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

1. Mar 17, 2010

### Serik

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) = $$\frac{\mu I}{\pi}$$ $$\frac{2\sqrt{2}}{L}$$

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| = $$\frac{\mu I}{4\pi}$$ $$\frac{1}{r}$$ sin$$\theta$$d$$\theta$$

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 = $$\frac{L}{\sqrt{2}}$$.

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 = $$\frac{L}{\sqrt{2}}$$, yields the final answer when multiplied by 4 (taking into account all contributions to the center of the square).

|B| = 4($$\frac{\mu I}{4 \pi}$$ $$\frac{2 \sqrt{2}}{L}$$)

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.