$ – Alex Jun 19 '18 at 19:15
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Don't try to visualize it as one wave traveling perpendicular to other. Just think in terms of the changing components of the magnetic and electric fields.

Say a magnetic field increases.Then there is a change in the electric field in the perpendicular direction which in turn creates a circulation of the magnetic field in perpendicular direction .So the variation of one field causes the other to vary and the wave propagates forward. Thus it is not actually two waves perpendicular to each other but the propagation of an electric field and the a perpendicular magnetic field together...it is just that we have found the variation of the magnitude of the fields to be in the form sine and cosine waves .

If you want a physical example think of perpendicular waves in two ropes held close together.The waves will travel with out interacting with each other (in fact this is not a complete picture as electromagnetic waves travel because the electric and magnetic fields interact with each other).

Em wave is just varying magnetic and electric fields perpendicular to each other and unlike mechanical waves the varying fields don't superimpose.

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There is just one four vector field $A^\mu = A^0,\vec A$. From this you can find $\vec E$ and $\vec B$ by $\vec E = -\partial_t \vec A - \nabla A^0$ and $\vec B = \vec \nabla \times \vec A$. In the absence of charges for a monochromatic wave $\vec E$ has the same polarisation as $\vec A$ and is 90 degrees out of phase with it.

By the way, $\vec E$ and $\vec B$ are strictly not vectors. $\vec E$ is odd under time inversion. $\vec B$ is invariant under space inversion so it is a pseudovecor or axial vector.

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