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XY plane as the interface between two media

  1. Nov 19, 2017 #1
    1. The problem statement, all variables and given/known data
    The xy-plane serves as the interface between two different media. Medium 1 (z < 0) is filled with a material whose µr=6, and medium 2 (z > 0) is filled with a material whose µr=4. If the interface carries a current (1/µ0)ˆy (y- hat) mA/m, and B2 = 5ˆx (x-hat) + 8ˆz (z-hat) mWb/m2 , find H1 and B1.

    2. Relevant equations

    H = 1/μ0 B - M.
    Habove - Hbelow = Kf x ^n (n-hat)


    3. The attempt at a solution

    I know I am not doing this right because it seems to simple and my linear algebra isn't the best (I forgot some of the concepts) but this is what made sense to me:

    H2 = 1/(μ04) [5 ^x + 8 ^z]

    Sub that into the second equation.

    so

    1/(μ04) [5 ^x + 8 ^z] - H1 = 1 ^y
    =>
    1/(μ04) [5 ^x + 8 ^z] - 1/(μ06)B2 = 1 ^y

    I change it into vector coordinates

    1/μ0 (5/4, 0, 8/4) - 1/μ0 (x/6, -6/6, z/6) = 1 ^y

    Then solve for x and z.

    I get H1 = 1/(μ0) (7.5/6 ^x - 1 ^y + 12/6 ^z)
    and B1 = 7.5 ^x - ^y + 12^z


    Am I going about this right or am I all the way in Mars right now?
     
  2. jcsd
  3. Nov 19, 2017 #2

    TSny

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    Homework Helper
    Gold Member

    Your result for H2 is correct, but it isn't clear how you got it. It does not follow immediately from either of your equations in the "Relevant equations" section.

    This isn't right. Your relevant equation ##\mathbf H_{\rm above} - \mathbf H_{\rm below} = \mathbf K_f \times \hat {\mathbf n}## is not correct in general. (Even if it were correct, the right hand side does not have a direction in the ##\hat y## direction.)

    The correct equation is generally written as shown here https://en.wikipedia.org/wiki/Inter...terface_conditions_for_magnetic_field_vectors
    See the second equation.
     
  4. Nov 20, 2017 #3
    Equation 1 of my "Relevant Equations" is how I got H2. I just inserted B2 into the equation.

    And thank you! That link makes a lot of sense.

    I was wondering if I was going in the right direction in terms of subtracting the vectors to get zero. Because from the link you posted, I need to find B1 and then subtract it from B2 to get zero? Because from there I can find H2. Am going about this the right way?
     
  5. Nov 20, 2017 #4

    TSny

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    Homework Helper
    Gold Member

    Your equation 1 has the magnetization M. So, it is not immediately obvious how you dealt with this term. After substituting B2 into the equation, what did you do with M?

    I would need to see the details of your attempt. The equation to use is ##\hat {\mathbf n} \times \left( \mathbf H_2 - \mathbf H_1\right) = \mathbf K##, where ##\hat {\mathbf n}## is the unit normal vector pointing from medium 1 to medium 2. Consider separately the x and y components of this equation.
     
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