# Why are the electric and magnetic fields in phase in EM waves?

Consider a radio wave emanating from a dipole antenna. Close to the antenna, I would think the electric and magnetic fields would be 90 degrees out of phase (i.e. if the strength of the magnetic field is due to the rate of change in electric field, the maximum magnetic field strength should be between the peaks of the electric field when current flow is ). But as the wave radiates, the electric and magnetic field are thought to be in phase. How and why does this transition come about? Is it due to the inductance/capacitance of free space?

(I still don't get this, even though similar questions have been answered in the past)

• You may want to link to those similar questions. – shai horowitz Apr 8 at 18:53
• Maybe you have to look at near field and far field. Near the scattering object fields are different respect to the far field region. – Mark_Bell Apr 8 at 20:21

Since the times of maximum charge separation (electric field) and maximum current (magnetic field) are out-of-phase in a oscillating dipole, the question makes perfect sense.

The answer is to look at the exact dipole solution:

$${\bf E}=\frac 1{4\pi\epsilon_0}\Big[ \frac{\omega^2}{c^2r}(\hat{\bf r}\times{\bf p})\times\hat{\bf r}+ \big[ \frac 1 {r^3}-\frac{i\omega}{cr^2} \big](3\hat{\bf r}[\hat{\bf r}\cdot{\bf p}]-{\bf p}) \Big]e^{\frac{i\omega r}c}e^{-i\omega t}$$

$${\bf B}=\frac {\omega^2}{4\pi\epsilon_0rc^3} (\hat{\bf r}\times{\bf p})\big( 1-\frac c {i\omega r} \big) e^{\frac{i\omega r}c}e^{-i\omega t}$$

The Poynting vector is the cross product of these terms, and the parts that go as $$1/r^2$$ are in phase, so for the electric field that is:

$${\bf E}\propto \frac 1 r[(\hat{\bf r}\times{\bf p})\times\hat{\bf r}]$$

and for the magnetic field is:

$${\bf B}\propto \frac 1 r (\hat{\bf r}\times{\bf p})$$

which should make sense for a plane wave when $$r\omega \gg c$$.

The out-phase parts that make complete sense when considering charge and current in a simple dipole antenna fall faster than $$1/r^2$$, and is thus non-radiative near-field.

Consider an oscillating electric charge (part of an oscillating current). The moving charge introduces a transverse distortion into its (preexisting) radial electric field (strongest when moving most rapidly). As part of a current it also produces a magnetic field (wrapped around the direction of motion), also strongest when the charge is moving most rapidly. These two interacting fields move away from the charge at a speed predicted by Maxwell's equations, with both being at their maximum at the same time.

• Thanks for the reply. My problem with this explanation is the following: If the magnetic field is due to a change in the strength of the E field, then the greatest rate of change in the E field is where it is zero (at this point the derivative of the E field is maximum). This is assuming the E field is a sine wave. If the maximum magnetic field occurs where the E field is zero, this would put the two fields would be 90 degrees out of phase. This is the puzzle. – DMJ Apr 10 at 2:44
• The strength of the B field near the charge depends on the speed of the charge. According to Tipler, as the wave propagates, it is the rate of change of of B with distance (not B itself) that is proportional to the rate of change of E with time. – R.W. Bird Apr 10 at 13:54
• R.W.: You might be on to something . . . In your model you talk about a moving charge. However, if we are observing an electromagnetic wave in space, I don't think there are any moving charges involved; its just a disturbance or oscillation in the E and B fields. Does your model work in this regard? Also, I'm not quite sure what you mean by the rate of change of B with distance. If B is described by a sine wave, its rate of change should be just the derivative of that wave and will be the same for any similar point in the cycle. – DMJ Apr 10 at 18:16
• At a given time, the strength of B changes from point to point. The moving charges are only at the source of the wave. – R.W. Bird Apr 10 at 19:53

The propagating wave has complex power density (Poynting vector) $$\mathbf S = \mathbf E \times \mathbf {\bar H}$$. Any phase shift between the fields $$\mathbf E$$ and $$\mathbf H$$ is reactive and does not correspond real power. Recall the analogous case of $$V\bar I$$ from circuit theory.

• I don't see how this answers the question. – my2cts Apr 8 at 23:09

I realize that the truly correct answer to this question is perhaps to simply say that if there is a wave solution to Maxwell's equations, then a necessary property of that solution is that the electric and magnetic fields will be at right angles to each other and will be in phase.

I was looking for a more intuitive type understanding. Please tell me if this makes any sense at all:

The "self-propagating" nature of an EM wave arises from the fact that a changing/collapsing electric field generates a magnetic field and vice versa. Basic experimentation in electromagnetism revealed that these fields are generated at right angles to each other. An important consideration in visualizing how these fields interact is to recognize that the fields themselves are travelling at the speed of light. Consider the changing (oscillating) electric field. It can't generate a magnetic field in front of itself: the magnetic field can't spread out in front of the EM disturbance because the EM field is already travelling at the speed of light--the maximum. The magnetic field would have to be generated at or behind the changing EM field. Let's assume its generated behind the EM field. If that were the case, then when that induced magnetic field collapsed, the EM field it generated would be in turn even further behind the original E field and the wave would slow or stop. So because the reciprocally generated fields can't be in front or behind each other, they would have to be parallel to each other (in phase) in order to generate a self-sustaining EM wave travelling at the speed of light.