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I recently wrote this answer, in which I said:

Radio waves are electromagnetic radiation. Electromagnetic radiation contains of two components, one electrical and one magnetic. These components create each other, as said above. The red magnetic field creates a blue electric field, which creates the next magnetic field, and so on.

enter image description here

I got this diagram from wikipedia, but my physics book and Jim Hawkins WA2WHV give the same diagram.

In the comments, a discussion followed:

Olin Lathrop: Your first diagram is wrong. The B and E fields are actually 90 degrees out of phase with each other, not in phase as the diagram shows. The energy is constantly sloshing back and forth between the E and B fields.

Keelan: Are you sure? Wikipedia and my physics book show different. The two fields should have a fixed ratio, I believe, which cannot happen when out of phase. One field is horizontal and the other vertical, there's a 90 degrees angle - the diagram is an attempt to show three dimensions.

Olin Lathrop: Hmm. I always understood them to be in quadrature, but I don't have time to look that up right now. This could be a case of one bad diagram blindly copied by lots of others. Where is the energy when both fields hit 0 in your diagram? In quadrature, the sum of the squares of each field's amplitude is a constant, which gives a good explanation of how the energy can persist. It sloshes back and forth between the two fields, but its total is always the same.

I follow Olin's logic and can't say myself why the fields would be in phase. So my question is: are the E and B fields of electromagnetic radiation in phase or not? How can one understand this?

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  • \$\begingroup\$ "Where is the energy when both fields hit 0 in your diagram? ". Elsewhere. It's not like the E and B fields are momentarily 0 everywhere. \$\endgroup\$ – MSalters Apr 5 '13 at 19:01
  • \$\begingroup\$ In the figure, Magnetic Field is in XY plane whereas the electric field is in YZ plane. (Assuming Z pointing upwards) So, isn't there a phase difference of 90 degree shown? Please correct if I'm wrong. \$\endgroup\$ – Enthusiast Jul 20 '13 at 9:56
  • \$\begingroup\$ Look at physics.stackexchange.com/questions/461393/… there is a picture with 90° phase within E and B waves. In this situation we can have "resulting E and B" fields not in phase but these are due to the addition of two EM waves - both with its E and B in phase - but traveling in opposite direction \$\endgroup\$ – Alejadro Xalabarder Apr 18 at 18:41
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The complete derivation from Maxwell's equations fills entire college-level textbooks, and is too involved to get into here.

But when considering radiation from an antenna (a current flowing in a linear conductor), it boils down to the fact that there are several distinct components to both the E (electric) and H (magnetic) fields around the antenna. For the H field, there is one component that is proportional to 1/r2 and another that is proportional to 1/r. For the E field, there are three: a 1/r3 component, a 1/r2 component and a 1/r component.

The 1/r3 term is the dipole electrostatic field, which represents the energy stored in a capacitive field. Similarly, the 1/r2 term represents the energy stored in an inductive field. This represents the "self inductance" of the antenna conductor, in which the magnetic field produced by the current induces a "back EMF" on the conductor itself. Only the 1/r term represents energy that is actually carried away from the antenna.

Near the antenna, where the 1/r3 and 1/r2 components dominate, the phase relationship between E and H is complicated, and these fields do indeed store energy in the manner that Olin describes, and return energy back to the antenna itself.

However, in the "far field" (e.g., more than 10 wavelengths away from the antenna), the 1/r components of the fields dominate, creating the propogating electromagnetic plane wave, and these components are indeed in phase with each other.

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    \$\begingroup\$ I'd have liked to have seen a bit more of an explanation about 1/r^2 for both E and H fields. \$\endgroup\$ – Andy aka Apr 5 '13 at 16:38
  • \$\begingroup\$ the main question was if a plane electromagnetic wave (showed in the picture) has its components E B in phase or not. What happens in an antenna is just another topic. \$\endgroup\$ – Alejadro Xalabarder Apr 18 at 16:52
  • \$\begingroup\$ @AlejadroXalabarder: Not really. You can't launch an electromagnetic wave without an antenna of some sort. The "plane wave" is just a simplified view of what happens in the "far field" of an antenna. This is why I tried to make the connection between what Olin said and what the OP was reading. \$\endgroup\$ – Dave Tweed Apr 18 at 17:01
  • \$\begingroup\$ @Dave: we do have planes waves everywhere as you for sure know, light for example. You are talking about what is happening in radio antennas which is a particular case of EM generation. But also in this case, close to the antenna, the Maxwell equations are valid so E and B are in phase as well even if, in this particular case, we don't have plane waves. Actually the question is general, for all EM waves, only that is much easier to see the phase issue using the most common and real plane waves. \$\endgroup\$ – Alejadro Xalabarder Apr 18 at 17:52
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The impedance of free space is constant. Its value is proportional to the ratio of E and H.

It is a resistive quantity which means E and H must rise and fall in magnitude together.

Wikipedia: - enter image description here

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    \$\begingroup\$ This is the key ... I'll just add a little detail. E X B will still give a direction in the case of a time phase difference, the key point is that value will be complex (real and imaginary components) - i.e. have "storage". A purely real quantity will be resistive. \$\endgroup\$ – placeholder Apr 5 '13 at 14:17
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The confusion stems from the fact that they (the instantaneous electric and magnetic vector fields) are 90 degrees apart in space, not in time. That is to say:

\$ E \cdot B = 0\$ , and \$ E \times B \$ is the direction of propagation (the poynting vector).

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Suppose we're given an electric field propagating in the \$\hat{z}\$ direction, \$\vec{E} = \hat{x}E_0 \cos\left(\omega t - kz\right)\$. Maxwell's curl equation relating the electric and magnetic fields is given by, $$ \nabla \times \vec{E} = -\frac{\partial}{\partial t}\mu\vec{H} $$ This essentially relates the spatial derivatives of the electric field to the time derivative of the magnetic field. If we look at this equation, in order to find \$\vec{H}\$ we'll have to take the space derivative of \$\vec{E}\$ and then to find \$\vec{H}\$ we will have to integrate that same function over time, so essentially we end up with the same time-harmonic function we began with. Due to the negative sign on the time derivative of \$\vec{H}\$ we end up with the electric and magnetic fields being technically 180\$^\circ\$ out of phase, but they aren't in the same plane in space either.

Basically, diagrams like the one linked in the question can be nice for visualizing the fields in space, and if you look carefully you can see the field phasing. Looking at the equations can be just as revealing though, and if you go through the math Maxwell will give you the answer.

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To quote the Wikipedia:

The electric and magnetic parts of the field stand in a fixed ratio of strengths in order to satisfy the two Maxwell equations that specify how one is produced from the other. These E and B fields are also in phase, with both reaching maxima and minima at the same points in space (see illustrations). A common misconception is that the E and B fields in electromagnetic radiation are out of phase because a change in one produces the other, and this would produce a phase difference between them as sinusoidal functions (as indeed happens in electromagnetic induction, and in the near-field close to antennas).

Electromagnetic wave

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  • \$\begingroup\$ As a side note, the direction of these fields determines the polarisation of the signal. If you interchanged the axes of the E and B fields, some types of antenna would fail to pick up the signal until you rotated the antenna by 90 degrees. (Or some types of sunglasses would fail to transmit the signal) \$\endgroup\$ – Brian Drummond Apr 5 '13 at 14:32
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Voltage is not dependent on magnetic field but on its rate of change. Therefore induced voltage is highest when magnetic field is zero, when its derivative is highest.

For constant energy in an EM wave, we need the magnetic component and the electric component of voltage to be 90 degrees out of phase : thus we need the effect of the magnetic field to be greatest when the electric field is 0; this happens when the fields themselves are in phase.

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  • \$\begingroup\$ Aren't you confusing the induced voltage in an antenna with the E field of the electromagnetic wave? In vacuum, B=k̄/c☓E (with k̄ the wave direction) \$\endgroup\$ – MSalters Apr 5 '13 at 18:59
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Yes they are in phase or -180° phase as pointed by "Captainj2001" when using the Maxwell equation to demonstrate it.

I actually learned it with the wrong 90° phase between \$\vec{E}\$ and \$\vec{B}\$ (or \$\vec{H}\$), but now I am convinced after following the reasoning of Maxwell equation.

Another way of seeing the need of \$\vec{E}\$ and \$\vec{B}\$ for being in phase is that is the only way to keep \$\vec{E} \times \vec{B}\$ (Pointing vector) in the same direction. And this has to happen since the Pointing vector is always aligned with the propagation of the wave.

So for instance if the wave has only \$Ex\$, \$By\$ components and the propagation is in positive \$\vec{z}\$ direction, then \$\vec{E} \times \vec{B} = \hat{z}Ex By\$ can only have this direction if \$Ex\$ and \$By\$ are either both positives or both negatives. And the last conditions may only happen if \$\vec{E}\$ and \$\vec{B}\$ are in phase as can be easily checked in the initial picture.

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