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

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SUMMARY

The discussion focuses on demonstrating that the forces between two current loops, described by the equation F_{12}=\frac{\mu I_{1}I_{2}}{4 \pi s^{2}} d\textbf{r}_{1}\times(d\textbf{r}_{2}\times \textbf{s}}, adhere to Newton's Third Law. The user attempts to show that F_{12}=-F_{21} by setting up a coordinate system and using the BAC CAB rule. The integration process leads to a result of zero, prompting questions about the correctness of the approach. A suggestion is made to interchange the loops to demonstrate the sign reversal of the integrand without integration.

PREREQUISITES
  • Understanding of electromagnetism, specifically the forces between current loops.
  • Familiarity with vector calculus and the BAC CAB rule.
  • Knowledge of Newton's Third Law of Motion.
  • Ability to perform line integrals in a coordinate system.
NEXT STEPS
  • Study the derivation of the force between current loops using Biot-Savart Law.
  • Learn about the mathematical properties of vector cross products.
  • Explore the implications of Newton's Third Law in electromagnetic contexts.
  • Investigate the process of interchanging variables in integrals to analyze symmetry.
USEFUL FOR

Physics students, educators, and researchers interested in electromagnetism and the mathematical principles governing forces between current-carrying conductors.

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Homework Statement


show that the forces between two current loops obey Newtons 3rd law.

Homework 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

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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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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