Question

A thin glass rod of radius Rand length Lcarries a uniform surfacecharge $\mathbf{\delta }$ .It is set spinning about its axis, at an angular velocity $\mathit{\omega }$.Find the magnetic field at a distances $\mathit{s}\mathbf{\gg }\mathit{R}\mathbf{}$ from the axis, in the xyplane (Fig. 5.66). [Hint:treat it as a stack of magnetic dipoles.]

Short answer

The magnetic field at a distances $s\gg R$from the axis of a thin glass rod of radius R, length Lcarrying a uniform surface charge $\mathrm{\delta }$ and spinning about its axis at an angular velocity $\omega$is localid="1658485044736" $-\frac{{\mu }_0\sigma \omega R^{3}L}{4{\left(s^2+{\left(\frac{L}{2}\right)}^2\right)}^{3/2}}\hat{z}$.

Step-by-step solution

Given data

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

Magnetic field due to a dipole

The magnetic field from a dipole $\overrightarrow{m}$ is

$\overrightarrow{\mathrm{B}}=\frac{{\mathrm{\mu }}_0}{4\mathrm{\pi }}\frac{\mathrm{m}}{{\mathrm{r}}^3}(2\mathrm{cos\theta }\hat{\mathrm{r}}+\mathrm{sin\theta }\hat{\mathrm{\theta }})$ ……. (1)

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),

$$\begin{gathered} \overrightarrow{\mathrm{B}}=\frac{{\mathrm{\mu }}_0}{4\mathrm{\pi }}2\mathrm{m}\underset{0}{\overset{\raisebox{1ex}{$\mathrm{L}$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}{\int }}\frac{\left(2\mathrm{cos\theta }\hat{\mathrm{r}}+\mathrm{sin\theta }\hat{\mathrm{\theta }}\right)}{{\mathrm{r}}^3}\mathrm{dz} \\ =\frac{{\mathrm{\mu }}_0}{4\mathrm{\pi }}2\mathrm{m}\underset{0}{\overset{\raisebox{1ex}{$\mathrm{L}$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}{\int }}\frac{\left(2\mathrm{cos\theta }\left(\mathrm{cos\theta }\hat{\mathrm{z}}\right)+\mathrm{sin\theta }\left(-\mathrm{sin\theta }\hat{\mathrm{z}}\right)\right)}{{\mathrm{r}}^3}\mathrm{dz} \\ =\frac{{\mathrm{\mu }}_0}{4\mathrm{\pi }}2\mathrm{m}\underset{0}{\overset{\raisebox{1ex}{$\mathrm{L}$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}{\int }}\frac{\left(3{\cos }^2\mathrm{\theta }-1\right)}{{\mathrm{r}}^3}\mathrm{dz}\hat{\mathrm{z}} \end{gathered}$$

From the figure

$$\begin{gathered} \sin \theta =\frac{s}{r} \\ z=-scot\theta \\ dz=\frac{s}{{\sin }^2\theta }d\theta \end{gathered}$$

The magnetic moment is

$m=\pi \sigma \omega R^3$

Substitute these in the magnetic field equation to get

$$\begin{gathered} \overrightarrow{\mathrm{B}}=\frac{{\mathrm{\mu }}_0}{4\mathrm{\pi }}2{\mathrm{\pi \sigma \omega R}}^3\underset{\mathrm{\pi }2}{\overset{{\mathrm{\theta }}_{\mathrm{m}}}{\int }}\left(3{\cos }^2\mathrm{\theta }-1\right)\frac{{\sin }^3\mathrm{\theta }}{{\mathrm{s}}^3}\frac{\mathrm{s}}{{\sin }^2\mathrm{\theta }}\mathrm{d\theta }\hat{\mathrm{z}} \\ =\frac{{\mathrm{\mu }}_0{\mathrm{\sigma \omega R}}^3}{2{\mathrm{s}}^2}\underset{\mathrm{\pi }2}{\overset{{\mathrm{\theta }}_{\mathrm{m}}}{\int }}\left(3{\cos }^2\mathrm{\theta }-1\right)\mathrm{sin\theta d\theta }\hat{\mathrm{z}} \\ =\frac{{\mathrm{\mu }}_0{\mathrm{\sigma \omega R}}^3}{2{\mathrm{s}}^2}{\mathrm{cos\theta }}_{\mathrm{m}}\left(1-{\cos }^2{\mathrm{\theta }}_{\mathrm{m}}\right)\hat{\mathrm{z}} \\ =\frac{{\mathrm{\mu }}_0{\mathrm{\sigma \omega R}}^3}{2{\mathrm{s}}^2}{\mathrm{cos\theta }}_{\mathrm{m}}{\sin }^2{\mathrm{\theta }}_{\mathrm{m}}\hat{\mathrm{z}} \end{gathered}$$

But the maximum angle is given by

$$\begin{gathered} \sin {\theta }_m=\frac{s}{\sqrt{s^2+{\left(\frac{L}{2}\right)}^2}} \\ \cos {\theta }_m=\frac{-{\displaystyle \frac{L}{2}}}{\sqrt{s^2+{\left(\frac{L}{2}\right)}^2}} \end{gathered}$$

Substitute these to get

role="math" localid="1658486298306" $$\begin{gathered} \overrightarrow{B}=\frac{{\mu }_0\sigma \omega R^3}{2s^2}\frac{-{\displaystyle \frac{L}{2}}}{\sqrt{s^2+{\left(\frac{L}{2}\right)}^2}}\frac{s^2}{\left(s^2+{\left(\frac{L}{2}\right)}^2\right)}\hat{z} \\ =-\frac{{\mu }_0\sigma \omega R^{3}L}{4{\left(s^2+{\left(\frac{L}{2}\right)}^2\right)}^{3/2}}\hat{z} \end{gathered}$$

Thus, the net field is role="math" localid="1658486286935" $-\frac{{\mu }_0\sigma \omega R^{3}L}{4{\left(s^2+{\left(\frac{L}{2}\right)}^2\right)}^{3/2}}\hat{z}$.