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ElectroMagnet Wire Gauge

Which Gauge?

  • 32

    Votes: 0 0.0%
  • 30

    Votes: 0 0.0%
  • 28

    Votes: 0 0.0%

  • Total voters
    0
  • Poll closed .
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Njguy

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I am looking for advice on choosing wire gauge for an electromagnet I am creating.

If I choose 32 gauge I will need 729ft of wire and it will produce 2170 loops of wire. This equates to 119ohms of resistance or 1.2 watts at 12 volts.

If I choose 30 gauge I will need 466ft of wire and it will produce 1400 loops of wire. This equates to 47ohms of resistance or 3.06 watts at 12 volts.

28 gauge follows the same patter with less loops and 8watts at 12 volts.

I really am not sure which to choose from. Any suggestions? Thanks guys.
 
The magnetic strength is determined by the ampere-turns (A-T) so, for example, 100 turns at 1A gives the same magnetic field as 200 turns at 1/2A. So you want to design with the maximum number of ampere-turns while not dissipating too much power.

In your example, 32 gauge at 12 V would give 12/119 *2740 =276 A-T while 30 gauge will give you 12/47*1400 = 357A-T. So the 30 gauge magnet will give you a 29% higher magnetic field at the expense of 255% more power as compared to the 32 gauge magnet.
 
One thing though, I have always read that the more loops, the stronger the electromagnet will be. If I go by your measurement and choose 28 gauge, there will be less loops but more amperes per loop. This is a bit confusing.
 
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Hi there NJguy,


I happen to be from NJ too :)

As you are finding out, the more 'loops' (usually called 'turns') of wire you have the more ampere turns you have, and the more ampere turns you have the stronger the magnetic field, and the stronger the magnetic field the stronger the magnet is and can pick up heavier objects. But what else you are finding out is that with a certain voltage (like 12v) the resistance of the wire limits the current, and the current is part of the ampere turns which is what gives the magnet its strength.

What you may not have discovered yet however is that all batteries and power supplies have a particular internal resistance. That means that when you connect your magnet to a battery it's not only the resistance of the wire that limits the current, it is also the internal resistance of the battery which limits current. It may not seem like much, but it plays an important role in the selection of the wire.

The selection of the wire is an optimization problem, and if you search around this site you'll find a thread that talks about how to go about doing this optimization because we had an extensive talk about it a while back. Luckily, the results are much more simple than it would seem at first because the optimization process results in a wire size who's resistance (when wound on the coil form) is exactly equal to the internal resistance of the battery.

There's a catch here however, and that is that the internal resistance of the battery might be very low and you might not want to have to use that heavy wire, or the battery might be small compared to its internal resistance. In this case you'd be using wire that has total resistance higher than the internal battery resistance so the magnet would always be run sub optimally.

A simplified formula for the force is:
F=K*N^2*I^2

where K depends on the form size and some other things. Note that with a limited form size you cant add more than a certain number of turns of a given diameter.

Rewriting that formula like this:
F=K*(N*I)^2

reveals that if we increase N and decrease I by the same proportion, the force F will stay the same. For example, if we start out with 100 turns and a current of 2 amps and set K=1 so we can try to optimize, we get:
F=1*(100*2)^2=40000 scaled units of force.
Now if we halve the diameter we can use 4 times as many turns because that much more will fit on the form now, but the resistance will increase by 4 times as well because the area goes down by a factor of 25 percent, so we end up with 1/4 of the current too which brings us to:
F=1*(400*2/4)^2=40000

which is the same as before.

The catch here is that we consider the battery internal resistance to be much less than the coil resistance, but as the total coil resistance comes down we get more and more current so the force would increase, but only up to the point where the coil resistance equals the battery internal resistance.
So the bottom line then is to use as low a resistance as possible without going lower than the battery internal resistance, but in cases where the battery internal resistance is very low compared to any coil we can wind conveniently it wont matter too much. You only have to think about the wire size and the current level, if the wire can handle the current. It's good to go a little low here to keep dissipation low. A very rough guide would be if there is 1 square inch of surface area for every watt the temperature rise will be 60 degrees C, but more square inches per watt results in a lower temperature rise.
 
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