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Given an air-core coil of known inductance and a given frequency of applied voltage, you can predict the coil reactance, the current, and the voltage/current phase relationship.

What effect, if any, will a permanent magnet placed at one end of the coil have on the predicted properties of the coil?

Additionally, if the reactance for the given frequency is altered by the magnet, can the applied frequency be changed such that a new reactance (in the presence of the magnet) can be matched to the original reactance without the magnet.

The former paragraph is at the core of what I'm trying to get at, because I want to know if the magnet will have any kind of transformer effect (parasitic or otherwise) on the oscillating coil. Will the presence of a magnet simply change the inductance of the coil, or will it cause other behaviors (similar to those seen when two coils share mutual inductance)? Or is there something else altogether I am missing?

The arrangement of coil and magnet is end-to-end as the field polarities would be aligned along their axes like this:

 coil <--> magnet
_//////_  [N | S]

The arrangement would also allow for the magnet to slide into the coil, thereby creating a sort of magnet-core coil.

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    \$\begingroup\$ A permanent magnet, or hard magnet is already magnitized very close to saturation so it can't take any flux swing on top of that. It's also called hard because it resists being magazined/unmagnitized in the first place. \$\endgroup\$ – winny Dec 27 '16 at 8:10
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The reactance of the coil will change and there are two opposing mechanisms at play: -

  • The permeability of the magnet will tend to increase inductance
  • The conductivity of the magnet will cause eddy currents that tend to decrease inductance

To what extent that either increases or decreases the coil's inductance is guesswork and will vary at different frequencies. I've seen situations where reactance/inductance doesn't change at all.

Another effect is that the eddy currents in the magnetic material will cause the magnet to take power from the AC circuit i.e. losses will increase.

At high frequencies the magnet will act like a plate of a capacitor and thus there will be a resonance effect. Again, without detail of the experiment this is hard to quantify.

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  • \$\begingroup\$ "Another effect is that the eddy currents in the magnetic material will cause the magnet to take power from the AC circuit i.e. losses will increase." I hadn't considered conductivity. Thank you. This is effectively similar to the loading from the secondary in a transformer and is exactly the type of thing I wanted to be aware of. Your answer is excellent! \$\endgroup\$ – JamesHoux Dec 28 '16 at 5:58
  • \$\begingroup\$ Accepted. Can you explain the capacitor plate effect? Would it mimic a capacitor in parallel with the coil? \$\endgroup\$ – JamesHoux Dec 28 '16 at 16:23
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    \$\begingroup\$ At high frequencies there is detrimental capacitance between each turn of an inductor and this is increased by a conducting "plate" because it "bridges" all windings and thus increases the net capacitance. It would mimic a parallel capacitor. \$\endgroup\$ – Andy aka Dec 28 '16 at 17:03
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If it is assumed that the magnet is non-coductive and has permeability and permitivity the same as air the only effect is a slight resistance increase due to current crowding caused by the hall effect.

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  • \$\begingroup\$ I wasn't aware of this. Thank you very much! +1 :) \$\endgroup\$ – JamesHoux Dec 28 '16 at 16:19
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The fixed magnetic field will not have much of an effect unless you are getting near the saturation flux of any magnetic materials in the coil such as a core. I think ferrite magnets have a high enough permeability to act as a core but Nd and some other magnets are close to nothing so to get an effect you would need to add the magnet to an inductor with a core. Those are a rarely used but are called permanent magnet inductors (PMIs) and the magnet is added to offset the magnetic field of a large DC bias current in the inductor to prevent the core saturating. It lets you make the core smaller without core losses and I think there are other benefits in efficiency or noise or something too.

I think they would be useful in DC-DC converter inductors where there is a very high bias current but I think they might not use them because having a closed magnetic path is important for energy storage and EMI so there is nowhere to put the magnet. If anyone knows more it's something I've been wondering about.

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  • \$\begingroup\$ a smaller core need less copper to circle it, so lower resistance would be one of the other benefits. \$\endgroup\$ – Jasen Oct 21 '18 at 19:26
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If you use steel wire (not forbidden by your question), then the magnetic field will bias the iron's magnetic domains very strongly to one side and minor signal currents will not be able to cause flipping of the domains.

Since any domain flipping injects noise into the circuit, we do not want that flipping to occur in our low-noise circuits.

With various connectors using STEEL pieces for strength, such as RCA audio plugs and jacks, a permanent magnet has its uses therein.

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