Calculating flux densities via induction - diyAudio
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Old 12th September 2005, 11:23 PM   #1
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Default Calculating flux densities via induction

Hi,

Okay I have a quick quiz question!

I have coils of wire formed into inductors. Usually the inductors act as a sense coils, picking up changes in magnetic flux. I'd like to be able to make a rough guess at the actual flux density that the coils are detecting rather than just quoting a floating value of induced voltage differential.

Properties I know about the inductors, their;
Inductances
Iron / steel cored
Lengths
Number of turns
Cross-sectional areas, although they're not circles
Frequency of the flux changes they're sensing
Range of voltages created across it during use
Approximate DC resistance
Q factor
That the flux density has a permanent value around a few hundred to a thousand gauss and is changing in a roughly sinusoidal way over the inductors

From some combination of those factors, preferably the simplest, I'd like to be able to take a guess at the actual flux density changes.

Actually, if anyone knows of any Excel sheets for magnetic calculations that'd be excellent!
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Old 13th September 2005, 06:28 PM   #2
Limhes is offline Limhes  Orkney Islands
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You don't look familiar with Maxwell's beautiful equations, with which you can solve this perfectly well:

With a closed loop you can measure the flux change per time unit through the loop. The formula is as follows:

Click the image to open in full size.

If you now assume that the flux doesn't change over the core's surface, then you can change the first integral into a simple multiplication, leading to:

E*S = -dF/dt

In which E is the electrical field in the conductor, equal to V'/epsilon (V' = voltage on one turn; epsilon = dielectric constant of the core), S is the cross-sectional area and F is flux. With an N-turn coil, you get a voltage of V = N*V'.
Now you can calculate the flux change with:

dF/dt = - (V*S) / (N*epsilon)

In which V is the voltage you actually measure. Ofcourse, with a simple Miller integrator or something you would be able to calculate 'real' flux values, assuming you know the starting flux. Makes a nice summer holiday project

I have no Spice at hand, so here's some Paint art:

Click the image to open in full size.

You'll still need some proper timing and decharching circuitry, but the idea is there...
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Old 13th September 2005, 06:52 PM   #3
Limhes is offline Limhes  Orkney Islands
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Okay, I'm not very familiar with the forum yet, so I'll try to put the images right this time.

One of Maxwell's equations:
Click the image to open in full size.

And the state of the art schematic:
Click the image to open in full size.

-------------------------
So you can't edit your posts and the censoring mechanism even works on URLs?
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Old 13th September 2005, 07:21 PM   #4
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Thanks for your help Limhes,

I've seen these formulas before but since maths isn't one of my best subjects I always need to check, otherwise I end up following mistakes round for days on end!

By dieletric constant do you mean the electronic capacitive dielectric constant or the magnetic permeability of the core? It's just that you mentioned core rather than coil dielectric.

Also, does it matter if the cross sectional area of the coil is distorted from a perfect circular loop. Say if it were a square form.

What would the units be with this, area in metres squared and flux in Teslas?
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Old 13th September 2005, 08:08 PM   #5
Limhes is offline Limhes  Orkney Islands
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Quote:
Originally posted by eeka chu
[B]By dieletric constant do you mean the electronic capacitive dielectric constant or the magnetic permeability of the core? It's just that you mentioned core rather than coil dielectric.
Ah, I think I mean the coil dielectric, ie the dielectric constant (most of the time it's the greek letter epsilon). It's just that I don't know the terms in english
Quote:
Also, does it matter if the cross sectional area of the coil is distorted from a percect circular loop. Say if it were a square form.
No. Maxwell's equations are perfect for every case you can ever imagine . So you can perfectly use a square...
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Old 13th September 2005, 09:22 PM   #6
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The voltage generated in a loop of wire in volts, equals the -rate of change of the flux through the loop (delta BA).


If you have 10 turns,, 1 meter square, and the flux changes at 1 tesla per second, the voltage will be -10 volts.

1 cm by 1 cm...area=10e-4 meters. 1 tesla per second...delta BA is 10e-4 Tesla meter2/ second..

one turn, -100 uvolts..100 turns, -10 millivolts.

Smaller the radius of the coil, more accurate it will show the field due to field divergence, but the lower the voltage produced. Tradeoffs, always tradeoffs..

John
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Old 13th September 2005, 10:19 PM   #7
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Would that be;

turns * area * (delta flux / time) = delta voltage?
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Old 14th September 2005, 09:16 AM   #8
Limhes is offline Limhes  Orkney Islands
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Ah, wait, I am going to correct myself though

I have used the formula in a wrong way, sorry! I just took the first I could find using google, which was in some strange form I think.

Using it correctly leads to:

V = - dF/dt * epsilon / l

In which dF/dt is the flux change per second, epsilon same as above and l is the total conductor length in the inductor (i.e. in which the electric field is present). The cross-sectional area is of no importance since the flux density is already transformed into flux.

The flux change then equals:

dF/dt = - V*l / epsilon
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Old 19th September 2005, 02:13 PM   #9
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Quote:
Originally posted by eeka chu
Would that be;

turns * area * (delta flux / time) = delta voltage?
Yes.

Area is in square meters

delta flux/time is tesla per second

delta voltage should just be voltage..

Cheers, John
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