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Im trying to design a maximum strength electromagnet. I have a soft iron core that is 1.3" (33mm) round by 1.9" (50mm) long and I'm using 12 volts of input and my wire gauge is 18awg. Unfortunately I don't have the proper program to achieve this task. I was wondering if there was anyone with the know how that could help. Thanks Warren.

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  • \$\begingroup\$ What does "maximum strength" mean in this case? Ability to pick up metal objects stuck to one end? A U-shaped magnet touching the object on both ends is stronger, for instance. en.wikipedia.org/wiki/… \$\endgroup\$ – endolith Oct 15 '10 at 14:14
  • \$\begingroup\$ This thread mentions some software tools used to optimize electromagnets: eng-tips.com/viewthread.cfm?qid=137215&page=1 \$\endgroup\$ – endolith Oct 15 '10 at 15:16
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Assuming this is a DC application, wind as many turns as you have room for.

The reason being, magnetization is the product of the number of turns and the current through them. If you double the number of turns, however, while keeping the wire gauge constant, you double the resistance and thereby get half the current from your fixed voltage source; the turns*current product comes out being the same. What's different, though, if you double the number of turns, is that the power that the coil will draw will be half.

So if you keep adding turns, the magnetization doesn't change, but the current drawn goes down, and consequently the power required to run the coil decrease, too.

(If you're intending to use AC though, even 50Hz or 60Hz, the above argument doesn't apply since it disregards inductance issues.)

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  • \$\begingroup\$ Can you show the equations or at least a reference? \$\endgroup\$ – endolith Oct 15 '10 at 11:52
  • \$\begingroup\$ B is proportional to N * I (en.wikipedia.org/wiki/Solenoid#Quantitative_description), so this seems correct, that magnetization doesn't change with different numbers of turns. So for a given mass of copper, as wire diameter goes to zero, magnetization stays constant as current goes to zero? What's the limiting factor? \$\endgroup\$ – endolith Oct 15 '10 at 16:06
  • \$\begingroup\$ 'magneto motive force' is the magnetic equivalent of voltage, and is proportional to current times number of turns, times some other stuff that depends on the geometry of the coil. the amount of flux you actually get depends on the magnetic reluctance in the magnetic circuit. \$\endgroup\$ – JustJeff Oct 15 '10 at 20:55
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    \$\begingroup\$ @endolith -- consider splitting your wire lengthwise, and then putting the two halves in series; the mass of copper stays constant; the resistance of each piece doubles b/c it has half the cross section; the total resistance quadruples, b/c you have two of these 2X resistance pieces in series; so if you put twice the voltage on, you get half the current, the same magnetization, and the same power. It's kind of like characteristic impedance; you can have the same magnet with more voltage and less current, or vice versa. \$\endgroup\$ – JustJeff Oct 15 '10 at 20:58
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    \$\begingroup\$ So wind the hell out of it, is what your saying... Not quite the answer I was looking for, there must be a point of saturation where the windings stop producing and start "costing" more $$$. Current draw is not really a concern, I'm guesstimating that it would probably be about 1.5-3 amps. I would like to optimize so as not to waste wire or money. \$\endgroup\$ – Warren Oct 16 '10 at 1:34
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The core dimensions in one think. The other is the dimensions of the available form of the coil, and the temperature limit of the coil (state above or includes ambient). With this three plus the wire diameter and available DC voltage and current, you can calculate the maximum AmperTurns you can get from this coil. With the core that you have, the maximum coil diameter will be 66mm and 50mm long. So in this case the maximum turns with awg 18 will be around 530 turns (or about 32 m of wire) and you can apply 2,5A to have 1300At or 0,022Tesla at the core tip. And your coil will keep at 35oC plus ambient. However if you make the winding on the same form but with awg 28, then you will get about 1400 At. A question is how do you know that you have a piece of a real "soft iron"? Do you have a picture or a cut of this?

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