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Magnetic field vector due to linear conductor

  1. Sep 13, 2015 #1
    1. The problem statement, all variables and given/known data
    Through linear conductor flows current I, with direction shown in the picture. Axis where conductor is placed is common edge of three areas with different ferromagnetic materials. They form angles θ1, θ2, θ31 + θ2 + θ3 = 2π). If space is filled with homogeneous materials with permeabilities μ1, μ2, μ3 respectively, find intensity of magnetic field vector for arbitrary point in the space due to linear conductor.

    slika za forum.jpg

    2. Relevant equations
    Generalized Ampere's law.

    3. The attempt at a solution

    This is my attempt:

    Considering the direction of the current, it's obvious that intensity of magnetic field vector depends on the distance from the linear conductor. Since the vector B is normal to the boundary of two materials, from the boundary conditions we have that B is same for any point with same distance from the conductor regardless of the material, which means that H is different in different materials.

    Using generalized Ampere's law we have:

    H*dl=∫H*dl*cos(H,dl)= H11*r + H22*r + H33*r

    ∑I=I


    If we express H2 and H3 as B/μ2 and B/μ3 respectively we get the expression for H1 and we can do similar thing for H2 and H3.

    so,

    H1=(I-B*r(θ22 + θ33))/θ1*r

    Is this correct, and is this reasonable approach?
     
  2. jcsd
  3. Sep 13, 2015 #2

    mfb

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    You still have H1 and B, one of them has to disappear for the final result.

    The approach looks fine so far.
     
  4. Sep 14, 2015 #3
    I know what you mean, i forgot to express B in final formula, which i would do by expressing H1, H2 and H3 in first formula using B divided by corresponding permeability that way i could easily find B since it's the same for any point regardless of material.
     
  5. Sep 15, 2015 #4

    rude man

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    Not sure what the OP said in post 3, but the expression for Hi (or Bi) should have nothing but I, θ1, θ2, θ3 and r in it. No H's or B's. If this is what the OP meant I have no additioal comment.
     
  6. Sep 16, 2015 #5
    I wasn't clear enough, sorry, anyway, this is what i meant:

    Screenshot_3.jpg
     
  7. Sep 16, 2015 #6

    rude man

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    Fine! I thought that's what you meant but wasn't sure.
     
  8. Sep 16, 2015 #7

    mfb

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    The last expression for H1 can be simplified significantly (e.g. starting from the expression for B).
     
  9. Sep 22, 2015 #8
    I know, i should get much "smaller" expression, i could do that, but the most important thing for me was to see if i solved it correctly.
     
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