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Magnetic flux through triangular ring

  1. Nov 9, 2013 #1
    1. The problem statement, all variables and given/known data
    Very long wire carrying current I is surrounded by a brass ring of a triangular cross section. (figure attached)

    Show that ψ =
    Code (Text):

    μ° I h
    ------   (b - a ln ((a+b)/b)
      2∏b
    2. Relevant equations
    A = (μ°I/2∏ * ln x) az (according to one of the solutions, where x is the distance between wire and surface)

    ψ = ∫A.dl



    3. The attempt at a solution
    First attempt:
    My first approach was to integrate the magnetic vector potential over the volume of the the ring, I defined the distance from the wire to the surface as:
    d(x,z) = a + b(z/h) + x {0<= z<= h and 0<=x<=b(1 - z/h)

    where
    z is the rise in the z axis
    x is the distance through the surface of the ring (i.e. distance penetrated through the surface)


    then I took ψ as a triple integral of
    z from 0 to h
    x from 0 to b(1-z/h)
    θ from 0 to 2∏,

    of °I/2∏ * ln d(x,z)) az * dxdydθ

    but I got a wrong answer

    Second attempt
    I took only the surface in consideration, i.e. removing all mention of θ. Still couldnt reach the original formula.

    I spent almost 4 hours trying to figure this out. I looked up an online solution (out of genuine interest to know how the problem is solved, not academic dishonesty) and after a lengthy process they reach a different formula and claim that this is the solution!

    This is really puzzling me because it looks rather simple, like I'm missing a basic formula or something.

    Any help is greatly appreciated! Thanks!
     

    Attached Files:

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  2. jcsd
  3. Nov 9, 2013 #2

    TSny

    User Avatar
    Homework Helper
    Gold Member

    Hello, Ngineer.

    Note that finding "the total number of magnetic flux lines in the ring" is equivalent to finding the total magnetic flux through the triangular cross-section of the ring.

    So, use the expression for the magnetic field B of a wire and integrate B over the triangular cross-section.

    The answer given doesn't appear to me to be correct. You would expect the answer to go to zero as b goes to zero while keeping a and h constant.

    [EDIT: I believe the b in the denominator of the logarithm should be a.]
     
    Last edited: Nov 9, 2013
  4. Nov 9, 2013 #3
    Yeah. I also just did it using B=μ°I/2∏r and doing a double integral with z on the inside going from 0 to (h/b)(r-a) and r going from a to a+b. The result I got was nearly the same as expected except that I also found the denominator inside the logarithm is a and not b.
     
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