Magnetic Moment of a Charged, Rotating Sphere Problem

In summary, we are trying to show that for a solid spherical ball with mass m rotating about an axis through its center with a charge q uniformly distributed on the surface, the magnetic moment μ is related to the angular momentum L by the equation μ=(5q/6mc)L. To do this, we need to use the equations μ=IA/c and I sphere =(2/5)mr^{2}, and perform a volume integral to account for the total charge of the sphere. This can be achieved by using a coordinate (such as angle or z) along the rotation axis and finding dI (current per coordinate) and the area as a function of this coordinate. The result can be obtained through a 1-dimensional integral. It
  • #1
San1405
4
0

Homework Statement



Show for a solid spherical ball of mass m rotating about an axis through its center with a charge q uniformly distributed on the surface of the ball that the magnetic moment μ is related to the angular momentum L by the relation

μ=(5q/6mc)L

Homework Equations



μ=IA/c

I sphere =(2/5)m[itex]r^{2}[/itex]

The Attempt at a Solution


So I have been at this problem for a few days and haven't gotten very far. I know I need to do a volume integral to account for the total charge of the sphere. But I just got the normal volume of a sphere and that didn't get me the relation. I also think I need to integrate dI with respect to dq and da. But I am not really sure on how to go about that. I am sorry for the lame attempt. I really am just lost. Any help would be really appreciated! Thank you!
 
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  • #2
The charge is on the surface only. It is convenient to use some coordinate (angle or z) along the rotation axis, and find dI (current per coordinate) and the area as function of this coordinate. The result follows from a 1-dimensional integral.
 
  • #3
Thank you so much for the reply!

I am a little confused. I understand that I should be using a dσ. So I get something like

A=∫[itex]r^{2}[/itex]sinσdσ.
then
dI=(2/5)m[itex]r^{2}[/itex]dσ

This is confusing me because then the pi won't cancel out and I am missing my q. I think I need to add a dq and flip one of my integrals. Am I on the right track?

Thank you again!
 
  • #4
The area needs a factor of pi, and I have no idea why you put 2/5m in the equation for the current. This is independent of the mass, and the interior.
 
  • #5
Thank you again for replying!

Ok, so for the area, I am getting ∏[itex]r^{2}[/itex] after doing the integral.

as for dI, I put the (2/5)m in because that is what I of a sphere equals but then I realized I am looking for dI.
Thus,
dI=∫[itex]r^{2}[/itex]dσ
and
I=∫(2/5)m[itex]r^{2}[/itex]dI

I am still not sure of the q. Would I be able to use q/(2r)dr=dq and put that I to the equstion of dI=[itex]r^{2}[/itex]dq? I tried it but it got me nowhere too...

Sorry for the trouble.
 
  • #6
I think you are mixing two different things here.

- the moment of inertia, coming from the mass. You already have the formula there, you don't need to calculate it. This has nothing to do with the current.
- the current, coming from the charge on the surface. This has nothing to do with the mass.
Both happen to have I as symbol, which is a bit confusing.

I am still not sure of the q
Can you calculate the surface charge density?
 

1. What is the magnetic moment of a charged, rotating sphere?

The magnetic moment of a charged, rotating sphere is a measure of the strength of its magnetic field. It is a vector quantity that describes the orientation and strength of the magnetic field generated by the sphere's rotation and charge.

2. How is the magnetic moment of a charged, rotating sphere calculated?

The magnetic moment of a charged, rotating sphere can be calculated using the formula μ = (2/3)Qωr^2, where μ is the magnetic moment, Q is the charge of the sphere, ω is its angular velocity, and r is its radius.

3. What factors affect the magnetic moment of a charged, rotating sphere?

The magnetic moment of a charged, rotating sphere is affected by the charge and size of the sphere, as well as its angular velocity. The direction of rotation also plays a role in determining the orientation of the magnetic moment.

4. What is the significance of the magnetic moment of a charged, rotating sphere?

The magnetic moment of a charged, rotating sphere is important in understanding the behavior of magnetic materials and in various applications such as magnetic resonance imaging (MRI) and particle accelerators.

5. How does the magnetic moment of a charged, rotating sphere differ from that of a stationary charged sphere?

A stationary charged sphere has a magnetic moment of zero because it does not have any angular momentum. However, a charged, rotating sphere has a non-zero magnetic moment due to its angular momentum and charge, resulting in a magnetic field that can be measured and manipulated.

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