# coil gauge

Discussion in 'Electronic Basics' started by Ken O, Jul 16, 2006.

1. ### Ken OGuest

HI again,

I read on the internet that depnding on the gaude size lets 18, it will take
more amp to generate the same EM strenght as with a 30 gauge. considering
there are the same weight of magnet wire to do the coil.
I am not finding much information on this ( different gauge size Vs coil )
on the internet.

Ken

2. ### John FieldsGuest

---
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.

3. ### Ken OGuest

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

4. ### HKJGuest

http://www.miscel.dk/MiscEl/miscelAirCoil.html to calculate the coils.

5. ### Ken OGuest

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

ken

6. ### JamieGuest

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.

7. ### John FieldsGuest

---
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"

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'

---
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:

http://www.powerstream.com/Wire_Size.htm

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  