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I am learning about loop antennas at the moment. I read that the response of a small loop Antenna is proportional to the rate of change of magnetic flux through the loop which causes a uniform current distribution in the loop. This would be Faraday's law. However, then we got a large loop (wavelength is as big as the circumference); how do we describe the interaction? Let's say, there is an incident electromagnetic wave going through a large loop. Can we still use Faraday's law to describe the response and is it the same response as in the case of a small loop?

This might a stupid question. I read in some books that small loops can be described using Faraday's law but they don't use Faraday's law to describe big loops.

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    $\begingroup$ Faraday's laws apply to all antennas and electric things. The large loop is large enough that (unlike the small loop) opposite sides of the antenna are opposite phases. It more or less works the same as a half wave dipole, except that it is a full wave loop. Note the radiation pattern of a large loop is opposite that of a small loop. $\endgroup$
    – user10489
    Feb 3, 2022 at 2:50
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    $\begingroup$ Not a stupid question at all. Faraday's law is one of 4 "Maxwell" equations that completely describe interactions with electric and magnetic fields. I'll try to come up with a good answer for you. $\endgroup$
    – AG5CI
    Mar 3, 2022 at 14:53

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In short, yes: Faraday's law describes what's happening for the large loop antenna as well. A transverse EM wave induces a current in the loop. It's Faraday all the way down for all EM/antenna interaction for that matter.

An aside: you can think of a full wave loop as a stretched-out folded dipole.

  • Both have maximum current nodes at the feed point and the point directly opposite the feed point
  • Both have relative nulls coming off the "ends" (in the loop the "end" is the current minimum ~1/4λ from the feed point in each direction)

Feedpoint impedance for a full wave loop is about half of the 300Ω impedance of a folded dipole... but that's from memory and I'll come back and correct this after I get a moment to check.

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