Magnetic Moment of Uniformly Charged Rotating Sphere

In summary, the small sphere has a magnetic moment of 2/5-2/3 and its' angular momentum is L=M\vec{v}\times\vec{r}
  • #1
latentcorpse
1,444
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A small sphere is uniformly charged throughout its' volume and rotating with constant angular velocity [itex]\omega[/itex]. Show it's magnetic moment is given by
[itex]m=\frac{1}{5}Q \omega a^2[/itex]. the question doesn't say but I'm assuming a is the radius.

Anyway, so far I have:

[itex]m=\frac{1}{2} \int_V dV (\mathbf{r \wedge J})[/itex]

but [itex]\mathbf{J}= \rho \mathbf{v} = \rho \mathbf{\omega \wedge r}[/itex]

and so [itex]m=\frac{\rho}{2} \int_V r^2 \mathbf{\omega} - (\mathbf{r \cdot \omega})\mathbf{r}[/itex]

but i don't really know where to go from here?
 
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  • #2
Choose a coordinate system and carry out the integration. The easiest choice is probably spherical coordinates, with omega pointing in the z-direction.
 
  • #3
ok so [itex]\int (\mathbf{r \cdot \omega})\mathbf{r} dV=\int r \omega \cos{\theta} r^2 \sin{\theta} dr d \theta d \phi \frac{\mathbf{\hat{r}}}{r}=\omega \int_0^a r^2 dr \int_0^{\pi} \cos{\theta} \sin{\theta} d \theta \int_0^{2 \pi} d \phi \mathbf{\hat{r}}[/itex]
giving
[itex]\omega \frac{a^3}{3} 2 \pi \int_0^{\pi} \frac{1}{2} \sin{2 \theta} d \theta \mathbf{\hat{r}}= \frac{2}{3} \pi a^3 \left[ -\frac{1}{4} \cos{\theta} \right]_0^{\pi} \mathbf{\hat{r}}[/itex]

which is

[itex]\omega \frac{a^3}{12} \pi (-[-1-1])=\frac{\omega \pi a^3}{6} \mathbf{\hat{r}}[/itex]
 
  • #4
and the first integral is [itex]\omega \int_0^a r^4 dr \int_0^{\pi} \sin{\theta} d \theta \int_0^{2 \pi} d \phi \mathfb{\hat{z}}=\omega \frac{2a^5}{5} 2 \pi \mathbf{\hat{z}}=\frac{4 \omega \pi a^5}{5} \mathbf{\hat{z}} [/itex]
 
  • #5
reckon I've messed up somewhere though because they shouldn't have directions should they?
 
  • #6
latentcorpse said:
reckon I've messed up somewhere though because they shouldn't have directions should they?

Magnetic moment is a vector, so it most definitely should have a direction.

But you have messed up somewhere; is [itex]\hat{r}[/itex] really position independent? Because that's what you are assuming when you pull it out of the integral.
 
  • #7
ok. not sure what to do with it then?
 
  • #8
Cartesian unit vectors are position independent, so rewrite [itex]\hat{r}[/itex] in terms of them.

[tex]\hat{r}=\sin\theta\cos\phi\hat{x}+\ldots[/tex]
 
  • #9
ok. did that. both the [itex]\hat{x},\hat{y}[/itex] bits dropped out.

and i got

[itex]\omega \int_0^a r^2 dr \int_0^{\pi} \sin{\theta} \cos^2{\theta} d \theta \int_0^{2 \pi} d \phi \mathbf{\hat{z}}=-\frac{\omega a^3}{3} \int_{1}^{-1} (-u^2) du 2 \pi \mathbf{\hat{z}}[/itex] where [itex]u=\cos{\theta}[/itex]
which simplifies to [itex]\frac{2 \omega \rho \pi a^3}{3} \mathbf{\hat{z}}[/itex] when you multiply in the factor of [itex]\frac{\rho}{2}[/itex] and so overall we get

[itex]m=\left( \frac{2 \omega \rho \pi a^5}{5} -\frac{2 \omega \rho \pi a^3}{3} \right) \mathbf{\hat{z}}[/itex] which doesn't quite give me what i want. there must still be a mistake somewhere i guess.

and i assume [itex]Q=\fac{4}{3} \pi \rho a^3[/itex]
 
  • #10
What is 2/5-2/3?:wink:

Edit why does your second terms have a^3 instead of a^5?
 
Last edited:
  • #11
kool. i got it.

the next bit is to find the angular momentum L of the sphere of mass M and verify that [itex]\mathbf{m}=\frac{Q}{2M}\mathbf{L}[/itex]

i was wanting to use the formula I'm meant to be verifying for finding L so since that's out the window, I was guessing [itex]\mathbf{L}=M(r^2\mathbf{\omega}-(\mathbf{r \cdot \omega})\mathbf{r}[/itex] but that's not getting me anywhere!
 
  • #12
Use the definition of angular momentum: [tex]\vec{L}=M\vec{v}\times\vec{r}[/tex]

P.S. \mathbf doesn't always show up too well here, so you might want to switch to \vec

Edit: [tex]\vec{dL}=\rho_M\vec{v}\times\vec{r}dV[/tex] where [tex]\rho_M=\frac{M}{\frac{4}{3}\pi a^3}[/tex]:wink:
 
Last edited:
  • #13
do i sub in for v before i integrate?
 
  • #14
did you sub in v before integrating when inding m?:wink:
 
  • #15
yes. so is it basically going to be the same integral again?
 
  • #16
You tell me...
 

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

The magnetic moment of a uniformly charged rotating sphere is defined as the product of the charge of the sphere and the angular velocity of its rotation.

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

The magnetic moment of a uniformly charged rotating sphere can be calculated using the formula μ = Qω, where μ is the magnetic moment, Q is the charge of the sphere, and ω is the angular velocity of rotation.

3. What is the relationship between the magnetic moment and the magnetic field produced by a uniformly charged rotating sphere?

The magnetic moment of a uniformly charged rotating sphere is directly proportional to the magnetic field produced by the sphere. This means that as the magnetic moment increases, the magnetic field strength also increases.

4. How does the magnetic moment of a uniformly charged rotating sphere affect its interaction with external magnetic fields?

The magnetic moment of a uniformly charged rotating sphere determines the strength of its interaction with external magnetic fields. A larger magnetic moment will result in a stronger interaction with external fields.

5. Can the magnetic moment of a uniformly charged rotating sphere be changed?

Yes, the magnetic moment of a uniformly charged rotating sphere can be changed by altering either the charge or the angular velocity of rotation. For example, increasing the charge or the speed of rotation will result in a larger magnetic moment.

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