Measuring magnetic field

Hi guys:

I'm working on an application where I need a simple DC coil that provides a particular magnetic field strength. I have to experiment to determine just how strong the field needs to be. So I have wound an arbitrary coil to experiment with. I will vary the voltage on it until I experimentally achieve the necessary field in my application.

Now when I know that this coil is producing the right field, how can I measure it? If I can get a measurment of the field intensity, I can wheel-and-deal mathematically with different voltages, wire gauges and dimensions of a coil so that I can design one to meet the field requirement.

Is there a simple way to do this without expensive equipment? I suppose I could take apart my arbitrary test coil and calculate what the field must have been, working backwards, but is there a more direct, emperical way? I want to be able to say, "My application requires ______ teslas".

Don Mechanical Engineer

Reply to
eromlignod
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I don't think I explained myself very well. I'm *applying* DC voltage

*to* the coil to produce a magnetic field, like an electromagnet. The coil has a ferrous core. I just need to measure how strong this field is (polarity does not matter) so that I can reproduce the same field with a coil of different design.

I guess I could rent a gaussmeter, but it seems a shame just to take one simple measurement.

Don

Reply to
eromlignod

A coil produces voltage in proportion to the rate of change of flux passing through the hole. So, unless you have a way to spin the sensor coil around to convert the DC field to AC, it is not going to give you a signal. You might get a linear hall effect sensor like: and calibrate it with a Helmholtz coil. There are simple formulas that relate the field strength in the center of the coil versus the number of turns and the amperes passing through those turns.

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Reply to
John Popelish

If wildly high accuracy isn't needed, how about using the earth's magnetic field, your coil's field and a suitably placed compass?

Orient your coil so that it is at right angles to magnetic north, position the compass, and see how large the deflection is from "true north." For example, if your field, at a given distance was equal to the earth's field then the compass would be deflected

45 degrees. Using deflections near 0 degrees or 90 degrees would be less accurate but if you could arrange to have a deflection that was somewhere between those then this might give you an estimate of your field, be easy to repeat, and inexpensive.

You could also buy hall effect sensors, small solid state devices that are available giving an output linearly proportional to the magnetic field they are in. There are some available that are non linear or that just switch on and off, so watch the specs.

You might also search Ebay for gauss meter or gaussmeter, sometimes there are relatively cheap meters available there.

Reply to
Don Taylor

You might want to take a look at at the Honeywell/Microswitch SS94 series of linear hall effect sensors. They have them to directly measure static fields from +-500 up to +-5 Kilo-Gauss. They provide a ratiometric output. A DVM and regulated power source is all you really need to get a decent reading. The price is around $15 ea. They are a little hard to find in unit quantities, but can be found.

Mike

Reply to
Mike

Andy writes;

This is a very innovative idea and I think Don's approach is a good way for a home experimenter to check out his Helmholtz coil, if "approximate" accuracy is good enough.

There are websites which show the earth's magnetic field over the globe, with fair accuracy, and these will give you what number, in microTeslas, to use for your location.

If you can't find it, use 50 microTeslas, which is a pretty good round number for the earth's magnetic field in the US.

When playing with the compass, turn the coil's current up much higher than needed to make sure the needle will go 90 degrees to the "non energized coil" reading, so that you know you are at right angles to the earth's field. The magnetic deviation changes a BUNCH throughout the US, so an eyeball alignment may not be particularly good. Then , just back off the coil current till you get 45 degrees, and the coil field AT THE COMPASS will be equal to the earth field. Then, you can use this as a calibration point...... When you compare it to the equations ( B = uH, etc) it should work out very close to 50 microTeslas. If it is way off, then you probably used meters instead of feet or furlongs instead of fortnights, or some other such dumb error that we all do when we do it for the first time.........

Anyway, I ran an Electromagnetics Lab at Raytheon for a long time, and learned to have confidence in those weird calculation of turns, amperes, diameters and areas..... all that stuff really works, tho it's REAL EASY to make a mistake in using the wrong units.......

Good luck, and kudos to Don for suggesting a simple way to avoid buying expensive equipment.....

Andy in Texas

Reply to
Andy

Hall sensors will work for measuring the field in the air. Measuring the field inside the iron is not an easy matter, and the interaction of an iron-core electromagnet with a nearby ferromagnetic object is quite complicated--you may not be able to transfer the results to a different coil design. The basic problem is that the magnetization in the object induces a magnetization in the iron core, which induces magnetization in the object, and so on...and it's all nonlinear.

Cheers,

Phil Hobbs

Reply to
Phil Hobbs

Might be tricky, but waving another sense coil at a specified speed through the field at a specified distance and rotation, will give you a relative measurment for the other coil.

greg

Reply to
GregS

Don,

I thought your self-tuning piano was further along than this. Or is this a new project?

-Robert Scott Ypsilanti, Michigan

Reply to
Robert Scott

Hi Robert:

Yeah, this is it. It has been on the back-burner for over a year while QRS worked on another invention of mine (guitar-related). We are back in full force on the piano now. I already have over-sized experimental sustaining coils in the prototype, but now I need to trim them to a production-worthy cost. So I'm going to experiment to see what the minimum field I need for sustaining is, and then design the cheapest coil I can. Since there are over 200 of them in a piano, I hope I can keep them under $1.

Don Kansas City

Reply to
Don A. Gilmore

I'd calculate it.

Rene

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Reply to
Rene Tschaggelar

If you know the turns and the coil and core dimensions, as well as the core permeability, the calculation is pretty straightforward, giving Tesla/A or something like it. Just read off the current when the field's on the dot, this should give you enough data to calculate an optimised coil. The formulas are out on the net somewhere if you don't already have a handbook.

- YD.

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Reply to
YD

Of course. But it is much easier for me to measure and vary the

*voltage* in my experiment. What current is passing through my test coil is immaterial to me if I empirically measure the field anyway. And if I calculate the field, it is still simple to do using the voltage.

Don Kansas City

Reply to
eromlignod

I had a feeling this was coming. I neglected to mention that it was low-frequency AC to avoid the inevitable admonishing pep talk about impedance. The highest frequency that I will encounter is about 4000 Hz, the lowest is about 27 Hz.

Don Kansas City

Reply to
eromlignod

Well, here's my latest setup. The sustainers are all independent. Each piano string passes just above a tiny optical fork sensor (photo-interrupter). The coil is really just an electromagnet, like in a relay. When activated, it pulls the steel string down into the beam of the fork sensor. The fork sensor interrups the current to the coil.

So it acts sort of like a relay buzzer. Each time the string is pulled into the beam, the switch turns the coil back off and releases the magnetic pull on it. Then it bounces back out of the beam and the cycle repeats.

My large coil takes enough current that I need an intermediate transistor, since the little fork switch can't handle it alone. I would like to reduce the power consumption of the coil so that I can switch it with the sensor directly. Obviously, I would also like to reduce the coil cost.

Don Kansas City

Reply to
eromlignod

1) It is the *current* that determines the magnetic field. 2) Look in a good physics textbook on electromagets and magnetic fields; you will find equations to calculate the magnetic field in the center of the coil.
Reply to
Robert Baer

Yabbut, you said elsewhere that

1) there are over 200 coils, and

2) you aren't worried about the current.

ISTM you're gonna have to start worrying about "minimal" impedance effects per coil adding up and affecting the common power supply for all those coils. You don't want your piano sounding like a motorboat with a drunken pilot.

Mark L. Fergerson

Reply to
Mark Fergerson

Precisely. And make the length of the coil equal or longer than it's diameter. Did that once and I could almost calibrate my gaussmeter with it.

Meindert

Reply to
Meindert Sprang

eromlignod wrote: (snip)

Are you using the pull of one end of a solenoidal coil? If so, you can greatly increase the force by putting the coil along side the string, and use the poles at both ends to attract the string. This gets rid of a lot of air in the magnetic field path. Even better would be to use a small E core lamination stack with the center pole and two other end poles all lined up along the string.

Reply to
John Popelish

Almost any size coil can be made to sustain oscillation, depending on what circuit is used for feedback, on how close you can position the coils to the piano strings, and depending on what stability you want. Since you are multiplexing many coils to one sustaining circuit, you can afford to put more cost into the sustaining circuit if it means a reduction in cost of the coil. But if the desgn of the sustaining circuit is already cast in concrete, then I understand your search for the lowest-cost compatible coil.

-Robert Scott Ypsilanti, Michigan

Reply to
Robert Scott

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