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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. 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)

    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:

    • prob.png
      File size:
      113.5 KB
  2. jcsd
  3. Nov 9, 2013 #2


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    Gold Member
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    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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