91Ó°ÊÓ

The surface charge density of shell is given as:

$\mathrm{ÒÏ}=\frac{Q}{\left(\begin{array}{l}4\\ 3\end{array}{R}^3\right)}$

Here, $Q$is the charge on the shell and $R$is the radius of the shell.

The magnetic dipole moment of sphere is given as:

$\backslash begin\left\{aligned\right\}dm\&=\backslash frac\left\{4\right\}\left\{3\right\}\backslash pi\backslash rho\backslash omegar^\left\{4\right\}dr\backslash \backslash m\&=\backslash frac\left\{4\right\}\left\{3\right\}\backslash pi\backslash rho\backslash omega\backslash int\_\left\{0\right\}^\left\{R\right\}r^\left\{4\right\}dr\backslash \backslash m\&=\backslash frac\left\{4\right\}\left\{3\right\}\backslash pi\backslash rho\backslash omega\backslash left(\backslash frac\{R^\left\{5\right\}\left\}\right\{5\}\backslash right)\backslash end\left\{aligned\right\}$

Substitute all the values in the above equation.

$m=\frac{4}{3}\mathrm{Ï€}\left(\frac{Q}{\left({}^4\mathrm{Ï€}{R}^3\right)}\right)\mathrm{Ó\neg }\left(\frac{R^5}{5}\right)$

$m=\frac{1}{5}Q\mathrm{Ó\neg }{R}^2$

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

(b)

Consider the formula for the magnetic field of the sphere.

$B_a=\frac{{\mathrm{μ}}_0}{4\mathrm{π}}\left(\frac{2m}{R^3}\right)$

Substitute all the values in the above equation.

$\backslash begin\left\{aligned\right\}\&B\_\left\{a\right\}=\backslash frac\{\backslash mu\_\{0\left\}\right\}\{4\backslash pi\}\backslash left(\backslash frac\{2\backslash left(\backslash frac\{1\left\}\right\{5\}Q\backslash omegaR^\{2\}\backslash right)\left\}\right\{R^\left\{3\right\}\}\backslash right)\backslash \backslash \&B\_\left\{a\right\}=\backslash frac\{\backslash mu\_\{0\left\}\right\}\{4\backslash pi\}\backslash left(\backslash frac\{2Q\backslash omega\left\}\right\{5R\}\backslash right)\backslash end\left\{aligned\right\}$

Therefore, the average magnetic field within sphere is also$\backslash frac\{\backslash mu\_\{0\left\}\right\}\{4\backslash pi\}\backslash left(\backslash frac\{2Q\backslash omega\left\}\right\{5R\}\backslash right).$

(c)

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

$A=\frac{{\mathrm{μ}}_0}{4\mathrm{π}}\left(\frac{m\sin \mathrm{θ}}{r^2}\right)$

Substitute all the values in the above equation.

$\backslash begin\left\{aligned\right\}\&A=\backslash frac\{\backslash mu\_\{0\left\}\right\}\{4\backslash pi\}\backslash left(\backslash frac\{\backslash left(\backslash frac\{1\left\}\right\{5\}Q\backslash omegaR^\{2\}\backslash right)\backslash \sin \backslash theta\left\}\right\{r^\left\{2\right\}\}\backslash right)\backslash \backslash \&A=\backslash frac\{\backslash mu\_\{0\left\}\right\}\{4\backslash pi\}\backslash left(\backslash frac\{Q\backslash omegaR^\left\{2\right\}\backslash \sin \backslash theta\left\}\right\{5r^\left\{2\right\}\}\backslash right)\backslash end\left\{aligned\right\}$

Therefore, the vector potential at a point is$\frac{{\mathrm{μ}}_0}{4\mathrm{Ï€}}\left(\frac{Q\mathrm{Ó\neg }{R}^2\sin \mathrm{θ}}{5r^2}\right)$

Differentiate the expression for potential due to the spherical shell:

$\backslash begin\left\{aligned\right\}\&dA\_\left\{e\right\}=\backslash left(\backslash frac\{\backslash mu\_\left\{0\right\}\backslash rho\backslash omega\backslash \sin \backslash theta\left\}\right\{r^\left\{2\right\}\}\backslash right)\backslash left(\backslash bar\{r\}^\{4\}dr\backslash right)\backslash \backslash \&A\_\left\{e\right\}=\backslash left(\backslash frac\{\backslash mu\_\left\{0\right\}\backslash rho\backslash omega\backslash \sin \backslash theta\left\}\right\{r^\left\{2\right\}\}\backslash right)\backslash left(\backslash int\_\{0\}^\{R\}\backslash bar\{r\}^\{4\}dr\backslash right)\backslash \backslash \&A\_\left\{e\right\}=\backslash left(\backslash frac\{\backslash mu\_\left\{0\right\}\backslash rho\backslash omega\backslash \sin \backslash theta\left\}\right\{r^\left\{2\right\}\}\backslash right)\backslash left(\backslash frac\{R^\left\{5\right\}\left\}\right\{5\}\backslash right)\backslash end\left\{aligned\right\}$

Substitute all the values in the above equation.

$A_{\mathrm{θ}}=\left(\frac{{\mathrm{μ}}_0\left(\frac{Q}{\left(\frac{4}{3}\mathrm{Ï€}{R}^3\right)}\right)\mathrm{Ó\neg }\sin \mathrm{θ}}{r^2}\right)\left(\frac{R^5}{5}\right)$

$A_e=\frac{{\mathrm{μ}}_{0}Q\mathrm{Ó\neg }{R}^2\sin \mathrm{θ}}{20\mathrm{Ï€}{r}^2}$

Therefore, the exact potential outside sphere is$\backslash frac\{\backslash mu\_\{0\}Q\backslash omegaR^\{2\}\backslash \sin \backslash theta\}\{20\backslash pir^\{2\left\}\right\}.$

(e)

Consider the expression for field due to uniformly charged sphere:

$\backslash begin\left\{aligned\right\}\&B\_\{ai\}=B\_\left\{a\right\}\backslash \backslash \&B\_\{ai\}=\backslash frac\{\backslash mu\_\{0\left\}\right\}\{4\backslash pi\}\backslash left(\backslash frac\{2Q\backslash omega\left\}\right\{5R\}\backslash right)\backslash \backslash \&B\_\{ai\}=\backslash frac\{\backslash mu\_\{0\}Q\backslash omega\}\{10\backslash piR\}\backslash end\left\{aligned\right\}$

Therefore, the average magnetic field inside the sphere is$\backslash frac\{\backslash mu\_\{0\}Q\backslash omega\}\{10\backslash piR\}.$