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Griffiths dipole moment

  • Thread starter ehrenfest
  • Start date
1,996
1
[SOLVED] Griffiths dipole moment

1. Homework Statement
On page 244 of Griffiths E and M book, Griffiths defines the magnetic dipole moment as
[tex]\vec{m} = I \int d\vec{a} [/tex]. Then in problem 5.58, he asks us to calculate the dipole moment of a uniformly charged spinning solid sphere. How is that compatible with the definition he gave?
It seems like Griffiths derived the formula for the magnetic dipole moment only using line currents not surface or volume currents and then asks us to find the magnetic dipole moments of volume and surface currents. I don't understand how to do that!

2. Homework Equations



3. The Attempt at a Solution
 
Last edited:

Answers and Replies

pam
455
1
Doesn't G give the integral of rXj form for the mag moment?
 
1,996
1
What do you mean? All I see is [tex]
\vec{m} = I \int d\vec{a}
[/tex]

which is an integral but it does not work for volume or surface charges? So, I think that the answer is no he does not which is very annoying.
 
511
1
Griffiths has given the wxpression for the volume charge density J= rho*v

Integrate this to have I

Then calculate m
 

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