91影视

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:

$\mathrm{dm}=\frac{4}{3}{\mathrm{蟺蚁蝇谤}}^4\mathrm{dr}$

$\mathrm{m}=\frac{4}{3}\mathrm{蟺蚁蝇}{鈭�}_0^2鈥�{\mathrm{r}}^4\mathrm{dr}$

$\mathrm{m}=\frac{4}{3}\mathrm{蟺蚁蝇}\left(\frac{{\mathrm{R}}^5}{5}\right)$

Substitute all the values in the above equation.

$\mathrm{m}=\frac{4}{3}\mathrm{蟺}\left(\frac{\mathrm{Q}}{\left(\frac{4}{3}{\mathrm{蟺搁}}^3\right)}\right)\mathrm{蝇}\left(\frac{{\mathrm{R}}^5}{5}\right)$

$\mathrm{m}=\frac{1}{5}{\mathrm{蚕蝇搁}}^2$

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

Consider the formula for the magnetic field of the sphere.

${\mathrm{B}}_{\mathrm{惟}}鈭�\frac{{\mathrm{渭}}_0}{4\mathrm{蟺}}\left(\frac{2\mathrm{m}}{{\mathrm{R}}^3}\right)$

Substitute all the values in the above equation.

${\mathrm{B}}_{\mathrm{胃}}=\frac{{\mathrm{渭}}_0}{4\mathrm{蟺}}\left(\frac{2\left(\frac{1}{5}{\mathrm{蚕蝇搁}}^2\right)}{{\mathrm{R}}^3}\right)$

${\mathrm{B}}_{\mathrm{e}}=\frac{{\mathrm{渭}}_0}{4\mathrm{蟺}}\left(\frac{2\mathrm{Q蝇}}{5\mathrm{R}}\right)$

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

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

$\text{A=}\frac{{渭}_0}{4蟺}\left(\frac{msin鈦�胃}{r^2}\right)$

Substitute all the values in the above equation.

$\mathrm{A}=\frac{{\mathrm{渭}}_0}{4\mathrm{蟺}}\left(\frac{\left(\frac{1}{5}{\mathrm{蚕蝇搁}}^2\right)\sin 鈦�\mathrm{胃}}{{\mathrm{r}}^2}\right)$

$\mathrm{A}=\frac{{\mathrm{渭}}_0}{4\mathrm{蟺}}\left(\frac{{\mathrm{蚕蝇搁}}^2\sin 鈦�\mathrm{胃}}{5{\mathrm{r}}^2}\right)$

Therefore, the vector potential at a point is $\frac{{\mathrm{渭}}_0}{4\mathrm{蟺}}\left(\frac{{\mathrm{蚕蝇搁}}^2\sin 鈦�\mathrm{胃}}{5{\mathrm{r}}^2}\right)$.

Differentiate the expression for potential due to the spherical shell:

${\mathrm{dA}}_{\mathrm{e}}=\left(\frac{{\mathrm{渭}}_0\mathrm{蚁蝇sin}鈦�\mathrm{胃}}{{\mathrm{r}}^2}\right)\left({\stackrel{炉}{\mathrm{r}}}^4\mathrm{dr}\right)$

width="191">A胃=渭0蚁胃蝉颈苍鈦�胃r2鈭�0R鈥�r炉4dr

${\mathrm{A}}_{\mathrm{e}}=\left(\frac{{\mathrm{渭}}_0\mathrm{蚁蝇sin}鈦�\mathrm{胃}}{{\mathrm{r}}^2}\right)\left(\frac{{\mathrm{R}}^5}{5}\right)$

${\mathrm{A}}_{\mathrm{e}}=\left(\frac{{\mathrm{渭}}_0\mathrm{蚁蝇sin}鈦�\mathrm{胃}}{{\mathrm{r}}^2}\right)\left(\frac{{\mathrm{R}}^5}{5}\right)$

Substitute all the values in the above equation.

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

${\mathrm{A}}_{\mathrm{e}}=\frac{{\mathrm{渭}}_0{\mathrm{蚕蝇搁}}^2\sin 鈦�\mathrm{胃}}{20{\mathrm{蟺谤}}^2}$

${\mathrm{A}}_{\mathrm{e}}=\frac{{\mathrm{渭}}_0{\mathrm{蚕蝇搁}}^2\sin 鈦�\mathrm{胃}}{20{\mathrm{蟺谤}}^2}$

Therefore, the exact potential outside sphere is $\frac{{\mathrm{渭}}_0{\mathrm{蚕蝇搁}}^2\sin 鈦�\mathrm{胃}}{20{\mathrm{蟺谤}}^2}$.

Consider the expression for field due to uniformly charged sphere:

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

${\mathrm{B}}_{6\mathrm{i}}=\frac{{\mathrm{渭}}_0}{4\mathrm{蟺}}\left(\frac{2\mathrm{Q蝇}}{5\mathrm{R}}\right)$

${\mathrm{B}}_{6\mathrm{i}}=\frac{{\mathrm{渭}}_0\mathrm{Q蝇}}{10\mathrm{蟺搁}}$

Therefore, the average magnetic field inside the sphere is ${\mathrm{B}}_{6\mathrm{i}}=\frac{{\mathrm{渭}}_0\mathrm{Q蝇}}{10\mathrm{蟺搁}}$.