91Ó°ÊÓ

There is a thin glass rod of radius R, length Lcarrying a uniform surface charge $\mathrm{δ}$ and spinning about its axis at an angular velocity $\mathrm{Ó\neg }$.

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

$\stackrel{→}{\mathrm{B}}=\frac{{\mathrm{μ}}_0}{4\mathrm{Ï€}}\frac{\mathrm{m}}{{\mathrm{r}}^3}(2\mathrm{³¦´Ç²õθ}\hat{\mathrm{r}}+\mathrm{²õ¾\pm ²Ôθ}\widehat{\mathrm{θ}})$ ……. (1)

Here, ${\mathrm{μ}}_0$ 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),

$$\begin{gathered} \stackrel{→}{\mathrm{B}}=\frac{{\mathrm{μ}}_0}{4\mathrm{Ï€}}2\mathrm{m}\underset{0}{\overset{\raisebox{1ex}{$\mathrm{L}$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}{∫}}\frac{\left(2\mathrm{³¦´Ç²õθ}\hat{\mathrm{r}}+\mathrm{²õ¾\pm ²Ôθ}\widehat{\mathrm{θ}}\right)}{{\mathrm{r}}^3}\mathrm{dz} \\ =\frac{{\mathrm{μ}}_0}{4\mathrm{Ï€}}2\mathrm{m}\underset{0}{\overset{\raisebox{1ex}{$\mathrm{L}$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}{∫}}\frac{\left(2\mathrm{³¦´Ç²õθ}\left(\mathrm{³¦´Ç²õθ}\hat{\mathrm{z}}\right)+\mathrm{²õ¾\pm ²Ôθ}\left(-\mathrm{²õ¾\pm ²Ôθ}\hat{\mathrm{z}}\right)\right)}{{\mathrm{r}}^3}\mathrm{dz} \\ =\frac{{\mathrm{μ}}_0}{4\mathrm{Ï€}}2\mathrm{m}\underset{0}{\overset{\raisebox{1ex}{$\mathrm{L}$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}{∫}}\frac{\left(3{\cos }^2\mathrm{θ}-1\right)}{{\mathrm{r}}^3}\mathrm{dz}\hat{\mathrm{z}} \end{gathered}$$

From the figure

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

The magnetic moment is

$m=\mathrm{Ï€}\mathrm{σ}\mathrm{Ó\neg }{R}^3$

Substitute these in the magnetic field equation to get

$$\begin{gathered} \stackrel{→}{\mathrm{B}}=\frac{{\mathrm{μ}}_0}{4\mathrm{Ï€}}2{\mathrm{πσÓ\neg R}}^3\underset{\mathrm{Ï€}2}{\overset{{\mathrm{θ}}_{\mathrm{m}}}{∫}}\left(3{\cos }^2\mathrm{θ}-1\right)\frac{{\sin }^3\mathrm{θ}}{{\mathrm{s}}^3}\frac{\mathrm{s}}{{\sin }^2\mathrm{θ}}\mathrm{»åθ}\hat{\mathrm{z}} \\ =\frac{{\mathrm{μ}}_0{\mathrm{σÓ\neg R}}^3}{2{\mathrm{s}}^2}\underset{\mathrm{Ï€}2}{\overset{{\mathrm{θ}}_{\mathrm{m}}}{∫}}\left(3{\cos }^2\mathrm{θ}-1\right)\mathrm{²õ¾\pm ²Ôθ»åθ}\hat{\mathrm{z}} \\ =\frac{{\mathrm{μ}}_0{\mathrm{σÓ\neg R}}^3}{2{\mathrm{s}}^2}{\mathrm{³¦´Ç²õθ}}_{\mathrm{m}}\left(1-{\cos }^2{\mathrm{θ}}_{\mathrm{m}}\right)\hat{\mathrm{z}} \\ =\frac{{\mathrm{μ}}_0{\mathrm{σÓ\neg R}}^3}{2{\mathrm{s}}^2}{\mathrm{³¦´Ç²õθ}}_{\mathrm{m}}{\sin }^2{\mathrm{θ}}_{\mathrm{m}}\hat{\mathrm{z}} \end{gathered}$$

But the maximum angle is given by

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

Substitute these to get

role="math" localid="1658486298306" $$\begin{gathered} \stackrel{→}{B}=\frac{{\mathrm{μ}}_0\mathrm{σ}\mathrm{Ó\neg }{R}^3}{2s^2}\frac{-{\displaystyle \frac{L}{2}}}{\sqrt{s^2+{\left(\frac{L}{2}\right)}^2}}\frac{s^2}{\left(s^2+{\left(\frac{L}{2}\right)}^2\right)}\hat{z} \\ =-\frac{{\mathrm{μ}}_0\mathrm{σ}\mathrm{Ó\neg }{R}^{3}L}{4{\left(s^2+{\left(\frac{L}{2}\right)}^2\right)}^{3/2}}\hat{z} \end{gathered}$$

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