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# 2 Tesla Electromagnet?

I need ideas for a 2 tesla electromagnet. I am trying to make the strongest magnet possible without additional cooling from a fan..

I don't know whether or not I can use high voltage, and low amps, or High amps, and low voltage. Or even inbetween..

Also, throw me that calculation for the electromagnetic strength. And if you could be awesome, explain to me how to use that formula, as I am not mathematically inclined.

Thank you.

The remanent field of a neodymium magnet is typically about 1-1.4 tesla,

http://en.wikipedia.org/wiki/Neodymium_magnets

and permanent magnets do not require cooling fans.

You should seriously consider Nd magnets, if that will work for you.

If that's not going to do it for you, then you should take a look at this graph:

http://en.wikipedia.org/wiki/Saturation_%28magnetic%29

The actual graphic is here:

http://en.wikipedia.org/wiki/File:Magnetization_curves.svg

And looking at that made me wonder: Is steel really the best ferromagnetic material out there? Is there any other ferromagnet that saturates at higher value of B? And I found a link to this:

http://www.consult-g2.com/papers/paper11/paper.html

which includes a saturation curve for something called "vanadium-permendur", whatever that is.

But steel (of some kind) is almost as good, and probably easier to come by.

Now you might be wondering about B and H? What are those quantities? Well,

I think those refer back to the ideal long solenoidelectromagnet, described here:http://en.wikipedia.org/wiki/Solenoid

And the formula for the field inside that ideal air core solenoid is just

B = μ0*(N*I/l) = μ0*I*(N/l) = μ0*H

where l is the length of the long solenoid, N is the number of turns, and I is the electric current through each turn. Because in the mathematically ideal case this solenoid is infinitely long, that lowercase l just refers to a short length of the whole thing, and N is the number of turns in that short length, and the B field inside that length.

H is the product of both I and (N/l), and it is called the "applied field" or the "laboratory field", and basically what that means is H is sort of the result of electric currents flowing through wires which the experimenter controls directly. H has units of "ampere*turns/meter" in SI.

H is sort of thought to cause B. The experimenter throws a switch, current flows through some wires, forming H, which in turn causes B to show up in the space around the wires.

For an air core solenoid, the relationship H and B is just B=μ0*H.

But a solenoid with an iron core, its a lot more complicated. Sometimes you'll see the formula

B= μ0*μr*H

But that's kind of a lie, or only approximately true for that region before the ferromagnetic material saturates. The truth is in the saturation curve (B vs. H). The approximation is the straight line part of that curve starting at the origin (where H=0 and B=0) and extending through low values of H. And that's the meaning of those saturation curves I linked to previously.

From those curves you can see that the H needed to magnetically saturate steel is around 100 to 200 ampere turns per inch, and that sounds do-able. I mean I can sort of imagine a one inch length of electromagnet with 100 turns wrapped around that one inch length, with one ampere flowing through each turn,and it seems believable that the field inside the steel core (again in that 1 inch section) would be something in the range of a tesla or two. Supposedly you could do the same trick with 10 turns per inch, with 10 A of current, or 1 turn per inch and 100 A of current.

Regarding heat, that's just ohmic heating, and you can calculate that in terms of power as P = I

^{2}*R, where R=ρ*L/A, where this L is the total length of your wire. A is its cross-sectional area, and ρ is bulk resistivity. You'll probably be using copper magnet wire, and actually there are tables for resistance per unit length for different AWG sizes of copper wire, here:http://en.wikipedia.org/wiki/American_wire_gauge#Table_of_AWG_wire_sizes

Also it will turn out that the total volume of wire wrapped around the magnet will depend on its length and width. Assuming you pack the turns

squarelyon top of each other, then this total volume is L*d^{2}, where d is the diameter of the wire(and its insulation).Moreover you can tie these equations back to N, and (N/l), and H. Using Ohm's law, the total voltage across your magnet is V=I*R, and sort of putting all these equations together you can basically choose the DC voltage you want to run it at, and that determines the gauge of the wire. Or vice-versa. You pick a wire size, and then that determines the DC voltage. This is all assuming you had some target for H, in ampere turns per meter, or per inch, or whatever, and also for the total volume of the magnetic field itself.

Sorry for the hand-waving. I don't remember the exact formulas, but they are easy enough to derive.Regarding the long solenoid approximation, for solenoids that are not long, and for the places where the magnetic flux goes from steel to air, the field is lot lower in those places. As you'd expect, the field is strongest in the steel itself in the most central part of the core.

The Florida Magnet lab has watercooled magnets, with Florida-Bitter coils that can manage that kind of flux. They consume several kilowatts of electrical energy, and the same in cooling plant.

if you want to make a strong magnet, do a in between everything but about 5 amps. to make one, first make a small winding around anything that is magnetic. now cut the wire and wrap tape around it. now add a bigger coil and add cut the wire and add tape. keep doing this until you think it is enough. oh and make sure you always wind in one direction, not one coil it's clockwise and the next is counter clockwise... you should also note each one with the poles (+/-) so to start, maybe you should just make the windings clock wise and make the inner coil positive and the outer one negative. try to get 15 gauge copper wire (thin insulation) this might get hot but it's ok since you copper is insulated so just put it in a tub of water. hope this helps!

p.s i am 11 but i have very bad grammar

Did you even

tryusing Google or Wikipedia? The latter might even explain to you how to use that formula, but I can give you a hint -- it involves multiplication and division.I am not sure you're going to build a 2 tesla magnet given your implied level of expertise. The particle physics detector which I supervised for ten years (BaBar), had a 1.5 T solenoid. It was a superconducting niobium titanate design, with the "wires" (really giant square cross-section conductor) carrying 4.6 kA (yes, that's "kilo", as in "it's going to kilo you if you're not careful).

Yes.. FYI... I have already tried that! I wanted to back up what I have learned. I have accomplished a nearly .5 tesla coil. So why can I not build a 2 tesla coil?

And I already have calculations that I am using, so that's why I even asked this question. I am using 2 tesla's just for the heck of it, if I can figure out what I am doing.

The problem is that you're going to be limited by the structural strength of your materials, and by Ohm's law. The latter is probably the more limiting case, since the power dissipation goes like I

^{2}, while the field scales linearly with current.What I've read is that somewhere between 1.5 and 3 T is the limit for sustained fields using "regular" designs. Ultra-high field magnets are either pulsed, or have some special materials and engineering to deal with the stresses involved.