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coil gauge

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

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  1. Ken O

    Ken O Guest

    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 Fields

    John Fields Guest

    ---
    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 O

    Ken O Guest

    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. HKJ

    HKJ Guest

    As long as your are making air coils your can use
    http://www.miscel.dk/MiscEl/miscelAirCoil.html to calculate the coils.
     
  5. Ken O

    Ken O Guest

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

    ken
     
  6. Jamie

    Jamie Guest

    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 Fields

    John Fields Guest

    ---
    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'

    ---
    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
     
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