What is the magnetic dipole moment of the sphere?

  • #1
1,444
2

Homework Statement


A uniformly charged solid sphere of radius R carries a total charge Q, and is set spinning with angular velocity [itex] \omega [/itex] about the z axis. (a) What is the magnetic dipole moment of the sphere?


Homework Equations


[tex] \vec{m} = I \int d\vec{a} [/tex]

The Attempt at a Solution


having a lot of difficulty with this stuff

since we are talking about a solid sphere ... first find volume current density [itex] J = \rho v[/itex]
[tex] \rho = \frac{Q}{\frac{4}{3} \pi R^3} [/tex]
[tex] v = \omega \times r = \omega r sin\theta [/tex]

so [tex] J = \frac{Q}{\frac{4}{3} \pi R^3} \omega r sin\theta [/tex]
is this ok so far??

alright now to find the total current [itex] I = \int J \cdot da [/itex]

[tex] I = \int \frac{Q}{\frac{4}{3} \pi R^3} \omega R^3 \sin^2\theta d\theta d\phi [/tex]

this doesnt seem dimensionally correct since the radians do not cancel out ...

where have i gone wrong??? Is it the part of the angular momentum??

thansk for your help
 

Answers and Replies

  • #2
39
0
Hello,

There are different current for different point on the spining solid sphere.
The magnitude of current density [tex]J[/tex] you have is correct.
But the total magnetic moment should be
[tex]\vec{m}=\int d\vec{m}[/tex]
,where [tex]d\vec{m}[/tex] is a thin ring with radius [tex]r\sin\theta[/tex] rotating along z-axis.
The cross section area of the thin ring is [tex]dr\cdot rd\theta[/tex].
The effective current due to the thin ring is [tex]J(dr\cdot rd\theta)[/tex].
Therefore, [tex]d\vec{m}=\hat{z}J(dr\cdot rd\theta)\pi(r\sin\theta)^2[/tex].
You can substitute the magnetic moment element into the first integral above and find the aim.


Good luck
 
  • #3
Meir Achuz
Science Advisor
Homework Helper
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Use (in Gaussian units)
[tex]{\vec m}=\frac{1}{2c}\int{\vec r}\times{\vec j}d^3r[/tex]
 
  • #4
1,444
2
ok so since [tex] \vec{m} = \frac{1}{2} \int \vec{r} \times \vec{J} d\tau [/tex]
in SI units
[tex] J = \rho v = \rho (\omega \times r) [/tex]

[tex] m = \frac{1}{2} \int r \times \rho (\omega \times r) d\tau [/tex]
using the appropriate product rule

[tex] m = \frac{\rho}{2} \int \left(\omega(r\cdot r) - r(\omega \cdot r)\right) d\tau = \frac{\rho}{2} \int \left(\omega r^2 - r^2\omega\cos\theta)\right) d\tau [/tex]

the second term in the integrand is zero since [tex] \int_{0}^{\pi} \cos\theta\sin\theta d\theta =0 [/tex]

so we hav
[tex] m =\frac{\rho}{2} \int \omega r^2 d\tau = \frac{\rho\omega}{2} \int_{0}^{R} r^4 dr (4\pi) = \frac{4\pi\rho\omega}{2} \frac{R^5}{5} [/tex]
but [tex] \rho = \frac{Q}{\frac{4}{3} \pi R^3} [/tex]

so [tex] m = \frac{2\pi\omega R^5}{5} \frac{Q}{\frac{4}{3} \pi R^3} = \frac{3}{10} Q\omega R^2 [/tex]

the answer is supposed to be [tex] \frac{1}{5} Q\omega R^2[/tex]

maybe im missing some factors??
 
  • #5
Meir Achuz
Science Advisor
Homework Helper
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[tex]\oint d\Omega{\vec r}({\vec r}\cdot\omega)
=4\pi r^2\omega/3.[/tex]
 
Last edited:
  • #6
i think there is a mistake in tis integral. but i dont know where. XD
[tex]
m = \frac{\rho}{2} \int \left(\omega(r\cdot r) - r(\omega \cdot r)\right) d\tau = \frac{\rho}{2} \int \left(\omega r^2 - r^2\omega\cos\theta)\right) d\tau
[/tex]
 
  • #7
[tex]
m = \frac{\rho}{2} \int \left(\omega(r\cdot r) - r(\omega \cdot r)\right) d\tau = \frac{\rho}{2} \int \left(\omega r^2 - r^2\omega\cos\theta)\right) d\tau [tex]
 
  • #8
9
0
the correct answer is = 1/3*Q*R^2*w .... please correct your solution as soon as possible .
 
  • #9
the correct answer is = 1/3*Q*R^2*w .... please correct your solution as soon as possible .

I disagree. I get [tex]\vec m = \frac 15 Q \omega R^2 \hat k[/tex]. I see the mistake as having been made in the second integration above; it's NOT identically zero. The [tex]x[/tex] and [tex]y[/tex] components are because of the integration in [tex]\phi[/tex], but the [tex]z[/tex] component is nonzero. Don't forget that the first "r" that shows up in the second integration is [tex]\vec r[/tex], NOT [tex]|\vec r|[/tex]. Since we're working in spherical coordinates, this has three components with complicated expressions in terms of [tex]\theta[/tex] and [tex]\phi[/tex].
 

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