Field Intensity Formula: Calculating Bitter Magnet Strength and Temperature

[microfracture]

alright so i think I've got a pretty good handle on this, I am looking for a proof read though. any acurate input is greatly appreciated :D

Field intensity in teslas = (amps x turns)/coil area; ignoring coil length effects

meaning

B=μ0(Ni)/m^2

right? but my question is: don't i have to add the thickness of the plate too? or is that just for resistance...

ps, anyone know how to calculate how hot the magnet will get per amper? I am a little confused.
thankyou

 

[K^2]

Not sure which plate do you mean. If you mean thickness of the wires, yes, that will only affect resistance. The amount of heat generated is I²R, where R is the resistance of the coil. How hot that will actually get the magnet depends on how quickly you can get the heat out.

If you simply want to estimate how the temperature will change with current, heat is radiated at a rate proportional to T⁴. So if radiation is your primary way of cooling the magnet, you can expect temperature to increase as the square root of the current. The actual coefficient will depend on configuration of the magnet, air flow, etc. It's going to be very difficult to predict in advance.

 

[microfracture]

A bitter magnet is composed of stacked plates, not wire. I was asking if I have to take the plates thickness into consideration when calculating B field.

 

[K^2]

Ah, I see. No, it doesn't matter. What matters is how much current flows through a plane you cut along the solenoid, and that's just the current times number of turns it makes. Be careful that the number of turns the current makes is not equal to number of plates. You have to take the way insulator is stacked into account.

 

[microfracture]

Im going to use teflon sheets between the plates. At low voltages it should be fine. I should be more concerned with heat honestly, considering the high ampres ill be pumping into this thing

 

[K^2]

You have cooling pipes running through the plates? I'm assuming that's typically how it's done, though, I've never worked with these. If so, it might be possible to estimate the heat flow out of the thing, and therefore, get some sort of an idea of how hot it will get.

Edit: In fact, if you don't mind sharing details on geometry, it might make things easier. Copper plates, right?

 

[fleem]

If you are pulsing it, then you also need to consider the skin effect. If the pulse is fast enough and plates thick enough, they will repel all the field lines and none will go through the plates.

 

[uart]

alright so i think I've got a pretty good handle on this, I am looking for a proof read though. any acurate input is greatly appreciated :D

Field intensity in teslas = (amps x turns)/coil area; ignoring coil length effects

meaning

B=μ0(Ni)/m^2

right?

No, that doesn't look correct to me (not even dimensionally correct).

I really don't know anything about "bitter" electromagnets, but just googling it and looking at the construction I say it would be something more like :

B_teslas = μ0 n I, where "n" is the number of helical turns per meter.

So $n = \frac{\theta}{360 t}$, where t is the thickness of each disc+insulator (in meters) and $\theta$ is the angle of advance of the helix per disc (in degrees).

 

[microfracture]

That was what I was looking for. Thank you :D

 

[microfracture]

plate thickness: 0.0135 in
insolater thickness: 0.015 in
angle of asent: now I am not sure how to calculate angle, so i took a swing at it. thickness of plate plus insolater is 0.029in

0.029×0.0254 (inches to meter) = 0.0007366m = t
right?
we will call that the hypotnuse of a right triangle, then have 3in as the radi which is 0.0762m. get 0.08m as the top leg

and 0.55 degrees as the angle of advance.
so n=0.55/360(0.0007366) = 2.0741 rounded.
now we have n.
u0 is 1.2566371×10−6

B= 1.2566371×10−6(2.0741)I
so if i want to solve for I
I= [1.2566371(2.0741)]/B
right?