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Relationship between line current & surface current density

  1. Jul 24, 2017 #1
    1. The problem statement, all variables and given/known data
    In a configuration having axial symmetry about the z axis, a line current I
    flows in the −z direction along the z axis. This current is returned at the
    radii a and b, where there are uniform surface current densities Kza and
    Kzb , respectively. The current density is zero in the regions 0 < r < b, b <
    r < a and a < r.

    (a) Given that Kza = 2Kzb , show that Kza= I/π(2a + b).
    (b) Show that H is:
    -I/2πr for 0<r<b
    -Ia/πr(2a+b) for b < r < a

    2. Relevant equations

    I = lim|J|→∞ A→0S J ⋅da (1)
    2 π r K⋅ in≈ ∫S J⋅da (2)
    cH⋅ds = ∫S J ⋅da (3)
    Kza = 2Kzb

    3. The attempt at a solution

    To get Kza, I used ∫S J ⋅da on the region a < r.
    Because there is not current density on this region, the only contribution to the integral comes form the surface current densities, using (2):

    S J ⋅da = 2π( b Kzb + aKza )

    Substituting the value of Kza:

    S J ⋅da = 2πKzb( b + 2 a)

    As the problem is talking about Line current and surface current densities, I am assuming is a tick wire, but that is length is much bigger than its thickness. Therefore, (I think) I can equate (1) and (2)

    2πKzb( b + 2 a) = I

    Kzb = I/2π(2a + b)
    Kza = I/π(2a + b)


    For answer (b), On the region r < b, the only contribution to the magnetic field comes from -I, therefore:
    cH⋅ds = - I
    H = -I/2πr i for r < b

    For the region b < r < a, the contribution to the integral should come from the surface current density for r = b and I, therefore:

    H = (bKzb/r -I/2πr) iφ for b < r < a
    H = (bI/2π(a+b)r -I/2πr) iφ for b < r < a

    I don't understand why for b < r < a Kza should be taken into account, because r doesn't necessarily have to be near a
     
  2. jcsd
  3. Jul 26, 2017 #2

    rude man

    User Avatar
    Homework Helper
    Gold Member

    Having a hard time seeing the setup.
     
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