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LAN_403

Signification

Thu Feb 12 2015, 01:42PM

Registered Member #54278 Joined: Sat Jan 17 2015, 04:42AM
Location: Amite, La.
Posts: 367

I was wondering how many coil-winders (human) were familiar with the Brooks Coil. I recently learned of this 'inductor' and have been working with it for a while now. A Brooks coil is an inductor wound in such a manner as to produce maximum inductance for a given length of wire. Specifically, it is a coil wound on a circular spool of diameter 2c. The length of the winding is c. The thickness of the winding is also c! . Note the winding has a square cross section. The total OD of the coil is 4c. "c" is the Brooks constant.

I discovered that many of the inductance (L) and field magnetic strength (B) formulas I used, before I knew of the Brooks configuration, can now be greatly simplified--some to such a degree that their units acquire intuitive meaning!

Although there remains much to be said of the many Brooks formula variations, I am wondering about it's application in maximizing (or minimizing) quantities in B-fields combined with inductance characteristics of the coil. In particular the, as far as I know un-named unit, would describe "magnetic field per inductance" or: "Teslas per Henry" [T/H] should be very useful--and one which, if I did not cancel out something I shouldn't have, has the very interesting basic units of "Amperes per square meter" This is just CURRENT DENSITY! (NOTE: I edited this last bit--it erroneously read "Charge Density")

Sulaiman

Thu Feb 12 2015, 03:25PM

Registered Member #162 Joined: Mon Feb 13 2006, 10:25AM
Location: United Kingdom
Posts: 3140

I have wound multilayer coils of 'Brooks' dimensions but more frequently slightly different.
if you wind a multilayer coil with near Brooks coil ratios the result is quite similar,
but, since this is the electromagnetic projectile accelerators section I'm going to assume it's for a coilgun etc.,
if so then there will be a ferromagnetic core and the 'Brooks' ratios no longer are optimum.

I may try manipulating units just out of interest because Coulombs per square meter
does indeed seem a peculiar result.

But then in SI units, the Henry reduces to kg.(m/s.A)^2
which is far from intuitive to me.
I did notice that is kg x area/(coulombs squared)
so you are probably correct or close.

klugesmith

Thu Feb 12 2015, 04:20PM

Registered Member #2099 Joined: Wed Apr 29 2009, 12:22AM
Location: Los Altos, California
Posts: 1716

The Brooks coil came to my attention a few years ago when we were making a LISN (50 amp 3-phase), but for that project we ended up using single-layer solenoids.

Your mention of B-field formulas reminded me of a small web site "devoted to the vanishing art of practical magnet design without FEA (Finite Element Analysis)."

and in particular, the page called "On-axis fields due to round, finite, air core solenoids".
It's not too cumbersome to set up those formulas in a spreadsheet calculator program,
and is straightforward in computer algebra program like Mathcad.

As Sulaiman says, the field and inductance will be very different if a ferromagnetic core is present, such as in a typical coilgun. Accelerating force comes from the slope of inductance vs. position curve, and (as always) is in the direction which increases the inductance. One suitable and free FEA program, often mentioned on this forum, is FEMM.

Signification

Fri Feb 13 2015, 12:36AM

Registered Member #54278 Joined: Sat Jan 17 2015, 04:42AM
Location: Amite, La.
Posts: 367

First, as stated, I was "close" but not correct-a dumb mistake: A missed 's' in the denominator created a Coulomb/sec which is "CHARGE DENSITY" where if that 's' was noted, would have given Ampere/sec which is "CURRENT DENSITY (J)". So the unique (corrected) result we get here (concerning coils) is: B/H = A/(m^2) OR a Tesla per Henry = an Amp per square meter OR, in a coil **MAGNETIC FIELD STRENGTH PER INDUCTANCE EQUALS CURRENT DENSITY** This seems very intuitive when you think of an inductor.
---Now we just need a 'name' for the ratio "Tesla / Henry" in this situation.

Klugesmith, I still work without FEA, but it seems I am being forced into it--especially since my old programming languages like Quick-BASIC and GX-Graphics don't like windows (and I sure do miss 'Debug'). One little thing concerning Brooks I think worth mentioning: a couple of years ago I derived that formula in the page you sent for B on the axis of a multi-layer solenoid, the formula is somewhat simpler, but still WAY to messy to type. If you assume this to be a Brooks type, things **really** simplify...in particular in a multi-layer Brooks solenoid the maximum B (which is central axis / dead-center) is simply: Bmax=uo*N*i / 3.1w [Teslas]; where "w" is the *Brooks dimension*

Klugesmith wrote:
"The strip coil will behave exactly the same as a wire coil of the same metal, if they have the same ID, OD, length, and number of turns."
I have been giving your words plenty thought and have a question here: Concerning copper wire vs thin copper sheet, if just ONE turn is considered then we can use the ratio of the area of a circular cross section of wire of radius "a" vs a square cross sectional area of square wire of side length "2a". This will reveal the difference in a coil with air gaps (wire) vs one that is solid (sheet or square wire). Picture the circle inscribed in the square. Now since the area of the wire cross section is Pi*(a^2) and that of the square wire is (2a)^2 the ratio of areas is 0.785 (78.5%). This implies ~21.5% less 'metal' in the wire-wound coil than in the sheet coil. I calculate this percentage as MUCH smaller in coils of many turns of small gauge wire and the small gap between sheets, hence the quote: "The strip coil will behave exactly the same as a wire coil" Of course, I may (probably) be overlooking something very important that makes
your statement exact.

I have also given much thought to what you said next"...

"And as I've said many times before:
Within that coil volume, you can trade off the wire area against the number of turns, with NO effect on magnetic field strength and duration.
Provided that you make a complementary trade off between capacitance and voltage, and have the same stored energy."

I have been trying to create an illustrative example of this--I *REALLY* would like to get a working knowledge of what you say here. I based an illustrative situation --very loosely-- to an old thread entitled "Just finished my 3.5kJ cap bank & RL test" posted in this forum by "FastMHz" (2/23/06). I imagine the setup (mental experiment) like so:

Suppose you have a coil of 8 turns wrapped around a 1 inch form energized by a capacitor rated at 7200uF @ 700V = 1764 J. He now wants to 'double' the magnetic field and keep the current the same. To accomplish this goal, he winds a second coil on the same size form but at twice the winding length (16 turns), he adds another identical capacitor in series with the 1st one for a capacitor bank of 3600uF @ 1400V = 3528 J rating. NOW, his concluding statement is:

"...So the new setup has twice the coil, half the microfarads, twice the voltage...." he also states the current is kept the same as before.

I assume that his thinking is to double the capacitor energy he must halve the capacitance (since he doubled the voltage) based on the formula for the energy stored in a capacitor:

E=(1/2)*C*(V^2);

In summary: (assume both coils use identical core and wire)
1st coil: Eight turns of wire energized by a 7200uF @ 700V capacitor bank (E=1764 J)
2nd coil: (Required to give twice the B-field with the same current and twice as many turns) that comes to16 turns of this wire energized by a 3600uF @ 1400V capacitor bank (E=3528 J)

Klugesmith, is your statement above concerning **trade offs of wire vs turns in a certain volume AND capacitance vs voltage** in this case, NOT EQUAL to the 1x the initial energy but to 2x the energy, since the new requirement in this example is 2x the original B field? I am assuming this to be the case.

...HOLD THE PHONE...another thought just came to mind that changes things: Since the energy in the B field is proportional to the SQUARE of this field strength (B^2), and in the case above where the B Field is doubled--thus the energy quadrupled--, wouldn't this imply a capacitor bank with now twice the energy just assumed?? This would require a capacitor bank with V=1400, E= 3528 x 2 = 7056 J, Thus: C=7200uF. The final capacitor bank is 7200uF @ 1400V = 7056 J.

Sulaiman,
Here is how I make sense out of non-intuitive units: Take the "TESLA" for example. The units of a Tesla are based on the force of a charge with a velocity in a magnetic field and are usually written as [N*s/Q*m] or Newton seconds per Coulomb meter--this is NON-intuitive--to say the least!!. Divide this up as: force on a charge (N/Q) for a velocity (m/s) and you get: 'Newtons per Coulomb' PER meters per second [(N/Q) / (m/s)] which says that the faster the charge (Q) moves (m/s) in a magnetic field (Teslas), the greater the force, Newtons (N), it feels. So, it should be intuitive that in a magnetic field of one Tesla, one coulomb of charge feels a force of one Newton for every "meter per second" of velocity it moves.

For inductance (L) the unit of a Henry (H) is [Vs/A] or "Volt seconds per amp" Thinking as above: re-write this as : "Volts PER (Amps per second)". Briefly, if you have an inductance of one Henry in a circuit, then if you increase the current through the coil at "one Amp per second", there will be one Volt (steady) across the coil. So, for every Henry of inductance in a coil you will read a Volt across it for every Amp you drive through it per second (Amp per second).
...hope this helps...

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