--- On 22 Sep 1997 20:49:18 -0700 [email protected] wrote:
> Does anyone remember the old mil spec that had the design details for a
> 5 foot diameter Helmholtz coil ?
> Paul Signorelli
> [email protected]
>
> to: INT:[email protected]
> cc: INT:[email protected]
> INT:[email protected]
> INT:[email protected]
> INT:[email protected]
> INT:[email protected]
> INT:[email protected]
> INT:[email protected]
Paul:
I wonder how I got on this rather select list of "experts"?
No, I don't recall any old Mil-Standard for a 5 foot diameter Helmholtz
coil. There were, however, some requirements for magnetic field exposure where
you might have used a large Helmholtz coil.
One are was for Mil-Std-1399, Section 401, for shipboard equipment
which might be exposed when the ship is being degaussed. This called for
equipment to withstand a field which went from zero to 20 oersteds in one
second, then hold at 20 oersteds for 5 minutes, then decrease to zero in one
second. The field could be any type of coil.
The largest coil I ever used was a 10 foot long by 5 foot diameter
solenoid. This coil had very good field uniformity throughout its volume. By
contrast, a Helmholtz coil has decent field uniformity only in about the center
1/3 of its volume. A Helmholtz coil is, by definition, two coils spaced one
radius apart. Thus, a 5 foot diameter coil is two windings spaced 2.5 feet
apart. the "sweet spot" is only the center 10" x 10" x 10" volume. That's not
much volume for the whole size of the coil. That's why I used a solenoid
(single layer, air core) coil. PS: I did see a satellite test chamber, with
three sets of orthagonally mounted Helmholtz coils. Each coil was about 30 foot
diameter. It was used to null out the Earth's magnetic field so as to place a
satellite in a true zero gauss environment. The whole building sheltering the
coil was built of non-magnetic materials. The facility got scrapped due to
right-sizing.
EMCO makes a Helmholtz coil, their #6406, which has a 6 foot diameter
coil. It has a max current of 20 Amps, with 64 turn, 91.4cm radius coils. The
Helmholtz coil formula is:
H=[8.99 * 0.00001 * N * I] / R
so
H=[8.99 * 0.00001 * 64 * 20] / 91.4
=0.115072 / 91.4
=0.00125899 Teslas
Since 1 Tesla equals 10,000 Oersteds, this gives us 12.59 Oersteds
So, that means, if you want 20 Oersteds, you need to pump about 31.8 Amps
through that specific EMCO coil.
Now let's look at my above described solenoid coil, with 3.648 m
length, 0.762 m radius, and 720 turns of #12 wire. The solenoid coil formula is
(simplified for the field at mid-coil):
H= [2 * PI * N * I]/{1000 * [((L*L)/4) + (R*R)]^0.5}
= [2*PI*720*8.75]/{1000*[((3.648*3.648)/4) + (.762*.762)]^0.5}
= [39584] / {1000*[(3.327)+(0.5806)]^0.5}
= 39584 / 1977
= 20 Oersteds
(Whew! Lucky guess on the current, huh?)
So, the solenoid can generate 20 Oersteds with only 8.75 Amps, while the
Helmholtz needs 31.8 Amps. Surprisingly, a probe of this solenoid during
operation showed less than 5% field variation almost end to end.
The solenoid design proves is more efficient & easier to build than the
Helmholtz. Also yields a bigger uniform field area. Cons are more wire &
heavier construction.
--------------------------
Ed Price
[email protected]
Electromagnetic Compatibility Lab
Cubic Defense Systems
San Diego, CA. USA
619-505-2780
List-Post: [email protected]
Date: 10/01/97
Time: 13:47:41
--------------------------
