(a) Exploit the analogy with the electrical case:
E =\frac{1}{4 \pi \epsilon_{0}} \frac{1}{r^{3}}[3( p \cdot \hat{ r }) \hat{ r }- p ] \quad \text { (Eq. 3.104) }=-\nabla V, \quad \text { with } V=\frac{1}{4 \pi \epsilon_{0}} \frac{ p \cdot \hat{ r }}{r^{2}} (Eq. 3.102).
E _{ dip }( r )=\frac{1}{4 \pi \epsilon_{0}} \frac{1}{r^{3}}[3( p \cdot \hat{ r }) \hat{ r }- p ] (3.104)
V_{\text {dip }}(r, \theta)=\frac{\hat{ r } \cdot p }{4 \pi \epsilon_{0} r^{2}}=\frac{p \cos \theta}{4 \pi \epsilon_{0} r^{2}} (3.102)
B =\frac{\mu_{0}}{4 \pi} \frac{1}{r^{3}}[3( m \cdot \hat{ r }) \hat{ r }- m ] \quad(\text { Eq. } 5.89)=-\nabla U,(\text { Eq. } 5.67) .
B _{ dip }( r )=\frac{\mu_{0}}{4 \pi} \frac{1}{r^{3}}[3( m \cdot \hat{ r }) \hat{ r }- m ] (5.89)
B = –∇U (5.67)
Evidently the prescription is p / \epsilon_{0} \rightarrow \mu_{0} m : \quad U( r )=\frac{\mu_{0}}{4 \pi} \frac{ m \cdot \hat{ r }}{r^{2}} .
(b) Comparing Eqs. 5.69 and 5.87, the dipole moment of the shell is m =(4 \pi / 3) \omega \sigma R^{4} \hat{ z } (which we also got in Prob. 5.37). Using the result of (a), then, U( r )=\frac{\mu_{0} \omega \sigma R^{4}}{3} \frac{\cos \theta}{r^{2}} for r > R.
Inside the shell, the field is uniform (Eq. 5.70): B =\frac{2}{3} \mu_{0} \sigma \omega R \hat{ z }, \text { so } U( r )=-\frac{2}{3} \mu_{0} \sigma \omega R z+ constant. We may as well pick the constant to be zero, so U( r )=-\frac{2}{3} \mu_{0} \sigma \omega R r \cos \theta for r < R.
B =\nabla \times A =\frac{2 \mu_{0} R \omega \sigma}{3}(\cos \theta \hat{ r }-\sin \theta \hat{ \theta })=\frac{2}{3} \mu_{0} \sigma R \omega \hat{ z }=\frac{2}{3} \mu_{0} \sigma R \omega (5.70)
[Notice that U(r) is not continuous at the surface (r = R): U_{ in }(R)=-\frac{2}{3} \mu_{0} \sigma \omega R^{2} \cos \theta \neq U_{ out }(R)=\frac{1}{3} \mu_{0} \sigma \omega R^{2} \cos \theta . As I warned you on p. 245: if you insist on using magnetic scalar potentials, keep away from places where there is current!]
(c) B =\frac{\mu_{0} \omega Q}{4 \pi R}\left[\left(1-\frac{3 r^{2}}{5 R^{2}}\right) \cos \theta \hat{ r }-\left(1-\frac{6 r^{2}}{5 R^{2}}\right) \sin \theta \hat{ \theta }\right]=-\nabla U=-\frac{\partial U}{\partial r} \hat{ r }-\frac{1}{r} \frac{\partial U}{\partial \theta} \hat{ \theta }-\frac{1}{r \sin \theta} \frac{\partial U}{\partial \phi} \hat{ \phi } .
\frac{\partial U}{\partial \phi}=0 \Rightarrow U(r, \theta, \phi)=U(r, \theta) .
\frac{1}{r} \frac{\partial U}{\partial \theta}=\left(\frac{\mu_{0} \omega Q}{4 \pi R}\right)\left(1-\frac{6 r^{2}}{5 R^{2}}\right) \sin \theta \Rightarrow U(r, \theta)=-\left(\frac{\mu_{0} \omega Q}{4 \pi R}\right)\left(1-\frac{6 r^{2}}{5 R^{2}}\right) r \cos \theta+f(r) .
\frac{\partial U}{\partial r}=-\left(\frac{\mu_{0} \omega Q}{4 \pi R}\right)\left(1-\frac{3 r^{2}}{5 R^{2}}\right) \cos \theta \Rightarrow U(r, \theta)=-\left(\frac{\mu_{0} \omega Q}{4 \pi R}\right)\left(r-\frac{r^{3}}{5 R^{2}}\right) \cos \theta+g(\theta) .
Equating the two expressions:
-\left(\frac{\mu_{0} \omega Q}{4 \pi R}\right)\left(1-\frac{6 r^{2}}{5 R^{2}}\right) r \cos \theta+f(r)=-\left(\frac{\mu_{0} \omega Q}{4 \pi R}\right)\left(1-\frac{r^{2}}{5 R^{2}}\right) r \cos \theta+g(\theta) ,
or
\left(\frac{\mu_{0} \omega Q}{4 \pi R^{3}}\right) r^{3} \cos \theta+f(r)=g(\theta) .
But there is no way to write r^{3} cos θ as the sum of a function of θ and a function of r, so we’re stuck. The reason is that you can’t have a scalar magnetic potential in a region where the current is nonzero