# magnetic field modulation coil

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

1. ### Indra Dev SahuGuest

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

2. ### D from BCGuest

How about keep on turning loops until the gauss meter reaches 1G???
D from BC

3. ### Paul E. SchoenGuest

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)

Paul

4. ### J.A. LegrisGuest

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

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  