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Using Biot-Savart to solve for B at center of square

  1. Mar 17, 2010 #1
    1. The problem statement, all variables and given/known data

    Use the Biot-Savart equation derived in class for a long, straight wire to show that

    B (at the center of a square with sides L and current I) = [tex]\frac{\mu I}{\pi}[/tex] [tex]\frac{2\sqrt{2}}{L}[/tex]

    2. Relevant equations

    We derived the following in class for a long, straight wire and it was indicated that I should also use this to solve the problem.

    |dB| = [tex]\frac{\mu I}{4\pi}[/tex] [tex]\frac{1}{r}[/tex] sin[tex]\theta[/tex]d[tex]\theta[/tex]

    3. The attempt at a solution

    If the center of the square = 0 on the x-axis, then the distance along each triangle base = L/2. The center of the square is a distance of L/2 from each side. So r = [tex]\frac{L}{\sqrt{2}}[/tex].

    I spent a long time trying to figure this out, and the only way I got the correct answer was by using the same limits of integration as the straight wire example (0 to pi). Doing so makes the integral = 2, which, when put together with r = [tex]\frac{L}{\sqrt{2}}[/tex], yields the final answer when multiplied by 4 (taking into account all contributions to the center of the square).

    |B| = 4([tex]\frac{\mu I}{4 \pi}[/tex] [tex]\frac{2 \sqrt{2}}{L}[/tex])

    However, setting the limits of integration as 0 and pi doesn't make sense because the wire is not infinitely long, but I cannot figure out how else to show the final answer. So if my algebra is correct, I've got the answer, but it just doesn't seem right.
     
  2. jcsd
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