magnetic field modulation coil

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. Thanks Sincerely, Indra

Reply to
Indra Dev Sahu
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How about keep on turning loops until the gauss meter reaches 1G??? D from BC

Reply to
D from BC

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 inches)


Reply to
Paul E. Schoen

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.

-- Joe

Reply to
J.A. Legris

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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