91影视

Consider r > R this is a pure dipole field.

Consider a "pure" magnetic dipole m.

Write the formula of scalarpotential energy for a 鈥減ure鈥� magnetic dipole.

$\mathbf{U}\mathbf{=}\mathbf{U}\left(r\right)$ 鈥︹赌� (1)

Here, role="math" localid="1657516668695" $\mathbf{U}\mathbf{\left(}\mathbf{r}\mathbf{\right)}$is potential energy of magnetic dipole.

Write the formula of scalar potential for the spinning spherical shell.

$\mathbf{U}\left(r\right)\mathbf{=}\mathbf{-}\mathbf{鈭�}\stackrel{\mathbf{鈬赌}}{\mathbf{B}\mathbf{.}}\stackrel{\mathbf{鈬赌}}{\mathbf{dz}}$ 鈥︹赌� (2)

Here,$\mathbf{B}$ is uniform magnetic field.

Write the formula of interior of a solid spinning sphere.

$\mathbf{B}\mathbf{=}\frac{{\mathbf{渭}}_{\mathbf{0}}\mathbf{蠔蚕}}{\mathbf{4}\mathbf{蟺搁}}\left[\left(1-\frac{3r}{5R^2}\right)\mathrm{肠辞蝉胃}\stackrel{铃�}{r}-\left(1-\frac{6r^2}{5R^2}\right)\mathrm{蝉颈苍胃}\stackrel{铃�}{胃}\right]$ (3)

Here, ${\mathbf{渭}}_{\mathbf{0}}$is permeability,$\stackrel{\mathbf{铃�}}{\mathbf{r}}$ is radius of spherical shell and Ris solid spinning sphere.

A comparison of the relationships between the electric and magnetic fields using the electric field.

$$\begin{gathered} \stackrel{鈬赌}{E}=\frac{1}{4{\mathrm{蟺蔚}}_0}\frac{1}{r^3}\left[3\left(\stackrel{鈬赌}{p.}\stackrel{鈬赌}{r}\right)\stackrel{铃�}{r}-\stackrel{鈬赌}{p}\right] \\ =-\stackrel{鈬赌}{鈭�}U \end{gathered}$$ 鈥︹赌� (4)

Determine the scalar potential.

$V=\frac{1}{4{\mathrm{蟺蔚}}_0}\frac{\stackrel{鈬赌}{p}.\stackrel{鈬赌}{r}}{r^2}$

Determine the magnetic field.

$$\begin{gathered} \stackrel{鈬赌}{B}=\frac{{渭}_0}{4\mathrm{蟺}}\frac{1}{r^3}\left[3\left(\stackrel{鈬赌}{m}.\stackrel{铃�}{r}\right)\stackrel{铃�}{r}-\stackrel{鈬赌}{m}\right] \\ =-\stackrel{鈬赌}{鈭�}U \end{gathered}$$ 鈥︹赌� (5)

Here, $U=U\left(r\right)$is the scalar potential for a 鈥減ure鈥� magnetic dipole.

Comparing the equations (4) and (5).

$\frac{\stackrel{鈬赌}{p}}{{蔚}_0}鈫�{渭}_0\stackrel{鈬赌}{m}$

Determine thepotential energy for a 鈥減ure鈥� magnetic dipole.

Substituterole="math" localid="1657519951106" $U\left(r\right)=\frac{{渭}_0}{4\mathrm{蟺}}\frac{\stackrel{鈬赌}{m}.\stackrel{铃�}{r}}{r^2}$ for into equation (1).

Therefore, the value of potential energy for a 鈥減ure鈥� magnetic dipole is .

$U\left(r\right)=\frac{{渭}_0}{4\mathrm{蟺}}\frac{\stackrel{鈬赌}{m}.\stackrel{铃�}{r}}{r^2}$

Determine the dipole moment of the shell is given by.

$m=\left(\frac{4\mathrm{蟺}}{3}\right)蠔未R^4\stackrel{铃�}{z}$

Then using the relation that

$U\left(r\right)=\frac{{渭}_0}{4\mathrm{蟺}}\frac{\stackrel{鈬赌}{m}.\stackrel{铃�}{r}}{r^2}$ 鈥︹赌� (6)

Substitute $\left(\frac{4\mathrm{蟺}}{3}\right)蠔未R^4\stackrel{铃�}{z}$for$\stackrel{鈬赌}{m}$into equation (6).

$$\begin{gathered} U\left(r\right)=\frac{{渭}_0}{4\mathrm{蟺}}\frac{\left(\left({\displaystyle \frac{4蟺}{3}}\right)蠔未R^4\stackrel{铃�}{z}\right)\left(\stackrel{铃�}{r}\right)}{r^2} \\ =\frac{{渭}_0蠔未R^4\left(\stackrel{铃�}{r}\cos 胃\right)\left(\stackrel{铃�}{r}\right)}{3r^2} \\ =\frac{{渭}_0蠔未R^4\cos 胃}{3r^2}Forr>R \end{gathered}$$

Inside the shell, the field is uniform and is given by

$B=\frac{2}{3}{渭}_0未蠔R\stackrel{铃�}{z}$

Determine the scalar potential.

Substitute $\frac{2}{3}{渭}_0未蠔R\stackrel{铃�}{z}$for$\stackrel{鈬赌}{B}$ into equation (2).

$$\begin{gathered} U\left(r\right)=-鈭�\left(\frac{2}{3}{渭}_0未蠔R\stackrel{铃�}{z}\right).\stackrel{鈬赌}{dz} \\ =-\frac{2}{3}{渭}_0未蠔RZ+\mathrm{co}na\tan t \end{gathered}$$

Integration constant can be taken as zero.

Then the required scalar potential is

$U\left(r\right)=-\frac{2}{3}{渭}_0未蠔Rr\cos 胃Forr<R$

Therefore, the value of required scalar potential for the spinning spherical shell for is .

$r>RisU\left(r\right)=-\frac{2}{3}{渭}_0未蠔Rr\cos 胃$

Determine the magnetic field inside the solid spinning sphere is

Substitute$-\stackrel{鈬赌}{鈭�}UforB=\frac{{渭}_0蠔Q}{4蟺R}\left[\left(1-\frac{3r^2}{5R^2}\right)\cos 胃\stackrel{铃�}{r}-\left(1-\frac{6r^2}{5R^2}\right)\sin 胃\stackrel{铃�}{胃}\right]$for into equation (3).

$$\begin{gathered} B=-\stackrel{鈬赌}{鈭�}U \\ \mathit{}\mathit{}\mathit{}\mathit{=}\mathit{-}\frac{\mathit{鈭�}\mathit{U}}{\mathit{鈭�}\mathit{r}}\stackrel{\mathit{铃�}}{\mathit{r}}\mathit{-}\frac{\mathit{1}}{\mathit{r}}\frac{\mathit{鈭�}\mathit{U}}{\mathit{鈭�}\mathit{胃}}\stackrel{\mathit{铃�}}{\mathit{胃}}\mathit{-}\frac{\mathit{1}}{\mathit{r}\mathit{}\mathit{s}\mathit{i}\mathit{n}\mathit{}\mathit{胃}}\frac{\mathit{鈭�}\mathit{U}}{\mathit{鈭�}\mathit{鈭�}}\stackrel{\mathit{铃�}}{\mathit{鈭�}} \end{gathered}$$

On comparison

role="math" localid="1657524010972" $$\begin{gathered} \frac{鈭�U}{鈭�鈭�}=0鈬�U\left(r,胃,蠒\right) \\ =U\left(r,胃\right) \\ \frac{1}{r}\frac{鈭�U}{鈭�胃}=\left(\frac{{渭}_0蠔Q}{4蟺R}\right)\left(1-\frac{6r^2}{5R^2}\right)\sin 胃 \end{gathered}$$

On integration, we have

role="math" localid="1657524715371" $U\left(r,胃\right)=\frac{1}{r}\frac{鈭�U}{鈭�胃}=\left(\frac{{渭}_0蠔Q}{4蟺R}\right)\left(1-\frac{6r^2}{5R^2}\right)\cos 胃+f\left(r\right)$$\left(\frac{{渭}_0蠔Q}{4蟺R}\right)\cos 胃+f\left(r\right)=g\left(胃\right)$ 鈥︹赌�

And

Performing integration, we have

role="math" localid="1657524380282" $$\begin{gathered} U\left(r,胃\right) \\ U\left(r,胃\right)=\left(\frac{{渭}_0蠔Q}{4蟺R}\right)\left(1-\frac{r^2}{5R^2}\right)r\sin 胃+g\left(胃\right) \end{gathered}$$role="math" localid="1657524478787" $-\left(\frac{{渭}_0蠔Q}{4蟺R}\right)\left(1-\frac{r^2}{5R^2}\right)r\cos 胃+g\left(胃\right)$ 鈥︹赌� (8)

Determine the interior of a solid spinning sphere.

Comparing the equations (7) and (8), we have

role="math" localid="1657524513898" $\left(\frac{{渭}_0蠔Q}{4蟺R}\right)\left(1-\frac{6r^2}{5R^2}\right)r\cos 胃+f\left(r\right)$

role="math" localid="1657524657198" $\left(\frac{{渭}_0蠔Q}{4蟺R}\right)r\cos 胃\left(1-\frac{6r^2}{5R^2}+1-\frac{r^2}{5R^2}\right)+f\left(r\right)=g\left(胃\right)$

$\left(\frac{{渭}_0蠔Q}{4蟺R}\right)r\cos 胃\left(\frac{5r^2}{5R^2}\right)+f\left(r\right)=g\left(胃\right)$

We are stuck since there is currently no indication on how to express as the sum of an -function and r-function.

This is due to the fact that a scalar magnetic potential cannot exist where the current is not zero.