Tesla coils generally use near-field energy transfer. But ...
The page cited is sloppy in its wording and it would better to say "near field" where it says "RF". But, (again) ...
The answer is necessarily not black and white.
"Near field" has a precise meaning BUT obtaining ONLY near field coupling in a given case is not certain.
There is a gradual transition from NF to FF and a boundary region where both can occur. Near Field essentially occurs where certain geometry dependant terms form a major or significant part in describing interactions between transmit and receive structures - these relate to cyclical transfer of energy between the antenna I and V and adjacent electrostatic and magnetic fields. As distance increases these terms become less significant until they can be ignored.
You'll always get "some of both" with "some" varying as you move away from the antenna.
(1) At distances of well under a wavelength there is substantial interaction between the electric and magnetic fields produced by the aerial and current and voltage in the antenna. Energy is transferred to and fro between fields and aerial throughout with losses caused by non idealities but no energy loss due to energy "leaving" the aerial structure. This close in zone is termed the "reactive zone" where power may be absorbed by a tuned load which has voltage and current induced in it and which then dissipates energy (ie has a resistive component). Coupling involving power transfer is magnetic.
(3) [note number] "RF communications or energy transfer occur at distances beyond several wavelengths form the"antenna" structure. Here the ratio if electric and magnetic coupling have "settled down" and any energy present is not coupled to the structure I & V so is "lost", whether "received" or not. One way of viewing this one is that the two aerials are geometrically distance and secondary terms which account for the filed coupling and which have a strong distance dependent component have become insignificant - the field has become essentially homogeneous over lengths of the order of the receiving antenna.
(2) At distances past about half a wavelength the "second order" terms on which pure NFC coupling depends start to get small and the field starts to become homogeneous. This is termed the "Fresnel zone" (the guy has his name all over) and there is a degree of non ideality in field coupling to the antenna.
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This wikipedia page on near and far field does a better than usual job of commenting.
Their summary section says, in part:
- The near-field is remarkable for reproducing classical electromagnetic induction and electric charge effects on the EM field, which effects "die-out" with increasing distance from the antenna (with magnetic field strength proportional to the inverse-cube of the distance and electric field strength proportional to inverse-square of distance), far more rapidly than do the classical radiated EM far-field (E and B fields proportional simply to inverse-distance). Typically near-field effects are not important farther away than a few wavelengths of the antenna.
Far near-field effects also involve energy transfer effects which couple directly to receivers near the antenna, affecting the power output of the transmitter if they do couple, but not otherwise. In a sense, the near-field offers energy which is available to a receiver only if the energy is tapped, and this is sensed by the transmitter by means of answering electromagnetic near-fields emanating from the receiver. Again, this is the same principle that applies in induction coupled devices, such as a transformer which draws more power at the primary circuit, if power is drawn from the secondary circuit. This is different with the far-field, which constantly draws the same energy from the transmitter, whether it is immediately received, or not.
Well if you would know the answer you could make a lot of money selling this technology.
I think that there will always be a large loss. But there will be savings on the car side. The battery doesn't need to be as large as normally as it can be charged continuously. This is also better for batteries, better lots of small charges than emptying it completely.
Best Answer
Of course it is possible in theory. But very very inefficient in practice.
Inductance of the primary coil (e.g.; A=10000mm\$^2\$ cross sectional area, \$\ell\$=100mm length, 10,000 turns) is
$$ L = \dfrac{\mu A N^2}{\ell} = \dfrac{(4\pi 10^{-7} \text{H/m}) (0.01\text{m}^2) (10000)^2}{0.1\text{m}} = 4\pi \text{H} = 12.5664\text{H}. $$
Resistance of the primary winding is
$$ R_p = \dfrac{2\pi\sqrt{\dfrac{A}{\pi}}N\rho_{cu}}{a} = \dfrac{2 \pi \sqrt{\dfrac{0.01\text{m}^2}{\pi}} (10000) (16.78 \times 10^{-9} \Omega\text{m})}{10^{-6}\text{m}^2} = 59.48 \Omega. $$
The RMS magnetizing current which will be wasted on the primary side will then be (ignoring the resistance of the wire)
$$ I_m = \dfrac{V_p}{Z_p} = \dfrac{V_p}{\sqrt{X_p^2 + R_p^2}} = \dfrac{V_p}{\sqrt{(2\pi f L)^2 + (59.48\Omega)^2}} = \dfrac{220\text{V}}{\sqrt{(2\pi (50\text{Hz}) (12.5664\text{H}))^2 + (59.48\Omega)^2}} = \dfrac{220\text{V}}{\sqrt{(3947.85\Omega)^2 + (59.48\Omega)^2}} = \dfrac{220\text{V}}{3948.30\Omega} = 55.72\text{mA} $$
which is low enough for most use cases.
Assume that you use copper wire of cross sectional area a=1mm\$^2\$ and the density of copper is d=8.96 g/cm\$^3\$. The mass of copper you need is (assuming that it fits in the given space)
$$ \text{m} = 2\pi\sqrt{\dfrac{A}{\pi}}aNd = 2 \pi \sqrt{\dfrac{0.01\text{m}^2}{\pi}}(10^{-6}\text{m}^2) (10000) (8960 \text{kg}/\text{m}^3) = 63.52 \text{kg}. $$
Note that, the same amount of copper you need on the secondary side if you want to get the same voltage level. If we take price of copper as p=6$/kg, total price of copper used will be $762.24.
Real power loss due to magnetizing current will be
$$ P_{\text{loss},m} = I_m^2R_p = (0.05572\text{A})^2(59.48\Omega) = 185 \text{mW}. $$
Real power loss when transferring 10A current will be
$$ P_{\text{loss},10A} = (10\text{A})^2(59.48\Omega) = 5.948 \text{kW}, $$
which means there won't be 220V on the secondary side due to heavy voltage drop on the primary side winding resistance. You need much bigger wire radius, more copper, more money!
You may do some optimization. For example, you may reduce number of turns, which will increase magnetizing current and reduce copper losses. But even at the optimum point, it will still be very inefficient.
Because of this, people don't transfer large power though air and they use cores for it.