Coils - Optimizing Turns vs. Current

Is the magnetic field intensity of a coil directly proportional to the current applied in all cases, or does the number of turns also factor in?

For example, if I feed a certain current through a thin wire of many turns, is the field equal to the same current through a few turns of thicker wire.

And any other considerations.

David King

On Tue, 11 Nov 2014 11:58:56 +1100, David King Gave us:

VOLTAGE makes the field.

Den tirsdag den 11. november 2014 01.59.04 UTC+1 skrev David King:

the field is ~ turns * current

To fit more turns in the same space you need to use thinner wire which can handle less current, so there is no magic bullet (other than superconductors)

-Lasse

If I understand your question, you want to know if larger gage wire will allow you to use fewer turns, which will result in the same weber (Tesla units) density?

My answer to that would be no, it's the number of turns and current. But let me add something to that, smaller gage wire will heat up due to resistance and can limit your current range.

Look up the weber/Tesla unit.

Jamie

If the coil is wound on a high-permeability core, flux will be proportional to N*I... modified by any core nonlinerity. But a given N*I product (measured in ampere-turns) will make nearly the same flux as you seesaw N against I.

If the coil is in free space, the flux density at any point in space is a more complex function of the measurement location and the winding geometry. Changing the number of turns necessarily changes the geometry some. For a fixed geometry and number of turns, flux at any point will be proportional to I.

There are formulas and online calculators for the simple free-space solenoid case.

What specific geometry did you have in mind?

--

John Larkin         Highland Technology, Inc 
picosecond timing   precision measurement  

jlarkin att highlandtechnology dott com 
http://www.highlandtechnology.com

Divided by an area.

George H.

** To produce a given magnetic field strength inside a copper wire coil requires a particular DC power to be applied and then dissipated as heat in the coil.

The coil can be wound from almost any gauge wire. Long as ALL the coil's dimensions are kept the same, then that power level will remain the same too. You see this with relay coils where the maker is able to supply identical looking relays with a host of coil voltages and current ratings - all having the same power.

If the wire is changed to another metal, the power level will increase in proportion to the resistance of the new metal compared to copper.

You rarely see another metal used to make coils for this reason.

... Phil

No it does not.

Ever hear of "amp-turns"?

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John Devereux

You are a complete and utter imbecile.

On Tue, 11 Nov 2014 08:46:23 +0000, John Devereux Gave us:

Of course.

What has to be on that coil for there to be current in it?

One designs using volts per turn as the constraining parameter for optimal operation. Saturation considerations and all that.

Keep it low, as in less than a couple volts per turn. Higher numbers mean higher failure event likelihood, especially on a scatter winds. The insulation on fine transformer wire isn't the same as the heavier gauges.

I have transformer designs where each layer was flat wound, and then taped before the next layer at 150 t per layer and up to ten layers in a little tiny core onto a little tiny bobbin. Vacuum baked and varnished with Dolph's.

So, *I* would say "Electron flow into a load creates a field on the conductor transporting said flow to said load."

Pressure (EMF) makes the flow, and the load determines how much demand (actual flow) occurs, but that is linear, resistive law.

Your imbecility is boundless.

So, as I understand, a 5,000T coil with with 12 Ohms DC resistance will emit a stronger field than a 2,500T coil with a 6 Ohm resistor in series, even though both pass the same current.

David King

Interesting 'twist' to design thinking. ...saying voltge makes the field.

current makes the field, BUT voltage makes the current, so in a REAL world which comes first? interesting.

Almost ALL of my designs are using air core, so saturation, nonlinearity, etc don't come into play much. But whenever I need a field strength, I've ALWAYS thought of CURRENT, then checked whether there was realistic volage to make that current.

However, in retrospect, I have thought in terms of voltage per turn in optimizing wireless power transfer, so actually was applying your 'voltage makes field' concept. ...interesting.

However, since I usually go back to BASIC equations to rederive just to make certain, I think I'll keep the premise that current makes field and voltage makes current, so voltage is a 'result' not a cause. Makes it easier to analyze those pesky SQUIDs.

[Snip]

Just curious. How did you optimize the wireless power transfer related to the voltage per turn?

Cheers

Klaus

yes.

download a free copy of femm 4.2 and join their user group. That PC Tool program is small and non-obtrusive to most systems and will provide hours of fun gaining understanding of all kinds of things relating to coilc - fields, resistance, skin effect, eddy currents, reflected fields, etc etc.

For examle, modeling your simple coils would tell you instantly what's going on.

It would also enable you to explore the features of a Helmhotz coil AND the modified form of the Helmholtz coils, where current creates rather accurate fields over a volume of space based upon construction parameters and current through the coil. And for one intrepid designer whose name escapes me, 'voltage' on the coil, but using voltage you have to watch out for the temperature effects on the wire changing the resistance, and therefore changing the current.

in the field generation. but very LONG story of how to do, AND [in my way of thinking] people are optimizing weak solutions, not considering what is actually going on and taking advantage of characteristics and thus end up fighting those characteristics rather than capitalizing on them.

The voltage is not the fundamental quantity, it depends on the resistance of the wire. It can be very high for thin wire or zero for a superconducting magnet. This becomes even more obvious at AC of course, e.g. V=L.dI/dt. It is the current that is more fundamental.

[...]

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John Devereux

Nice, I had not thought of it like that.

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John Devereux

I don't think so. N*I is constant in this example, so the field remains unchanged.