Can anyone explain Jackson's circular current loop formula?

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In summary, the conversation discusses a specific equation involving the cosine rule and vectors in 3D space. The equation is derived using coordinates and the distance formula. The speaker plans to try using the cylindrical coordinate system to better understand the equation.
  • #1
dingo_d
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Homework Statement



Every time I pick up Jackson I feel dumb :(

So in 5.5 chapter about Circular Current Loop, he says:

[tex]|\vec{r}-\vec{r}'|=[r^2+r'^2-2rr'(\cos\theta\cos\theta'+\sin\theta\sin\theta'\cos\phi')]^{1/2}[/tex]

So it's cosine rule, but how did he got to that sine/cosine thing? It should be the angle between r and r'. I tried drawing but to no avail :\

help...
 
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  • #2
in 3d space the vector r has coordinates

[tex] \bold r = (x,y,z) = (rsin(\theta)cos(\phi),rsin(\theta)sin(\phi),rcos(\theta))[/tex]

and vector r' has coordinates

[tex] \bold r' = (x',y',z') = (r'sin(\theta ')cos(\phi '),r'sin(\theta ')sin(\phi '),r'cos(\theta '))[/tex]

so basically you use that and the fact that

[tex]\left|\bold r - \bold r' \right| = \sqrt{ (x-x')^2 +(y-y')^2 +(z-z')^2}[/tex]

and that should be equivalent to what you have
 
  • #3
I thought of sth like that right now in the morning, only I thought of cylindrical coordinate system.

Altho that would get complicated in the z part.

I'll try this. Thank you :)
 

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