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magnetic field modulation coil

Discussion in 'Electronic Design' started by Indra Dev Sahu, May 6, 2007.

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  1. Dear all,
    I am trying to design a modulation coil that can produce 1 G magnetic
    field at the center of solenoid along its axis. The coil should be of
    radius 3.5mm and length 10mm. I am getting confusion for the number of
    turns.So I am looking for some suggestions for that.
  2. D from BC

    D from BC Guest

    How about keep on turning loops until the gauss meter reaches 1G???
    D from BC
  3. Not my area of expertise, but it appears that inductance is proportional to
    ampere-turns, so you must think about how much current you have available,
    and can easily handle. Then you can get an idea of wire size, wind the coil
    based on that, and then use the gauss meter as you raise the current.

    The approximate formula for inductance of a single layer air core coil is:

    L = ((N*A)^2)/(9A+10B) where A is radius and B is length (I think based on

  4. J.A. Legris

    J.A. Legris Guest

    Assuming the coil is significantly longer than it is wide, the formula
    is (in MKS units):

    NI = BL / u0 in MKS units,

    where N is the # of turns, I is the current in amps, B is the magnetic
    field in Tesla (= 10^4 Gauss), L is the length in meters and u0 is the
    permittivity of space (air)

    = (10^-4 T x 10^-2 m) / (4pi x 10^-7)
    = .79

    So, if you close-wind 40 turns of wire .25 mm thick, giving 10 mm
    length, you'll need 19 mA to get 1 Gauss.
  5. Guest

    Do you absolutely have to use a solenoid of those (rather small)
    dimensions? What is it intended to modulate?

    Helmholtz coils' fields are easy to calculate and provide an open
    center volume for easy access. They comprise two identical hoop-shaped
    coils spaced by their radius and connected in series.

    The formula for the field they make is

    B (in Gauss) = (.866 * mu * n * I ) / r

    mu = permeability of whatever's within the coil pairs' volume

    n = number of turns per coil

    I = current in Amperes

    r = radius in centimeters

    As a bonus, the field is nearly perfectly uniform in the central
    ninth or so of the volume between the coils.

    They do take up rather more room than solenoids for equivalent field
    strength, but solenoids don't produce uniform fields.

    Mark L. Fergerson
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