Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: 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
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook

Can you offer guidance or do you also need help?
Draft saved Draft deleted