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Bitter magnet help needed.

  1. Aug 19, 2011 #1
    alright so i think ive got a pretty good handle on this, im 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: dont 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? im a little confused.
    thankyou
     
  2. jcsd
  3. Aug 19, 2011 #2

    K^2

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    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.
     
  4. Aug 19, 2011 #3
    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.
     
  5. Aug 20, 2011 #4

    K^2

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    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.
     
  6. Aug 20, 2011 #5
    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
     
  7. Aug 20, 2011 #6

    K^2

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    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?
     
  8. Aug 20, 2011 #7
    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.
     
  9. Aug 20, 2011 #8

    uart

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    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 [itex] n = \frac{\theta}{360 t} [/itex], where t is the thickness of each disc+insulator (in meters) and [itex]\theta[/itex] is the angle of advance of the helix per disc (in degrees).
     
  10. Aug 20, 2011 #9
    That was what I was looking for. Thank you :D
     
  11. Oct 7, 2011 #10
    plate thickness: 0.0135 in
    insolater thickness: 0.015 in
    angle of asent: now im 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?
     
    Last edited: Oct 8, 2011
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