Question

A thin glass rod of radius $R$and length $L$carries a uniform surface charge $\sigma$. It is set spinning about its axis, at an angular velocity$\omega$. Find the magnetic field at a distances $s冈\mathrm{R}$from the axis, in the $xy$plane (Fig. 5.66). [Hint: treat it as a stack of magnetic dipoles.]

Short answer

The magnetic field at a distances $s$ 秱 from the axis of a thin glass rod of radius $R$, length $L$ carrying a uniform surface charge $\sigma$ and spinning about its axis at an angular

$$

-\frac{\mu_{0} \sigma \omega R^{3} L}{4\left(s^{2}+\left(\frac{L}{2}\right)^{2}\right)^{3 / 2}} \hat{z}

$$

velocity $\omega$ is

Step-by-step solution

Given data

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 $\omega$.

Magnetic field due to a dipole

The magnetic field from a dipole $m$ is

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

Magnetic field due to the glass rod

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

$$