Magnetic Field problem

1. Dec 12, 2004

rothrock98

I've been stuck on this problem from a few hours now. It seemed much easier when I started it.

- A square wire loop has sides 2 meters long, with a current 5 amps circulating in the loop. What is the magnetic field at the exact center of the loop?

I was going the direction of the law of Biot and Savart and integrating to get B = u0/4pi * integral( I*dl x r(hat)/r^2)

I just cant seem to get anywhere, any ideas?

2. Dec 12, 2004

dextercioby

Apply Biot-Savart-Laplace law correctly,takin into account that this problem admits symmetry.I state that the total magnetic field in the center of the loop is the sum of the 4 equal contributions coming from each side of the square.
The law is first given in the following mathematical formulation by Pierre Simon (Marquis de) Laplace:
$$d\vec{B}=\frac{\mu_{0}\mu_{r}I}{4\pi}\frac{d\vec{l}\times\vec{r}}{r^{3}}$$
,where "r" is the distance from the origin (chosen to be the center of the square) to a line element of the conductor "dl".
[tex]

3. Dec 12, 2004

rothrock98

Ok that clears some of it up thanks. Can dl be choosen to be any length?

4. Dec 12, 2004

dextercioby

No,not any.It is chosen as a vector line element along the side of the square with the sense equal to the sense of the current.
Integration by "dl" is made from 0 to "l",where "l" designates the length of the side of the square.

5. Dec 12, 2004

MatSci

You can derive a simpler formula from biot-savart by seeing that in a long straight wire, r = square root of l squared + R squared where l is the lenth of the wire and R is the distance from the wire. Also sin of the angle is R/r. subsituting these in and integrating with respect to s, assuming the wire is of infinite length, you can derive an easy formula to use.

6. Dec 12, 2004

rothrock98

so getting B = UoI/2πr for a simpler formula by doing MatSci method. Then substituting in giving values I got (4πx10^-7 * 5) / (2π * 1) = 1 x 10^-6. Take this times 4 for the summation yields 4 x 10^-6 T. This on the right track now? Thanks for all the help!

7. Dec 12, 2004

MatSci

Well actually we reduced biot-savart for an infinitely long wire. We don't wanna do this in this problem. So before integrating from 0 to infinite, you wanna integrate from -L/2 to L/2 where L is the length of each square side. This comes out to be (mui/2(pie)r) (L/(L^2 + 4R^2)^(1/2)). With the infinite wire, L went to infinite and you could essentially ignore R^2 to receive what you got. But in this case, L equals length of sides and R equals L/2. So the final in the box current is given by B=2mui/(pie)L2^(1/2) where L is the length of each side.