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Force due to three magnetic fields

  1. Jun 7, 2010 #1
    1. The problem statement, all variables and given/known data


    A wire comes in from in finity (left side) and forms a shape by going up a length A, over a lenght A and back down a length A.(forming three sides of a square that is attached to this infinite wire), then
    takes a turn and leaves the page opposite to the side it came in on, headed to infinity.

    There is a particle along the horizontal axis of this, a lenght 1/2 A from the left or right side of the shape, what is the magnetic field at that point,




    2. Relevant equations



    3. The attempt at a solution

    the answer is (sqrt 5)/5 * (Mu * I)/pi A But I have no idea how to get there
     
  2. jcsd
  3. Jun 7, 2010 #2
    Try Ampere's Law. [tex]\oint \vec{B} \cdot d\vec{l}=\mu_{0}I_{enclosed}[/tex]
     
  4. Jun 8, 2010 #3
    Would this qualify as high sym.?

    The way I was going to approach it was calculate the force on the particle due to the two verticle lengths of wires, since it would just be the force due to 1*2, then add the effect of the upper horizontal wire to that. But how do you calculate the first part? Can I use Ampere's law ?
     
  5. Jun 8, 2010 #4
    Superpose the fields from each of the sides of interest. The wire has cylindrical symmetry so you can use Ampere's law.
     
  6. Jun 8, 2010 #5

    collinsmark

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    Actually, Ampere's Law is not going to help with this one. Only infinitely long, straight wires have the appropriate cylindrical symmetry.

    But you can solve this by using Biot–Savart law, and integrating over the length of each straight-line segment. There is some symmetry involved, so you might not have to integrate over every segment. But you'll probably need to break up the problem into at least two integrals (3 integrals if you ignore the symmetry).
     
    Last edited: Jun 8, 2010
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