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This is a newbie question. I'm trying to understand what I think must be a very basic concept with an obvious answer, but I want to be sure.

Do plane wave electric and magnetic fields of equal strength at the same frequency and same waveform carry equal amounts of energy?

In other words, if a dipole and a loop antenna of equal length encounter plane waves in free space at the same frequency and with the same waveform, and signal source power levels are adjusted so that the electric field to which the dipole is exposed is equal in strength to the magnetic field to which the loop is exposed, and all other variables are equal, is the current induced in each antenna equal? Please assume ideal antennas that convert all of the energy they receive into current and that have no other properties that affect the results.

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  • \$\begingroup\$ antenna-theory.com/antennas/smallLoop.php \$\endgroup\$ – Sunnyskyguy EE75 Feb 20 '18 at 4:46
  • \$\begingroup\$ Thanks, Tony. The title of my question might be misleading. I'll update my question to reflect what I'm after. \$\endgroup\$ – dcorsello Feb 20 '18 at 5:08
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    \$\begingroup\$ An electromagnetic wave traveling in air or free-space has both an electric and a magnetic field. You cannot create a traveling wave without both fields. In a certain sense, both fields can be though of as having the same energy. It is not clear what you mean by antenna length. I think what you really want to ask is, what size loop antenna would deliver the same signal power to the load as a dipole antenna. \$\endgroup\$ – mkeith Feb 20 '18 at 5:10
  • \$\begingroup\$ Thanks, mkeith. Again, my choice of title was poor. I just updated both the title and the body of the question to shift the focus from the nature of antennas to the nature of electric and magnetic fields in plane waves. In a plane wave in free space, the electric component is 377 times as strong as the magnetic field (Z = E/H). My assumption, which I'm trying to validate, is that "field strength" is analogous to amount of energy. \$\endgroup\$ – dcorsello Feb 20 '18 at 5:26
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    \$\begingroup\$ The units don't mean anything. Comparing the numerical value of an electric field to the numerical value of a magnetic field is just totally arbitrary. For the same wave, the energy of the electric field at a certain point is the same as the energy of the magnetic field at that same point. Energy. Joules. Anything else is folly. \$\endgroup\$ – mkeith Feb 20 '18 at 5:46
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Do plane wave electric and magnetic fields of equal strength at the same frequency and same waveform carry equal amounts of energy?

An analogy: -

If I put 377 V RMS across a 377 ohm resistor the power would be \$V^2/R\$ = 377 watts and the current would be 1 amp. If I calculated power using current it would be \$I^2R\$ = 377 watts.

signal source power levels are adjusted so that the electric field to which the dipole is exposed is equal in strength to the magnetic field to which the loop is exposed

If both antennas have equal sized apertures (aka capture area) and are designed to be resonant at the incoming frequency then no adjustment is necessary. One will convert volts/m (E field) and the other will convert amps/m (H field) and both will produce the same power/signal output.

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  • \$\begingroup\$ Thanks. This is exactly what I wanted to know, but it's not the answer I was hoping for. I've been trying to understand what an engineer meant when he told me that the E field is 377 times stronger than the H field. Given your answer, I'm still not sure what he meant. Maybe this will help to clarify: Suppose we put a very thin, conductive barrier between the source and receiving antennas that attenuates the E field by 50 dB, but that doesn't attenuate the H field. How will this affect the power/signal output of the two receiving antennas in your answer above? \$\endgroup\$ – dcorsello Feb 20 '18 at 16:04
  • \$\begingroup\$ The so-called engineer appears to be wrong. You asked a similar question yesterday so I suggest you ask a new one based on that question. Proper diagrams would help. \$\endgroup\$ – Andy aka Feb 20 '18 at 16:51
  • \$\begingroup\$ Okay, thanks. Here's a quote from the email that sent me here looking for answers: "at far field conditions [...] the magnetic component is 377x smaller than the electric field component. But, [with a high-power signal], even the magnetic component can be significant." Maybe I misunderstood him. \$\endgroup\$ – dcorsello Feb 20 '18 at 17:22
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    \$\begingroup\$ @dcorsello I don't think the person sending you the email knows much or he's having a bad day. Both are as significant as each other! \$\endgroup\$ – Andy aka Feb 20 '18 at 17:27
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    \$\begingroup\$ @GavinR.Putland, yes, he was referring to Ohm’s Law for plane waves in free space. The problem with the way he phrased it was that it led me to believe that the E and H fields act independently, so that if you attenuate the E field, the H field continues to propagate, unaffected. I eventually got my head around the fact that the E and H fields in a plane wave are mutually dependent—they actually create one another. So a material that has a high impedance mismatch with air can attenuate both E and H fields by reflecting the E field, even if it has low permeability and provides no H shielding. \$\endgroup\$ – dcorsello Aug 11 '18 at 11:17
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An electromagnetic wave carries energy. That's the fundamental 'currency' of physics. This energy is carried in both electric and magnetic fields.

If we measure the amplitude of each field, and find their ratio is 377 ohms, this tells us as much about the specific units we have chosen to use to make the measurement as it does the fields themselves.

You can launch and intercept an EM wave with either an electric dipole, or a magnetic dipole. They both couple completely to the energy-carrying wave in the far field. They are different in the near field, where the non-travelling component of the fields is more nearly electric or magnetic respectively.

For any given antenna configuration, you can define an effective area (at a certain frequency, in a certain direction) that designates its coupling with the far field. In some types of antenna, for instance a parabolic reflector dish, what contributes to the 'area' is obvious. In other types, a Yagi array for instance, the 'area' increases as more elements are added in the direction of the wave. Different configurations will have different physical size to electrical area ratios, it's more the specific configuration than whether it's primarily electric or magnetic that controls this.

In the absence of losses, the power output from an antenna is equal to the energy it intercepts from the wave. For small antennae like a short electric dipole or small area magnetic loop, their terminal impedance will be high or low respectively, so if you restrict yourself to measuring their current output, the outputs will indeed be different. Instead, measure the power output into a matched impedance.

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