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Inductance

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steveB

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Another thought occurred to me about this. It seems that wire diameter is important for two reasons.

First, larger diameter helps make the higher resistance of the iron (as compared to copper) less of an issue in creating differences. Any test for inductance needs to be at a frequency well below the R/L frequency of the coil. In other words, when you charge the coil with a square wave voltage signal, you would like to see the current rise linearly, and not exponentially.

Also, greater diameter means more iron, which might make the effect (again, if there is an effect) more noticeable. More iron could cause more distortion of the fringe field.
 

Ratchit

Well-Known Member
Another thought occurred to me about this. It seems that wire diameter is important for two reasons.

First, larger diameter helps make the higher resistance of the iron (as compared to copper) less of an issue in creating differences. Any test for inductance needs to be at a frequency well below the R/L frequency of the coil. In other words, when you charge the coil with a square wave voltage signal, you would like to see the current rise linearly, and not exponentially.

Also, greater diameter means more iron, which might make the effect (again, if there is an effect) more noticeable. More iron could cause more distortion of the fringe field.

The time constant of a coil is L/R, not R/L. No matter what the time constant is, you are going to see a exponential rise of current when a constant voltage is applied. Inductance is independent of frequency unless the frequency is so high that secondary effects come into play.

Ratch
 

steveB

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The time constant of a coil is L/R, not R/L. No matter what the time constant is, you are going to see a exponential rise of current when a constant voltage is applied. Inductance is independent of frequency unless the frequency is so high that secondary effects come into play.

Ratch

Ratch, I said the R/L frequency, not the time constant. Frequency is the inverse of time constant. Strictly the curve will be exponential in shape, yes. But the curve is approximately linear in the beginning. An ideal inductor with zero resistance has a linear ramp, but no inductor is ideal. I'm pointing this out because if the iron inductor has too high a resistance, crude attempts to measure inductance can give the wrong answer. Basically, it's just easier to measure inductance at a frequency where it looks more like an ideal inductor. Otherwise, you need to know the resistance of it and be more careful. But, yes you can do it, so I should not have made it sound like you can't deal with the exponential shape, or higher frequency.
 
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Tony Stewart

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IF you recall the Electric and magnetic fields according the the right hand rule are at right angles to the direction of the wire. Thus the permeability only increases when the iron is at right angles to the flow of current. Since this geometry is inline with the iron, the geometry is poor. WOrse yet Iron has poor conductance so the ESR will be very high. Both properties make this a poor inductor.
 

dr pepper

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Interesting question, I've seen copper and ally windings but never iron.

Inductance at guess probably will be higher, however the coil probably will have some weird properties.
I see from your avatar your into music, is this for a guitar pickup, if you dont mind it not working its a good idea to test one, I'd expect the o/p to be higher and have a non linear freq response.
At least with a pickup saturation isnt likely to happen, unless you have a plectrum made from neodimium.
 

MrAl

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Hi,

I dont think this is that difficult of a question, although it is interesting.

The more basic question is, "what is the difference in inductance between a long straight wire made of copper wire and one made of steel wire".

The formula for a straight wire inductance will contain the permeability factor, therefore the permeability will change the inductance. Since the (relative) permeability of copper is about 1 and that for steel is 20 or above, i would expect the inductance of the steel wire to be 10 times greater or more.
I can not be as certain for steel as for nickle, which would give us an inductance about 20 times that of copper.

Also, as i am sure you guys have realized, the DC resistance will be quite a bit higher, but more importantly the skin effect will make the wire look like an open circuit (ha ha). Ok not that bad, but it will be a lot higher so the AC resistance will be way too high for anything other than a DC circuit.

Another view of the skin effect is the major conductor part will be more near the surface than inside, and since we have a ferromagnetic metal inside now we must expect the inductance to rise.
This is interesting because the cross sectional area isnt that big, but the change in inductance can still be 10 times higher.

We can look for the formula on the web. The inductance will not change as a linear factor with the permeability as it would with a regular magnetic core.

Inductors for a given application are sized for their inductance vs series resistance vs saturation level. Increase the series resistance by a factor of 5 and it takes the product out of the marketplace because nobody wants to use it, at least not in power applications, and for tuned circuits they probably want a high Q.

I also imagine the 'inductor' would saturate quite easily, which would lower the inductance. So it would have to be used at low currents only, perhaps very low.
 
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