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There is a sphere of radius $R$ which is charged uniformly by charge density $\rho$. It is rotating by $\omega \hat{z}$ about $z$ axis. Find the magnetic dipole.

I solve it in this way. Think of it in spherical coordinates and take infinitesimal ring on the sphere. Current density $J= \rho v = \rho \omega r \sin \theta$ Then infinitesimal magnetic dipole is $$(\pi r^2 \sin ^2 \theta)(\rho \omega r \sin \theta)(2 \pi r^2 \sin\theta d\theta dr)$$ where the first term is from the area enclosed by the ring, the second is from current density, and the third is from infinitesimal volume of the ring. Is it right? If it is wrong, where does the logic fail?

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You may know that current is defined as the total charge passing a certain area during a unit time. Therefore, \begin{align} dI &=\rho\times rdrd\theta\times2\pi r\sin\theta\times\frac{\omega}{2\pi} \\ &=\rho \omega r^2\sin\theta drd\theta \end{align} where second and third term constitute the volume of infinitesimal torus and the last term is a frequency in which the total charge in the torus rotate.

Also, magnetic dipole moment is well-known for the following: $$\mu=IA $$ where $A$ is the area enclosed by the path of current $I$.

Finally we obtain the total magnetic dipole moment: \begin{align}\mu&=\int A dI \\ &=\int \pi(r\sin\theta)^2 \rho \omega r^2\sin\theta drd\theta \\ &=\rho\omega\pi\int_0^\pi\int_0^R r^4\sin^3\theta drd\theta \\ &=\rho\omega\pi \frac{R^5}{5} \frac{4}{3} =\frac{4\rho\omega\pi R^5}{15}\end{align}

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