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The induced EMF in an antenna is often given as $E\cdot h$, where $h$ is the height of the antenna and $E$ is the electric field strength in V/m, but this does not seem to capture the power of the electromagnetic wave in the far field (i.e. the multiplication with the magnetic component related by the impedance of free space) and hence the correct transfer of power.

I read that the antenna automatically transforms the impedance of free space to the impedance of the antenna. Does this mean that the induced voltage is actually $$Eh\sqrt{\frac{Z_0}{Z_a}}$$ where $Z_0$ is the impedance of free space and $Z_a$ is the internal impedance of the antenna? If the antenna acts as a transformer, surely it would be.

And would I be right in thinking that the antenna acts as a transformer with a winding ratio that is actually determined by the impedances $$\frac {Z_0} {Z_a}$$ If this isn't the case then the power would be greater than the power taken from the electromagnetic wave, which doesn't make sense.

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Isn't induced EMF in an antenna dependent on the power of the electromagnetic wave?

Indeed it is not - for given antenna geometry, the induced EMF due to external wave depends only on electric field of the wave where the antenna is. The total induced EMF includes also self-induced EMF and the latter depends on behaviour of current in the antenna and its geometry (it's complicated). But neither of these are directly related to total power of the EM wave. For all practical situations, the available power or possibly absorbable power is much smaller than total power of EM wave.

The induced EMF in an antenna is often given as E⋅h, where h is the height of the antenna and E is the electric field strength in V/m, but this does not seem to capture the power of the electromagnetic wave in the far field (i.e. the multiplication with the magnetic component related by the impedance of free space) and hence the correct transfer of power.

Yes, induced EMF due to EM wave is line integral of EM wave's electric field; in case of straight line dipole antenna directed along the line of polarization of the wave, it is electric field of the wave at the antenna times length of the antenna. Also yes, this quantity does not capture available power, and neither it should be expected to.

Electric field or EMF of external EM wave for a given antenna geometry do not imply anything about available power that the antenna can draw from the EM wave. It tells us the maximum value of voltage that can be induced on the antenna terminals, and in practice this voltage is slightly lower due to the fact current has to be allowed to flow (the antenna is connected to a circuit). But it does not tell us induced current in the antenna.

The power drawn from the EM wave depends not only on external EMF, but also on details of electric current as function of position and time, that oscillates in the antenna. It is like with force and velocity in mechanics: if the current ("velocity") is high and in phase with the external electric field of the wave ("force"), then absorbed power can be high. This requires tuning antenna parameters and it results in condition of impedance matching/adaptation - in order to extract as much power as possible, impedance of the connected circuit has to be tuned to impedance of the antenna.

I read that the antenna automatically transforms the impedance of free space to the impedance of the antenna. Does this mean that the induced voltage is actually $$Eh\sqrt{\frac{Z_0}{Z_a}}$$ where $Z_0$ is the impedance of free space and $Z_a$ is the internal impedance of the antenna? If the antenna acts as a transformer, surely it would be.

Even if the formula for voltage would be correct, this would not be correct description in terms of physics. There is only single voltage on the antenna terminals, there is no transformation between two voltages going on like in HV-AC transformers. Yes, induced EMF has different value than induced voltage, but transformer does not transform EMF into voltage - it transforms one voltage to another voltage, using induced EMF. So the description in terms of transformer is not very physical.

There are other situations in physics where EMF and voltage magnitudes are not equal to each other - for example, real inductor with ohmic resistance. We do not say real inductor transforms EMF into voltage like a voltage tranformer with windings and core - we just accept that EMF and voltage are two different concepts, with possibly different values. It is only in special idealized circumstances they have the same magnitude (perfect conductor solenoid).

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  • $\begingroup$ Well I do get the fact that the electric field results in a fixed EMF that then generates a power, and the EMF and current isn't dependent on any input power. That's easy to visualise. But surely it has to be dependent on the input power of the electromagnetic field, i.e. the power is transferred, but I just don't know how it works out mathematically, especially when the current can be variable when EMF remains the same $\endgroup$
    – Lewis Kelsey
    Commented Jun 3, 2021 at 1:45
  • $\begingroup$ If you are after mathematical description of energy transfer between EM field and matter, search for local conservation of energy and momentum in EM theory. In macroscopic EM theory, this is in terms of Poynting vector and Maxwell tensor. The simple description is that energy is transferred continuously through space and can be absorbed wherever there is electric current present. In your example, the current in the antenna is "sucking away" part of ambient EM energy and converts it to energy in the circuit. $\endgroup$
    – Ján Lalinský
    Commented Jun 3, 2021 at 12:04
  • $\begingroup$ It seems to me that radiative coupling is much unlike capacitive and inductive coupling, where the other two can be modelled with a clearly defined circuit in terms of power transfer and ohms law. Inductive coupling is a transformer, and this isn't. I think the key is it's just an electric field, and the energy comes from the field. This induces a power based on the speed of EPE conversion (current), meaning the energy taken from the field is greater when the current is greater, surely there is some sort of constraint on the current so that the energy of the field isn't immediately depleted $\endgroup$
    – Lewis Kelsey
    Commented Jun 3, 2021 at 12:45
  • $\begingroup$ Wait is it something to do with the fact that the wave travels at the speed of light maybe, so the energy of the field at a point in time can't be depleted due to any high current in the circuit $\endgroup$
    – Lewis Kelsey
    Commented Jun 3, 2021 at 12:50
  • $\begingroup$ > radiative coupling is much unlike capacitive and inductive coupling -- "radiative coupling" is not a good term, the source of the EM wave may be very far away and is often immaterial. The effect of EM wave on the circuit is quite similar to "inductive coupling", in the sense that the work is done by induced (not Coulomb) electric field; in case of EM wave, this field is due to travelling wave whose source is far away, otherwise it is the same concept. $\endgroup$
    – Ján Lalinský
    Commented Jun 3, 2021 at 13:22
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For a linear electric dipole less than $\lambda/2$ long you are correct with your interpretation. It is incorrect for a magnetic dipole (ie., loop antenna) where the $Eh$ term is not really meaningful for $h$ as such is not very meaningful. But even then considering a dipole (electric or magnetic) antenna as an impedance transformer is very fruitful and can be used as basis to design non-directional antennas.

It is less fashionable to view a horn or some other aperture or phased array antennas that way but even those cannot work without proper impedance matching. The usual undergraduate view that an antenna is nothing but a physically large resonant circuit is completely misleading for it is not being resonant that matters but that besides being resonant the impedance of the circuit matches that of the amplifier or feed line of impedance $50[\Omega]$, the ratio of $V/I$ in the line, to the ratio $E/H$ of the free space that is just $120\pi [\Omega]$.

What is nice in aperture antennas (horns, reflectors, etc.) is that most of this matching is done implicitly by having a large aperture (several wavelengths across), and thus the ratio $E/H$ in the aperture of the horn, say, is already near $120\pi$. A "resonant" (tuned) antenna needs *explicit* matching in the sense that is its usual driving point impedance, can be very far from $50[\Omega]$ when the other side is terminated in $120\pi[\Omega]$, the shorter the antenna the smaller the impedance is, hence the need for tuning and impedance transformers.

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