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Showing forces between 2 current loops are in accord with newtons 3rd law

  1. Jul 19, 2008 #1
    1. The problem statement, all variables and given/known data
    show that the forces between two current loops obey newtons 3rd law.


    2. Relevant equations

    [tex] \textbf{F}_{12}=\frac{\mu I_{1}I_{2}}{4 \pi s^{2}} d\textbf{r}_{1}\times(d\textbf{r}_{2}\times \textbf{s}) [/tex]
    s is the distance between the elements of current
    3. The attempt at a solution
    So, i have to show that:
    [tex] \textbf{F}_{12}=-\textbf{F}_{21}[/tex]
    I will set all the scalar stuff equal to a constant k. i use the BAC CAB rule to get:
    [tex]\textbf{F}_{12}=k(d\textbf{r}_{2}(d\textbf{r}_{1}\cdot\textbf{s})-\textbf{s}(d\textbf{r}_{1}\cdot d\textbf{r}_{2}))[/tex]

    I set up my coordinate system so that my first loop lies at x = 0, and the other at x = 1, with equal radii, r = 1, which gives me the following equations:
    [tex]
    d\textbf{r}_{1}=(-sin(\theta),cos(\theta),0)d\theta[/tex]
    [tex]d\textbf{r}_{2}=(-sin(\phi),cos(phi),0)d\phi[/tex]
    [tex] \textbf{s}=\frac{\textbf{r}_1-\textbf{r}_{2}}{||\textbf{r}_1-\textbf{r}_{2}||}=\frac{(cos(\theta)-cos(\phi),sin(\theta)-sin(\phi),-1)}{||(cos(\theta)-cos(\phi),sin(\theta)-sin(\phi),-1)||}[/tex]
    Is this right? This will allow me to first integrate around the first loop and then the second. When I do the resulting integral i end up getting 0. Could someone please tell me why I am doing this incorrectly? Any help is greatly appreciated
     
  2. jcsd
  3. Jul 19, 2008 #2

    Dick

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    Science Advisor
    Homework Helper

    I don't think you have to integrate anything. Can't you just show that if you interchange 1<->2 (which means also changing s->-s) that the sign of the integrand reverses?
     
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