Griffiths dipole moment

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Homework Help Overview

The discussion revolves around the calculation of the magnetic dipole moment of a uniformly charged spinning solid sphere as presented in Griffiths' Electromagnetism textbook. Participants are exploring the compatibility of Griffiths' definition of magnetic dipole moment with the problem's requirements, particularly regarding the treatment of volume and surface currents.

Discussion Character

  • Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • Participants are questioning the applicability of Griffiths' integral definition of magnetic dipole moment, specifically whether it can be extended to volume or surface currents. Some are attempting to reconcile the provided definitions with the problem's context.

Discussion Status

The discussion is ongoing, with participants expressing confusion about the definitions and their application. Some guidance has been offered regarding the integration of charge density to derive current, but no consensus has been reached on the overall compatibility of the definitions with the problem.

Contextual Notes

There appears to be a concern regarding the limitations of Griffiths' definitions when applied to different types of currents, as well as the specific requirements of the homework problem. Participants are navigating these constraints without a clear resolution.

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[SOLVED] Griffiths dipole moment

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!

Homework Equations


The Attempt at a Solution

 
Last edited:
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Doesn't G give the integral of rXj form for the mag moment?
 
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.
 
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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