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Magnetic field from a square loop

  1. Nov 20, 2015 #1
    1. The problem statement, all variables and given/known data
    A wire is formed into the shape of a square of edge length L. Show that when the current in
    the loop is I, the magnetic field at point P a distance x from the center of the square along its axis is $$B=\frac{\mu_0 IL^2}{2\pi(x^2+L^2/4)\sqrt{x^2+L^2/2}}$$
    Capture2.png

    2. Relevant equations
    ##d\vec{B}=\frac{\mu_0}{4\pi}\frac{I\vec{ds}\times\hat{r}}{r^2}##

    3. The attempt at a solution
    Capture2.png

    ##d\vec{B}=\frac{\mu_0}{4\pi}\frac{I\vec{ds}\sin\theta}{r^2}## where ##\theta## is the angle between ##\hat{r}## and ##\vec{ds}## therefore ##\sin\theta=\sqrt{\frac{x^2+L^2/4}{z^2+x^2+L^2/4}}##
    From the symmetry of the problem it's clear that only the components parallel to ##x## will be leftover which seem to be given by ##\sin\phi## but in the solution they are given by ##\cos\phi## which doesn't make any sense to me.
     

    Attached Files:

  2. jcsd
  3. Nov 20, 2015 #2

    TSny

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    What angle does ##d\vec{B}## make to the x axis?
     
  4. Nov 20, 2015 #3
    Okay I got the angle it makes with the x axis from the most basic case where ##x=0##

    On one side of the x axis all angles must add up to ##\pi## therefore (where x is the angle between ##d\vec{B}## and the x axis):
    $$
    \pi=(\pi/2-\phi)+\pi/2+x\Longrightarrow x=\phi
    $$

    Thanks it makes sense now!
     
  5. Nov 20, 2015 #4

    TSny

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    Yes. Nice.
     
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