This is a pretty basic question. Assume that we have a uniformly charged solid sphere of radius $R$ with total charge $Q$ spinning with angular velocity $\omega$ about the $z$ axis. Therefore, the charge density is $\rho = \frac{3Q}{4\pi R^3}$. The dipole moment can be calculated using:
$\vec{m}=\frac{1}{2}\int \vec{r}\times\vec{J}d\tau$
in which $d\tau$ is the volume element and $\vec{J}$ is the current density. If we write $J=\rho \vec{\omega}\times \vec{r}$ and do the integration we will get the result. I can do this with no problem. However, here in the first question the author divides the sphere into several rings by first integrating over $\phi$. I suppose that the rings will look like something like this:
The volume element of the ring is $d\tau_{ring}=2\pi r^2dr\sin\theta d\theta$. Then the author writes that the magnetic dipole moment of each ring is given by:
$d\vec{m}_{ring}=\frac{1}{2}\int_{ring}\vec{r}\times\vec{J}d\tau = \frac{1}{2}dI\int_{ring}\vec{r}\times d\vec{l}=dI(\pi r^2 \sin^2\theta \hat{z})$
Can somebody tell me what is going on here? I am pretty sure that this is a pretty basic thing but for some reason I can't see the connection. We know that $\vec{r}\times\vec{J}=\vec{r}\times d\vec{l}$. Also we can write $dl=r\sin\theta d\phi$. From the geometry we know that the dipole will be in the $\hat{z}$ direction. So we can write:
$\frac{1}{2}dI\int_{ring}\vec{r}\times d\vec{l}=\frac{1}{2}dI\int_0^{2\pi}r.r\sin\theta \sin\alpha d\phi \hat{z}=dI\pi r^2 \sin\theta \sin\alpha \hat{z}$
where $\alpha$ is the angle between the vectors $\vec{r}$ and $d\vec{l}$. In order for these expressions to be equal we must have $\alpha=\theta$ but I can't see how that is possible. The unit vectors in the spherical coordinates are orthogonal. What am I doing wrong?
Edit: @yangxing844 gave the correct answer. In order to give a better description of his answer I updated the figure.
Best Answer
In an orthogonal coordinate system like the spherical coordinates system, $dr,d\theta,d\phi$ are always orthogonal to each other, therefor the $``\alpha"$ you mentioned acutally is always $\pi/2$, however, $r\times d l$ is a vector, during the integration, only its $`` \sin \theta"$ componet is reserved, the $``\cos \theta"$ componet will cancel out during the loop integration.