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I can't figure out the picture of resonance inductive coupling

I can image how magnetic inductive coupling works, It scatter magnetic around one coil and if second coil being near it then induce electric in second coil

It's so clear that the magnetic field line will weaken by distance

But when resonance come play the role. It like it can tether the invisible wire between coils and the portion of energy transfer increasing even the distance is more than meter

That's when I can't figure out what really happen. Some have mention quantum tunnelling still don't make it clear. Are there any clear explanation that's easy to picture like analogous to anything?

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  • $\begingroup$ In resonant coupling the Q of the two coils depends on their distance. If they are close together, Q is small but the coupling is large, when they are far apart, Q is large but the coupling is small. In a system in which the individual Qs of both the transmitter and the receiver coil were infinite, these two effects would offset each other perfectly at any distance. In reality the resonant coupling can only transfer significant amounts of power until the coupled Q approaches the free Q, after which the systems becomes lossy. $\endgroup$
    – CuriousOne
    Commented Apr 29, 2015 at 20:04
  • $\begingroup$ If you push a swing at right intervals, its amplitude will grow. If you push it at wrong intervals it won't (it can even decrease). Intuitively it's the same. Magnetic inductive coupling isn't as simple as that. Magnetic field of first coil must change all the time in order to induce current in the other. If the rate of change (frequency) is too high or too low, it's like pushing the swing at too short or too long intervals. But the amplitude can grow at right frequency. $\endgroup$
    – Azad
    Commented May 7, 2015 at 20:18
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    $\begingroup$ Quantum tunneling has nothing to do with this. $\endgroup$
    – DanielSank
    Commented May 8, 2015 at 0:16
  • $\begingroup$ @DanielSank academia.edu/6262117/… $\endgroup$
    – Thaina
    Commented May 8, 2015 at 9:28
  • $\begingroup$ @Thaina ah, perhaps I misunderstood OP. I though the question is asking about coupling between two modes via inductive coupling to an auxiliary resonant mode. OP is that what you're asking about or not? $\endgroup$
    – DanielSank
    Commented May 8, 2015 at 15:11

2 Answers 2

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What you seem to be referring to (a guess from your mention of quantum tunneling) is perhaps near-field magnetic induction.

Here's a summary - magnetic coils have a near-field region around them, the primary contriubtion to which involves minimal radiation of power. This region is dominated by the response of the coils (for example) to the fields, and is therefore a kind of evanescent wave (i.e. something that is present around a source only for a small distance).

On the other hand, we have the usual resonant inductive coupling, where two coils have the same natural frequency, and this maximizes the transfer of energy between them. The reason it is a maximum here is because when they have different natural frequencies, there are times when the natural restoring 'force' is out of phase (i.e. pulling oppositely) with the external 'force', effectively cancelling each other out. A simple (i.e. driven harmonic oscillator) treatment of this may be found here. It obviously still obeys the conservation of energy, and cannot transfer unreasonably large amounts of energy without limits. It's just that it is much more efficient that not having resonance.

When you combine the two, you simply have resonant coupling within the near-field regions. This is a good way to maximize the transfer of energy, simply because the near-field region is where the field is still fairly strong, and resonance is where maximum energy transfer occurs.

This obviously can't be used for energy transfer over very large distances; it's only that you can make the near-field region large enough for practical purposes. This is seemingly being implemented by some company.

Quantum tunneling occurs from a similar kind of "evanescent wave". Consider a particle in a finite potential well: while the solutions for the wavefunction are sinusoidal oscillations within the well, they decay rapidly with distance into the walls. However, when the wall is very thin, the wavefunction is still large enough at the other end, leading to a significant probability of finding the particle on the other side of the wall, even though it does not classically have the energy to cross the wall (see, for example, this picture). This has similarities with the "near-field" part (in the wall region), but not with the "resonance" part.

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  • $\begingroup$ Thanks for your answer. I can picture the image of this effect much more clearer $\endgroup$
    – Thaina
    Commented May 9, 2015 at 18:56
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The modeling of resonant systems whether they be mechanical, electrical or whatever by differential equations allows one to express them in terms of a transfer function . The transfer function relates an output (response) with an input (excitation). For resonant transfer functions, especially those with a high Q or quality factor there is significant amplification at the resonant (natural) frequency of the system. For two systems with matched resonant frequencies, there is amplification in transmitting the field or energy and amplification in receiving it.

So it's the matched resonant frequencies and the amplification factor that leads to this strong coupling and the ability to couple energy at great distance.

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