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Magnetic flux through a triangular loop?

  1. Jan 17, 2017 #1
    1. The problem statement, all variables and given/known data
    Determine the mutual inductance between a very long, straight wire and a conducting equilateral triangular loop

    2. Relevant equations
    I know how to do it but I'm stuck at the step of finding the total inductance through the triange.

    3. The attempt at a solution
    we know that the magnetic field due to an infinitely long wire is
    B = µI/2πr
    in order to find the magnetic flux through the triangle we need to find Φ = ∫B.dA
    we have B but what would dA be in this case
    the book says it's (2/√3) * (r - d) *dr
    why is that so?

    thank you
     

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  3. Jan 17, 2017 #2

    RUber

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    Look at the geometry for your triangle.
    Starting at the point closest to the wire, we will call that ##(d, 0)## , you will follow a counterclockwise path to ##(d+\frac{b\sqrt{3}}{2}, -\frac b2 )##, then from
    ##(d+\frac{b\sqrt{3}}{2}, -\frac b2 )## to ##(d+\frac{b\sqrt{3}}{2}, +\frac b2 )##, and finally from ##(d+\frac{b\sqrt{3}}{2}, +\frac b2 )## to ##(d,0 )##.
    Of course, you can parameterize each of those line integrals to get a form that is easier to work with.
     
  4. Jan 17, 2017 #3
    Ok that really helped to be honest.
    Thank you sir
    But that would be a line integral right?
    what if I wanted to cover the surface as in the area covered by the sides of the triangle
     
  5. Jan 17, 2017 #4

    RUber

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    Then, in that case, you would have x going from d to d+b/2 * sqrt(3) and y ranging from the lower line to the upper line (as a function of x).
    These bounds are no fun to work with, but will get you to the answer for the surface integral.
     
  6. Jan 17, 2017 #5
    Yeah I get ya...
    is there anyway to do it with a change in r since it covers both x and y directions...
    this is how it is in the solution manual perhaps if you take a look you could understand how it was derived
    I just couldn't figure it out
     

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  7. Jan 18, 2017 #6

    RUber

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    I see what they did there. Since the wire in infinitely long, B is only dependent on r, the distance from the line.
    So, if you slice up the equilateral triangle vertically, along lines with the same distance to the wire, you get
    ##\int\int B \, dA = \int B h(r) dr, ##
    where ##h(r)## is the height of the triangle at distance r.
    At r=d, the height is 0, and at r = ##d+b\sqrt{3}/2##, the height is b. So you can make a linear rule for the height that looks like
    ## h = (r-d)\frac{2}{\sqrt{3}} .##
    Check that against the reference points to make sure.
    So, now you can treat the integral in terms of one variable, r.
    ##\int B h(r) dr = \int B (r-d)\frac{2}{\sqrt{3}}\, dr ##
    The final answer applies a simplification in the natural log, but otherwise, you should be able to follow the rest from there with about 2 extra steps.
    I hope this helps.
     
  8. Jan 18, 2017 #7
    It's great man
    Thanks a lot
    You're the best
     
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