(a) Construct the scalar potential $\mathbf{U}\left(r\right)$for a "pure" magnetic dipole m.

(b) Construct a scalar potential for the spinning spherical shell (Ex. 5.11). [Hint: for$\mathbf{r}\mathbf{>}\mathbf{R}$this is a pure dipole field, as you can see by comparing Eqs. 5.69 and 5.87.]

(c) Try doing the same for the interior of a solid spinning sphere. [Hint: If you solved Pro b. 5.30, you already know the field; set it equal to $\mathbf{-}\mathbf{\nabla }\mathbf{U}$, and solve for U. What's the trouble?]

Consider r > R this is a pure dipole field.

Consider a "pure" magnetic dipole m.

Write the formula of scalarpotential energy for a “pure” magnetic dipole.

$\mathbf{U}\mathbf{=}\mathbf{U}\left(r\right)$ …… (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.

$\mathbf{U}\left(r\right)\mathbf{=}\mathbf{-}\mathbf{\int }\stackrel{\mathbf{\rightharpoonup }}{\mathbf{B}\mathbf{.}}\stackrel{\mathbf{\rightharpoonup }}{\mathbf{dz}}$ …… (2)

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

Write the formula of interior of a solid spinning sphere.

$\mathbf{B}\mathbf{=}\frac{{\mathbf{\mu }}_{\mathbf{0}}\mathbf{\varpi Q}}{\mathbf{4}\mathbf{\pi R}}\left[\left(1-\frac{3r}{5R^2}\right)\mathrm{cos\theta }\stackrel{⏜}{r}-\left(1-\frac{6r^2}{5R^2}\right)\mathrm{sin\theta }\stackrel{⏜}{\theta }\right]$ (3)

Here, ${\mathbf{\mu }}_{\mathbf{0}}$is permeability,$\stackrel{\mathbf{⏜}}{\mathbf{r}}$ 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.

$$\begin{gathered} \stackrel{\rightharpoonup }{E}=\frac{1}{4{\mathrm{\pi \epsilon }}_0}\frac{1}{r^3}\left[3\left(\stackrel{\rightharpoonup }{p.}\stackrel{\rightharpoonup }{r}\right)\stackrel{⏜}{r}-\stackrel{\rightharpoonup }{p}\right] \\ =-\stackrel{\rightharpoonup }{\nabla }U \end{gathered}$$ …… (4)

Determine the scalar potential.

$V=\frac{1}{4{\mathrm{\pi \epsilon }}_0}\frac{\stackrel{\rightharpoonup }{p}.\stackrel{\rightharpoonup }{r}}{r^2}$

Determine the magnetic field.

$$\begin{gathered} \stackrel{\rightharpoonup }{B}=\frac{{\mu }_0}{4\mathrm{\pi }}\frac{1}{r^3}\left[3\left(\stackrel{\rightharpoonup }{m}.\stackrel{⏜}{r}\right)\stackrel{⏜}{r}-\stackrel{\rightharpoonup }{m}\right] \\ =-\stackrel{\rightharpoonup }{\nabla }U \end{gathered}$$ …… (5)

Here, $U=U\left(r\right)$is the scalar potential for a “pure” magnetic dipole.

Comparing the equations (4) and (5).

$\frac{\stackrel{\rightharpoonup }{p}}{{\epsilon }_0}\to {\mu }_0\stackrel{\rightharpoonup }{m}$

Determine thepotential energy for a “pure” magnetic dipole.

Substituterole="math" localid="1657519951106" $U\left(r\right)=\frac{{\mu }_0}{4\mathrm{\pi }}\frac{\stackrel{\rightharpoonup }{m}.\stackrel{⏜}{r}}{r^2}$ for into equation (1).

Therefore, the value of potential energy for a “pure” magnetic dipole is .

$U\left(r\right)=\frac{{\mu }_0}{4\mathrm{\pi }}\frac{\stackrel{\rightharpoonup }{m}.\stackrel{⏜}{r}}{r^2}$

Determine the dipole moment of the shell is given by.

$m=\left(\frac{4\mathrm{\pi }}{3}\right)\varpi \delta R^4\stackrel{⏜}{z}$

Then using the relation that

$U\left(r\right)=\frac{{\mu }_0}{4\mathrm{\pi }}\frac{\stackrel{\rightharpoonup }{m}.\stackrel{⏜}{r}}{r^2}$ …… (6)

Substitute $\left(\frac{4\mathrm{\pi }}{3}\right)\varpi \delta R^4\stackrel{⏜}{z}$for$\stackrel{\rightharpoonup }{m}$into equation (6).

$$\begin{gathered} U\left(r\right)=\frac{{\mu }_0}{4\mathrm{\pi }}\frac{\left(\left({\displaystyle \frac{4\pi }{3}}\right)\varpi \delta R^4\stackrel{⏜}{z}\right)\left(\stackrel{⏜}{r}\right)}{r^2} \\ =\frac{{\mu }_0\varpi \delta R^4\left(\stackrel{⏜}{r}\cos \theta \right)\left(\stackrel{⏜}{r}\right)}{3r^2} \\ =\frac{{\mu }_0\varpi \delta R^4\cos \theta }{3r^2}Forr>R \end{gathered}$$

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

$B=\frac{2}{3}{\mu }_0\delta \varpi R\stackrel{⏜}{z}$

Determine the scalar potential.

Substitute $\frac{2}{3}{\mu }_0\delta \varpi R\stackrel{⏜}{z}$for$\stackrel{\rightharpoonup }{B}$ into equation (2).

$$\begin{gathered} U\left(r\right)=-\int \left(\frac{2}{3}{\mu }_0\delta \varpi R\stackrel{⏜}{z}\right).\stackrel{\rightharpoonup }{dz} \\ =-\frac{2}{3}{\mu }_0\delta \varpi RZ+\mathrm{co}na\tan t \end{gathered}$$

Integration constant can be taken as zero.

Then the required scalar potential is

$U\left(r\right)=-\frac{2}{3}{\mu }_0\delta \varpi Rr\cos \theta Forr<R$

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

$r>RisU\left(r\right)=-\frac{2}{3}{\mu }_0\delta \varpi Rr\cos \theta$

Determine the magnetic field inside the solid spinning sphere is

Substitute$-\stackrel{\rightharpoonup }{\nabla }UforB=\frac{{\mu }_0\varpi Q}{4\pi R}\left[\left(1-\frac{3r^2}{5R^2}\right)\cos \theta \stackrel{⏜}{r}-\left(1-\frac{6r^2}{5R^2}\right)\sin \theta \stackrel{⏜}{\theta }\right]$for into equation (3).

$$\begin{gathered} B=-\stackrel{\rightharpoonup }{\nabla }U \\ \mathit{}\mathit{}\mathit{}\mathit{=}\mathit{-}\frac{\mathit{\partial }\mathit{U}}{\mathit{\partial }\mathit{r}}\stackrel{\mathit{⏜}}{\mathit{r}}\mathit{-}\frac{\mathit{1}}{\mathit{r}}\frac{\mathit{\partial }\mathit{U}}{\mathit{\partial }\mathit{\theta }}\stackrel{\mathit{⏜}}{\mathit{\theta }}\mathit{-}\frac{\mathit{1}}{\mathit{r}\mathit{}\mathit{s}\mathit{i}\mathit{n}\mathit{}\mathit{\theta }}\frac{\mathit{\partial }\mathit{U}}{\mathit{\partial }\mathit{\varnothing }}\stackrel{\mathit{⏜}}{\mathit{\varnothing }} \end{gathered}$$

On comparison

role="math" localid="1657524010972" $$\begin{gathered} \frac{\partial U}{\partial \varnothing }=0\Rightarrow U\left(r,\theta ,\varphi \right) \\ =U\left(r,\theta \right) \\ \frac{1}{r}\frac{\partial U}{\partial \theta }=\left(\frac{{\mu }_0\varpi Q}{4\pi R}\right)\left(1-\frac{6r^2}{5R^2}\right)\sin \theta \end{gathered}$$

On integration, we have

role="math" localid="1657524715371" $U\left(r,\theta \right)=\frac{1}{r}\frac{\partial U}{\partial \theta }=\left(\frac{{\mu }_0\varpi Q}{4\pi R}\right)\left(1-\frac{6r^2}{5R^2}\right)\cos \theta +f\left(r\right)$$\left(\frac{{\mu }_0\varpi Q}{4\pi R}\right)\cos \theta +f\left(r\right)=g\left(\theta \right)$ ……

And

Performing integration, we have

role="math" localid="1657524380282" $$\begin{gathered} U\left(r,\theta \right) \\ U\left(r,\theta \right)=\left(\frac{{\mu }_0\varpi Q}{4\pi R}\right)\left(1-\frac{r^2}{5R^2}\right)r\sin \theta +g\left(\theta \right) \end{gathered}$$role="math" localid="1657524478787" $-\left(\frac{{\mu }_0\varpi Q}{4\pi R}\right)\left(1-\frac{r^2}{5R^2}\right)r\cos \theta +g\left(\theta \right)$ …… (8)

Determine the interior of a solid spinning sphere.

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

role="math" localid="1657524513898" $\left(\frac{{\mu }_0\varpi Q}{4\pi R}\right)\left(1-\frac{6r^2}{5R^2}\right)r\cos \theta +f\left(r\right)$

role="math" localid="1657524657198" $\left(\frac{{\mu }_0\varpi Q}{4\pi R}\right)r\cos \theta \left(1-\frac{6r^2}{5R^2}+1-\frac{r^2}{5R^2}\right)+f\left(r\right)=g\left(\theta \right)$

$\left(\frac{{\mu }_0\varpi Q}{4\pi R}\right)r\cos \theta \left(\frac{5r^2}{5R^2}\right)+f\left(r\right)=g\left(\theta \right)$

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.