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Normal/Tangential magnetic flux density problem

  1. Apr 23, 2010 #1
    [PLAIN]http://img232.imageshack.us/img232/7172/60746674.png [Broken]
    1. The problem statement, all variables and given/known data
    The picture attached to this question shows a magnetic slab with μr = 50. A thin conducting film (with μr = 1) lies on top of the slab and carries a surface current of 1.0[A/mm] directed out of the page. If magnitude of B1 = 0.01 [T] and θ1 = 10o, find magnitude of B2 and θ2.

    I apologize in advance for my bad recreation of the image in the text!

    2. Relevant equations

    1.) B1n = B2n
    2.) (1/μ1)B1t - (1/μ2)B2t = Js

    3. The attempt at a solution

    So, for starters, I am assuming μ1 = μo = (4*pi)e-7. Please let me know if I cannot make this assumption, but I believe I can. Because angles are involved that are not perfectly 90o, I have to include sines and cosines in the equations above. Using formula #1 --> (0.01)*cos(10o) = B2cos(θ2) = 0.009848. Using formula #2 to find θ2, I can plug in values to have:
    (1/μo)0.01*sin(10o) - (795774.72)B2sin(θ2) = 1000[A/m]

    Solving for θ2, I find that the angle is 90o. However, this is not the answer in the back of the book. The answer for θ2 in the back of the text is 86.2o and a magnitude of B2 to be 0.15[T]. I was also not receiving this value for B2. I have a feeling that my problem is that I cannot assume that μ1 = μo. Please help me! Thanks for all help in advance!
    Last edited by a moderator: May 4, 2017
  2. jcsd
  3. Apr 23, 2010 #2
    I believe that you have everything essentially correct with one small, but important, exception.

    You need to reverse the sign of the current density in your equation. If you look at the orientation, you have set up your equations as if the current was flowing into the page, rather than out of the page.
  4. Apr 23, 2010 #3
    uR=50 so u2 is not equal to u1.
  5. Apr 23, 2010 #4
    Actually, I see one more error. I think you multiplied [tex] \mu_0[/tex] by 50 squared (in region 2) rather than by just [tex]\mu_r=50 [/tex] as specified.
  6. Apr 25, 2010 #5
    Elect_eng - this still results in an angle of 90o.

    Antiphon - when did I mention this? Can I assume that u1 = uo? That was one of my original questions.

    Elect_eng - I carefully went back over my implementation and I still receive 90o
  7. Apr 25, 2010 #6
    Here is my implementation:

    1.) B1n = B2n
    2.) (1/μ1)B1t - (1/μ2)B2t = Js
    B1cosθ1 = B2cosθ2 = 0.01cos(10o) = 0.009848
    (1/μo)B1sinθ1 - (1/μr)B2sinθ2 = -Js
    (1/μo)B1sinθ1 + Js = (1/μr)B2sinθ2
    ro)B1sinθ1 + μrJs = B2sinθ2
    119092.41 = B2sinθ2

    [B2sinθ2 / B2cosθ2] = (119092.41/0.009848)
    tanθ2 = 12,093,056
    θ2 = tan-1(12,093,056)
    θ2 = 90o
  8. Apr 25, 2010 #7

    The permeability in region 2 is [tex] \mu_0 \mu_r[/tex] not [tex] \mu_r[/tex].
  9. Apr 25, 2010 #8
    That was the problem. Thanks for your help and persistence Elect_Eng!
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