(adsbygoogle = window.adsbygoogle || []).push({}); 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

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# Homework Help: Showing forces between 2 current loops are in accord with newtons 3rd law

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