91影视

There is a thin glass rod of radius $R$, length $L$ carrying a uniform surface charge $\sigma$ and spinning about its axis at an angular velocity $\omega$.

The magnetic field from a dipole $m$ is

Here, $\mu_{0}$ is the permeability of free space.

Let the field point be along $x$ with the origin at the center of the rod as shown below.

The $x$ components from dipoles in the positive $z$ direction will cancel those from the negative $z$ direction. The $z$ components will add up. The net field will thus be along $z$ . From equation (1),

$$

\begin{aligned}

\stackrel{B}{B} &=\frac{\mu_{0}}{4 \pi} 2 m \int_{0}^{L / 2} \frac{(2 \cos \theta \hat{r}+\sin \theta \hat{\theta})}{r^{3}} d z \\

&=\frac{\mu_{0}}{4 \pi} 2 m \int_{0}^{L / 2} \frac{(2 \cos \theta(\cos \theta \hat{z})+\sin \theta(-\sin \theta \hat{z}))}{r^{3}} d z \\

&=\frac{\mu_{0}}{4 \pi} 2 m \int_{0}^{L / 2} \frac{\left(3 \cos ^{2} \theta-1\right)}{r^{3}} d z \hat{z}

\end{aligned}

$$

From the figure

$$

\sin \theta=\frac{s}{r}

$$

$$

z=-s \cot \theta

$$