coil gauge

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The strength of the magnetic field will be determined by the number
of turns and the current in them, (Ampere-Turns) so the larger the
diameter of the wire, the fewer turns you\'ll be able to wind on a
given core and the more current you\'ll have to force through them in
order to equal the ampere-turns of a coil wound on the same core
with smaller diameter wire.

Hi John,

Let me rephrase what i am trying to ask. The Magnetic field is H =N * I / L where N is the number of turn, I = current, L= lenght So, if I am making a coil and only have an inch of space to put my coil. I can use 18 Gauge or 30 gauge. I can wound 50 turns with the 18 Gauge and

100 turn with the 30 gauge. The smaller the wire, the less current I can put, but the more turn the more magnetic field I am creating. So basically I see that either way, I will end up with the same magnetic field. Therefore, just fill up the space with either gauge wire, I'll get the same weight, and by putting the maximum current capacity of the wire I get he same magnetic field. ???

Ken

As long as your are making air coils your can use

to calculate the coils.

Its not an air coil, but I'd like to know what are the formulas and how to use them .

ken

the strength is governed by the number of turns * the Amps. look up Webers, Gauss etc... if you use smaller gauge wire, you will need to factor in the resistance losses.

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Real Programmers Do things like this.
http://webpages.charter.net/jamie_5

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18 AWG wire has a diameter of 0.0403", so you should be able to get:

1" n = -------- ~ 24 turns in a 1" space. 0.0403"

30 AWG wire has a diameter of 0.0103". so you should be able to get

1" n = -------- ~ 97 turns in a 1" space, 0.0103"

a ratio of about:

97t k = ----- ~ 4, 24t

Which means you'll have to pump about four times the current into the heavy coil to get the same H out of it as the light coil.

Assuming your coil has a diameter of 0.318" means that the length of wire used to wind the 18 AWG coil will be about 24", and the length of wire used to wind the 30 AWG coil will be about 97".

18 AWG wire has a resistance of 6.39 ohms/1000', so the resistance of that coil will be about:

2' * 6.39R R = ------------ ~ 0.013 ohm 1000'

30 AWG wire has a resistance of 103.2 ohms/1000', so the resistance of that coil will be about:

8.1' * 103.2R R = --------------- ~ 0.836 ohm 1000'

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18 AWG copper wire weighs 4.97 lb per 1000 feet, and 30 AWG weighs 0.3042 lb per 1000 feet, so your 2 foot length of #18 will weigh:

2' * 4.97lb m1 = ------------ = 0.00994 lb = 1000'

and your 8.1 foot length of #30:

8.1' * 0.3042lb m2 = ---------------- = 0.00246 lb = 1000'

So, the coil wound with #18 wire will be about four times heavier than the other coil and, from the table at:

I get maximum currents of 2.3A for 18 AWG and 0.142A for 30 AWG, so that ratio is:

2.3A k1 = -------- = 16.19, 0.142A

so if you pump 16 times more current through the heavy coil than through the light coil, and it only takes four times as much to get the same H out of both, you'll get four times the intensity out of the heavy coil if you pump the maximum current allowed through both.

Finally, in order to get the maximum H out of the light coil you'll need a voltage across it of:

E = IR = 0.142A * 0.836R ~ 0.12V

and to get the same field out of the heavy coil:

E = 0.568A * 0.013R ~ 0.0074V

-- John Fields Professional Circuit Designer