Gyromagnetic ratio

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quasar987
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Main Question or Discussion Point

Hi,

The gyromagnetic ratio is the ratio of the magnetic dipole moment to the angular momentum.

I really don't get how the fact that the gyromagnetic ratio of a rotating circular loop of mass M and charge Q is g = Q/2M implies that the gyromagnetic ratio of a uniform rotating spere is also g = Q/2M!
 

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  • #2
Tide
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Points to ponder:

(a) Does the gyromagnetic ratio of the rotating loop depend on the size of the loop?

(b) Can you determine the dipole moment of a rotating spherically symmetric charge distribution by summing over loops?
 
  • #3
quasar987
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Tide said:
Points to ponder:

(a) Does the gyromagnetic ratio of the rotating loop depend on the size of the loop?
No.

Tide said:
(b) Can you determine the dipole moment of a rotating spherically symmetric charge distribution by summing over loops?
This is exactly what's throwing me off! In principle, since the sphere could be considered as an infinity of circular loops of different sizes, its dipole moment is the sum of all those. But since the sphere is uniform, the ratio of its mass to its charge is the same at every point of it. Hence, the dipole moment of each loops is the same at Q/2M, making the total dipole of the sphere infinite!
 
  • #4
Dr Transport
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The dipole moment for a charged rotating sphere is a problem in Griffths, Wangsness and Jackson amongst other texts. It is not infinite, look in some these texts to get an idea on how to calculate it. I think that the easiest manner to go about it is to start with the vector potential for a rotating sphere and go on from there.
 
  • #5
quasar987
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Griffiths says:

David J. said:
What is the gyromagnetic ratio for a uniform spinnin sphere? [This requires no new calculation; simply decompose the sphere into infinitesimal rings, and apply the result of part (a)
The result of part (a) was of course that g = Q/2M for a ring.

So he seems to be saying that just by reasoning, we can determine that it is the same. I exposed my reasoning to you; what is flawed in it?
 
  • #6
quasar987
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I did it by integration.

Does anyone know what is the argument that allows one to conclude that the gyromagnetic ratio of any solid of revolution is Q/2M ?
 
  • #7
Tide
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You should have gotten a clue from the work you did for the sphere. The integrals for the magnetic moment and angular momentum cancel when you form the ratio! :)
 
  • #8
Gokul43201
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quasar987 said:
No.



This is exactly what's throwing me off! .... Hence, the dipole moment of each loops is the same at Q/2M, making the total dipole of the sphere infinite!
No, each loop only has some infinitesimal charge dQ and infinitesimal mass dM.
 
  • #9
quasar987
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Gokul43201 said:
No, each loop only has some infinitesimal charge dQ and infinitesimal mass dM.
Sure but I'm convinced that nevertheless, their ratio is the same:

[tex]\frac{dQ}{dM} = \frac{dQ/dV}{dM/dV} = \frac{\rho_q}{\rho_m} = \frac{Q}{M}[/tex]
 
  • #10
quasar987
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Tide said:
You should have gotten a clue from the work you did for the sphere. The integrals for the magnetic moment and angular momentum cancel when you form the ratio! :)
For the sphere ok. But that seemed almost like magic you know. There are all these constant that come in from integration and they end up being 1/2 at the end. How do I know it wasn't a big coincidence and that for another random revolution figure, it will be different?

Edit: Ok, now I see it from the integral, by generalizing to an arbitrary volume of revolution, that it will always give Q/2M. Thanks for the hint Tide.
 
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  • #11
Tide
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Quasar,

Any time! :)
 

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