(adsbygoogle = window.adsbygoogle || []).push({}); 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 sidesLand currentI) = [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 ofL/2 from each side. Sor= [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 withr= [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 wireis notinfinitely 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.

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# Homework Help: Using Biot-Savart to solve for B at center of square

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