What is the magnetic dipole moment of the sphere?

[stunner5000pt]

Homework Statement

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

Homework Equations

\vec{m} = I \int d\vec{a}

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 J = \rho v
\rho = \frac{Q}{\frac{4}{3} \pi R^3}
v = \omega \times r = \omega r sin\theta

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

alright now to find the total current I = \int J \cdot da

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

this doesn't 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

 

[variation]

Hello,

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


Good luck

 

[Meir Achuz]

Use (in Gaussian units)
{\vec m}=\frac{1}{2c}\int{\vec r}\times{\vec j}d^3r

 

[stunner5000pt]

ok so since \vec{m} = \frac{1}{2} \int \vec{r} \times \vec{J} d\tau
in SI units
J = \rho v = \rho (\omega \times r)

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

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

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

so we hav
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}
but \rho = \frac{Q}{\frac{4}{3} \pi R^3}

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

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

maybe I am missing some factors??

 

[Meir Achuz]

\oint d\Omega{\vec r}({\vec r}\cdot\omega)<br /> =4\pi r^2\omega/3.

 

[jerzyjavier]

i think there is a mistake in tis integral. but i don't know where. XD
<br /> 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 <br />

 

[jerzyjavier]

<br /> 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

 

[klp_l123]

the correct answer is = 1/3*Q*R^2*w ... please correct your solution as soon as possible .

 

[AxiomOfChoice]

the correct answer is = 1/3*Q*R^2*w ... please correct your solution as soon as possible .

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