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BH Magnetization Curve problem

  1. Jan 15, 2017 #1
    1. The problem statement, all variables and given/known data

    2. Relevant equations
    Reluctance = small "L"/mu*A

    3. The attempt at a solution
    I went the route of using B/H=Mu ....since we know that B=1.2Tesla's and Mu=4pi*10^-7 we arrive at our "magnetic field intensity "H" as 954,929.7 H"

    BUT if I am trying to find Reluctance... then we have to consider the air gap...Im getting lost because the mean length would small "L" + air gap? Right?

    This should be so simple but Im missing something...
    Utilizing the magnetization curve is also confusing for me..see picture
    Thanks for your help in advance, I am stuck, willing to ask for help, and I want to learn!

  2. jcsd
  3. Jan 15, 2017 #2
    We also have upload_2017-1-15_10-56-10.png
  4. Jan 15, 2017 #3


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    Hi Joe.. Welcome to PF!

    You can get the value of relative permeability of the core for B=1.2T from the magnetization graph.
    Once you have the relative permeability of the core, all you need to do is simplify the magnetic circuit.
  5. Jan 15, 2017 #4

    I think then, if I go across from 1.2T, the Relative Perm = 6000 ?
  6. Jan 15, 2017 #5


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  7. Jan 15, 2017 #6
    Ok..So that means MUr=Mu/Mu(zero) = 6000/4pi*10^-7 =4.712x10^-4 which = Mu
    I can now go into my Reluctance formula Fancy "R" =small "L" / Mu*A to figure my Reluctance for the core

    Lcore - 52cm which converts to 0.0052 m/4.712x10^-4 x 0.0018 = 0.001986 Reluctance of Lcore
    Lgap -14cm which converts to 0.0014 m/4.712x10^-4 x .0018 = 0.005348 Reluctance of Lgap

    Add them together for Total series Reluctance = 0.001986+0.005348 = 0.007334 Total Reluctance
  8. Jan 15, 2017 #7
    flux x reluctance= number of turns x current

    flux x reluctance/number of turns =current

    Then I know the current
  9. Jan 15, 2017 #8
    once I know the current, I can multiply against the 64 ohms resistance to tell me the voltage across the gap?
    In my head this seems way too easy!
  10. Jan 15, 2017 #9
    OK=nexy questions, now my brain wheels are turning...

    Since this magnetic circuit is a series circuit, is the gap creating a Positive top part of the core, and a negative bottom part of the core! This would explain why the voltage drop is across the gap.. Am I right?

  11. Jan 15, 2017 #10


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    Right. You can see that the mmf drop Φxs is very high across the air gap.
  12. Jan 15, 2017 #11
    I see.. Yes Voltage is the Battery "E" which is applied across the winding's at 64 Ohms..

    The question for the problem= upload_2017-1-15_13-45-25.png
    Since I know the current now, and I have the resistance, I now know the voltage at the battery
  13. Jan 15, 2017 #12


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    I didn't check your earlier calculations but your steps are all correct. Check the reluctance values again. I think air gap reluctance should be very high.
  14. Jan 15, 2017 #13
  15. Jan 15, 2017 #14


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    That doesn't look right. Voltage and current are too small.
    No. You should use the absolute permittivity μ, which is μ0μr.
    This was making your reluctances very small.
  16. Jan 17, 2017 #15

    rude man

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    First thing is to realize that the permeabilty of space is << any permeability in your iron or whatever your core comprises. So compute the reluctance of the air gap, then flux x reluctance = mmf = Hi.

    If they want you to include the effects of finite core reluctance, shame on them! :smile:
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