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Magnetic Field problem

  1. Dec 12, 2004 #1
    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. jcsd
  3. Dec 12, 2004 #2


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    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:
    [tex] d\vec{B}=\frac{\mu_{0}\mu_{r}I}{4\pi}\frac{d\vec{l}\times\vec{r}}{r^{3}} [/tex]
    ,where "r" is the distance from the origin (chosen to be the center of the square) to a line element of the conductor "dl".
  4. Dec 12, 2004 #3
    Ok that clears some of it up thanks. Can dl be choosen to be any length?
  5. Dec 12, 2004 #4


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    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.
  6. Dec 12, 2004 #5
    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.
  7. Dec 12, 2004 #6
    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!
  8. Dec 12, 2004 #7
    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.
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