A uniformly charged solid sphere of radius R carries a total charge Q, and is set spinning with angular velocity w about the z axis.

(a) What is the magnetic dipole moment of the sphere?

(b) Find the average magnetic field within the sphere (see Prob. 5.59).

(c) Find the approximate vector potential at a point (r, B) where r>> R.

(d) Find the exact potential at a point (r, B) outside the sphere, and check that it is consistent with (c). [Hint: refer to Ex. 5.11.]

(e) Find the magnetic field at a point (r, B) inside the sphere (Prob. 5.30), and check that it is consistent with (b).

The surface charge density of shell is given as:

$\mathbf{p}\mathbf{=}\frac{\mathbf{Q}}{\left({\displaystyle \frac{4}{3}}{R}^3\right)}$

Here, $\mathrm{Q}$is the charge on the shell and $\mathrm{R}$is the radius of the shell.

The magnetic dipole moment of sphere is given as:

$\mathrm{dm}=\frac{4}{3}{\mathrm{\pi \rho \omega r}}^4\mathrm{dr}$

$\mathrm{m}=\frac{4}{3}\mathrm{\pi \rho \omega }{\int }_0^2 {\mathrm{r}}^4\mathrm{dr}$

$\mathrm{m}=\frac{4}{3}\mathrm{\pi \rho \omega }\left(\frac{{\mathrm{R}}^5}{5}\right)$

Substitute all the values in the above equation.

$\mathrm{m}=\frac{4}{3}\mathrm{\pi }\left(\frac{\mathrm{Q}}{\left(\frac{4}{3}{\mathrm{\pi R}}^3\right)}\right)\mathrm{\omega }\left(\frac{{\mathrm{R}}^5}{5}\right)$

$\mathrm{m}=\frac{1}{5}{\mathrm{Q\omega R}}^2$

Therefore, the magnetic dipole moment of sphere is $\frac{1}{5}{\mathrm{Q\omega R}}^2$.

Consider the formula for the magnetic field of the sphere.

${\mathrm{B}}_{\mathrm{\Omega }}-\frac{{\mathrm{\mu }}_0}{4\mathrm{\pi }}\left(\frac{2\mathrm{m}}{{\mathrm{R}}^3}\right)$

Substitute all the values in the above equation.

${\mathrm{B}}_{\mathrm{\theta }}=\frac{{\mathrm{\mu }}_0}{4\mathrm{\pi }}\left(\frac{2\left(\frac{1}{5}{\mathrm{Q\omega R}}^2\right)}{{\mathrm{R}}^3}\right)$

${\mathrm{B}}_{\mathrm{e}}=\frac{{\mathrm{\mu }}_0}{4\mathrm{\pi }}\left(\frac{2\mathrm{Q\omega }}{5\mathrm{R}}\right)$

Therefore, the average magnetic field within sphere is also $\frac{{\mathrm{\mu }}_0}{4\mathrm{\pi }}\left(\frac{2\mathrm{Q\omega }}{5\mathrm{R}}\right)$.

Consider the formula for the vector potential due to dipole moment:

$\text{A=}\frac{{\mu }_0}{4\pi }\left(\frac{msin\theta }{r^2}\right)$

Substitute all the values in the above equation.

$\mathrm{A}=\frac{{\mathrm{\mu }}_0}{4\mathrm{\pi }}\left(\frac{\left(\frac{1}{5}{\mathrm{Q\omega R}}^2\right)\sin \mathrm{\theta }}{{\mathrm{r}}^2}\right)$

$\mathrm{A}=\frac{{\mathrm{\mu }}_0}{4\mathrm{\pi }}\left(\frac{{\mathrm{Q\omega R}}^2\sin \mathrm{\theta }}{5{\mathrm{r}}^2}\right)$

Therefore, the vector potential at a point is $\frac{{\mathrm{\mu }}_0}{4\mathrm{\pi }}\left(\frac{{\mathrm{Q\omega R}}^2\sin \mathrm{\theta }}{5{\mathrm{r}}^2}\right)$.

Differentiate the expression for potential due to the spherical shell:

${\mathrm{dA}}_{\mathrm{e}}=\left(\frac{{\mathrm{\mu }}_0\mathrm{\rho \omega sin}\mathrm{\theta }}{{\mathrm{r}}^2}\right)\left({\overline{\mathrm{r}}}^4\mathrm{dr}\right)$

width="191">Aθ=μ0ρθsin⁡θr2∫0R r¯4dr

${\mathrm{A}}_{\mathrm{e}}=\left(\frac{{\mathrm{\mu }}_0\mathrm{\rho \omega sin}\mathrm{\theta }}{{\mathrm{r}}^2}\right)\left(\frac{{\mathrm{R}}^5}{5}\right)$

${\mathrm{A}}_{\mathrm{e}}=\left(\frac{{\mathrm{\mu }}_0\mathrm{\rho \omega sin}\mathrm{\theta }}{{\mathrm{r}}^2}\right)\left(\frac{{\mathrm{R}}^5}{5}\right)$

Substitute all the values in the above equation.

$\left.{\mathrm{A}}_{\mathrm{o}}={\left(\begin{array}{c}\mathrm{Q}\\ \left.3^{\mathrm{\pi }}{\mathrm{R}}^3\right)\end{array}\right)}^2\right)\left(\begin{array}{c}\mathrm{osin}\mathrm{\theta }\\ \left({\mathrm{R}}^5\right.\\ 5\end{array}\right)$

${\mathrm{A}}_{\mathrm{e}}=\frac{{\mathrm{\mu }}_0{\mathrm{Q\omega R}}^2\sin \mathrm{\theta }}{20{\mathrm{\pi r}}^2}$

${\mathrm{A}}_{\mathrm{e}}=\frac{{\mathrm{\mu }}_0{\mathrm{Q\omega R}}^2\sin \mathrm{\theta }}{20{\mathrm{\pi r}}^2}$

Therefore, the exact potential outside sphere is $\frac{{\mathrm{\mu }}_0{\mathrm{Q\omega R}}^2\sin \mathrm{\theta }}{20{\mathrm{\pi r}}^2}$.

Consider the expression for field due to uniformly charged sphere:

${\mathrm{B}}_{\mathrm{aj}}={\mathrm{B}}_{\mathrm{a}}$

${\mathrm{B}}_{6\mathrm{i}}=\frac{{\mathrm{\mu }}_0}{4\mathrm{\pi }}\left(\frac{2\mathrm{Q\omega }}{5\mathrm{R}}\right)$

${\mathrm{B}}_{6\mathrm{i}}=\frac{{\mathrm{\mu }}_0\mathrm{Q\omega }}{10\mathrm{\pi R}}$

Therefore, the average magnetic field inside the sphere is ${\mathrm{B}}_{6\mathrm{i}}=\frac{{\mathrm{\mu }}_0\mathrm{Q\omega }}{10\mathrm{\pi R}}$.