Can anyone explain Jackson's circular current loop formula?

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SUMMARY

The discussion focuses on Jackson's circular current loop formula presented in Chapter 5.5, specifically the expression for the distance between two points in 3D space, represented as |\vec{r}-\vec{r}'|. The formula incorporates the cosine rule and involves angles θ and φ, which relate to the spherical coordinates of the vectors r and r'. The participants clarify that the sine and cosine terms arise from the geometric relationships between these vectors in spherical coordinates.

PREREQUISITES
  • Understanding of spherical coordinates and their representation in 3D space.
  • Familiarity with vector mathematics and operations.
  • Knowledge of the cosine rule in geometry.
  • Basic grasp of electromagnetic theory as it relates to current loops.
NEXT STEPS
  • Study the derivation of the cosine rule in spherical coordinates.
  • Learn about vector operations in 3D space, focusing on subtraction and distance calculation.
  • Explore Jackson's Classical Electrodynamics, particularly sections on current loops.
  • Investigate applications of circular current loops in electromagnetic theory.
USEFUL FOR

Students of physics, particularly those studying electromagnetism, as well as educators and anyone seeking to understand the mathematical foundations of current loops in 3D space.

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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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
 
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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