91Ó°ÊÓ

The charge density is$\mathrm{ÒÏ}$ .

The uniform polarization is$P$ .

The integral is $∫\frac{\widehat{r}}{r^2}d{\mathrm{τ}}^{\text{'}}$

The inside the radius of sphere is $r$and outside the radius of the sphere is$R$ .

The expression for electric field due to uniformly charged sphere is given as follows.

$\mathit{E}\mathbf{=}\mathit{ÒÏ}\left[\frac{1}{4\mathrm{Ï€}{\mathrm{Î\mu }}_0}∫\frac{\widehat{r}}{r^2}d{\mathrm{Ï„}}^{\text{'}}\right]$ …… (1)

The expression for the potential due to uniformly charged sphere is given as follows.

$\mathit{V}\mathbf{=}\mathit{P}\mathbf{â‹\dots }\left[\left(\frac{1}{4\mathrm{Ï€}{\mathrm{Î\mu }}_0}∫\frac{\widehat{r}}{r^2}d{\mathrm{Ï„}}^{\text{'}}\right)\right]$ …… (2)

The expression for the vector potential due to uniformly magnetized sphere is given as follows.

$\mathit{A}\mathbf{=}\frac{{\mathbf{μ}}_{\mathbf{0}}}{\mathbf{4}\mathbf{π}}∫\frac{M×\widehat{r}}{r^2}d{\mathrm{τ}}^{\text{'}}$ …… (3)

The volume of the sphere is given by,

$∫d{\mathrm{τ}}^{\text{'}}=\frac{4}{3}\mathrm{π}{r}^3$

The inside potential due to uniformly charged sphere is given by,

$\begin{array}{l}{V}_{in}=Pâ‹\dots \left[\frac{1}{4\mathrm{Ï€}{\mathrm{Î\mu }}_0}\frac{\widehat{r}}{r^2}\frac{4}{3}\mathrm{Ï€}{r}^3\right]\\ V_{in}=Pâ‹\dots \left(\frac{1}{3{\mathrm{Î\mu }}_0}r\right)\\ V_{in}=\frac{1}{3{\mathrm{Î\mu }}_0}Pr\end{array}$

The outside potential due to uniformly charged sphere is given by,

$\begin{array}{l}{V}_{out}=Pâ‹\dots \left[\frac{1}{4\mathrm{Ï€}{\mathrm{Î\mu }}_0}\frac{\widehat{r}}{r^2}\frac{4}{3}\mathrm{Ï€}{R}^3\right]\\ V_{out}=Pâ‹\dots \left(\frac{1}{3{\mathrm{Î\mu }}_0}\frac{R^3}{r^2}\widehat{r}\right)\\ V_{out}=\frac{1}{3{\mathrm{Î\mu }}_0}\frac{R^3}{r^2}P\widehat{r}\end{array}$

The vector potential due to uniformly magnetized sphere inside the sphere is given by

$\begin{array}{l}{A}_{in}=\frac{{\mathrm{μ}}_0}{4\mathrm{π}}∫\frac{M×\widehat{r}}{r^2}d{\mathrm{τ}}^{\text{'}}\\ A_{in}=\frac{{\mathrm{μ}}_0}{4\mathrm{π}}\frac{M×\widehat{r}}{r^2}×\frac{4}{3}\mathrm{π}{r}^3\\ A_{in}=\frac{{\mathrm{μ}}_0}{3}(M×r)\end{array}$

The vector potential due to uniformly magnetized sphere outside the sphere is given by

$\begin{array}{l}{A}_{out}=\frac{{\mathrm{μ}}_0}{4\mathrm{π}}∫\frac{M×\widehat{r}}{r^2}d{\mathrm{τ}}^{\text{'}}\\ A_{out}=\frac{{\mathrm{μ}}_0}{4\mathrm{π}}\frac{M×\widehat{r}}{r^2}×\frac{4}{3}\mathrm{π}{R}^3\\ A_{out}=\frac{{\mathrm{μ}}_0}{3}\frac{R^3}{r^2}(M×\widehat{r})\end{array}$

Hence the inside and outside potential of uniformly magnetized sphere is$\frac{1}{3{\mathrm{Î\mu }}_0}Pr$ and $\frac{1}{3{\mathrm{Î\mu }}_0}\frac{R^3}{r^2}P\widehat{r}$ respectively. Inside and outside vector potential uniformly magnetized sphere is $\frac{{\mathrm{μ}}_0}{3}(M×r)$and $\frac{{\mathrm{μ}}_0}{3}\frac{R^3}{r^2}(M×\widehat{r})$ respectively.