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

[Serik]

Homework Statement

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

Homework 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$$\theta$$

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.

 

[anathema]

Thank you for your post. I understand your confusion about using the same limits of integration as the straight wire example. However, in this case, we are not dealing with an infinitely long wire, but rather a wire with a specific length (L) and a current (I). Therefore, it is important to consider the limits of integration in a different way.

To solve this problem, we can use the Biot-Savart equation that you have provided: |dB| = \frac{\mu I}{4\pi} \frac{1}{r} sin\thetad\theta. We can break this equation down into two parts: the constant term \frac{\mu I}{4\pi} and the term \frac{1}{r} sin\thetad\theta.

First, let's consider the constant term. This represents the strength of the magnetic field at any point along the wire. Since we are dealing with a wire of length L, we can rewrite this term as \frac{\mu I}{4\pi L}. This is because the magnetic field strength will decrease as we move away from the wire, and the length of the wire (L) represents the distance over which this decrease occurs.

Now, let's consider the term \frac{1}{r} sin\thetad\theta. This term represents the contribution of each small segment of the wire to the magnetic field at the center of the square. As you correctly pointed out, the distance from the center of the square to each side is r = \frac{L}{\sqrt{2}}. However, we also need to consider the angle \theta, which represents the angle between the segment of wire and the line connecting the segment to the center of the square. In this case, \theta will vary depending on the location of the segment along the wire. Therefore, we need to integrate this term over the entire length of the wire.

So, the integral that we need to solve is:

|B| = \int_0^L \frac{\mu I}{4\pi L} \frac{1}{\frac{L}{\sqrt{2}}} sin\theta d\theta

This integral can be simplified to:

|B| = \frac{\mu I}{4\pi L} \frac{2\sqrt{2}}{L} \int_0^L sin\theta d\theta