Question 6.28: A magnetic dipole m is imbedded at the center of a sphere (r...

In view of Eq. 6.33, there is a bound dipole at the center: m _{b}=\chi_{m} m . So the net dipole moment at the center is m _{\text {center }}= m + m _{b}=\left(1+\chi_{m}\right) m =\frac{\mu}{\mu_{0}} m . This produces a field given by Eq. 5.89:

J _{b}= \nabla \times M = \nabla \times\left(\chi_{m} H \right)=\chi_{m} J _{f}                            (6.33)

B _{ dip }( r )=\frac{\mu_{0}}{4 \pi} \frac{1}{r^{3}}[3( m \cdot \hat{ r }) \hat{ r }- m ]                                    (5.89)

\underset{\text { dipole }}{ B _{\text {center }}}=\frac{\mu}{4 \pi} \frac{1}{r^{3}}[3( m \cdot \hat{ r }) \hat{ r }- m ] .

This accounts for the first term in the field. The remainder must be due to the bound surface current (K _{b}) at r = R (since there can be no volume bound current, according to Eq. 6.33). Let us make an educated guess (based either on the answer provided or on the analogous electrical Prob. 4.37) that the field due to the surface bound current is (for interior points) of the form \underset{\text { curface }}{ B _{\text {surfant }}}=A m (i.e. a constant, proportional to m). In that case the magnetization will be: 

M =\chi_{m} H =\frac{\chi_{m}}{\mu} B =\frac{\chi_{m}}{4 \pi} \frac{1}{r^{3}}[3( m \cdot \hat{ r }) \hat{ r }- m ]+\frac{\chi_{m}}{\mu} A m .

This will produce bound currents  J _{b}= \nabla \times M = 0 , as it should, for 0 < r < R (no need to calculate this curl—the second term is constant, and the first is essentially the field of a dipole, which we know is curl-less, except at r = 0), and

K _{b}= M (R) \times \hat{ r }=\frac{\chi_{m}}{4 \pi R^{3}}(- m \times \hat{ r })+\frac{\chi_{m} A}{\mu}( m \times \hat{ r })=\chi_{m} m\left(-\frac{1}{4 \pi R^{3}}+\frac{A}{\mu}\right) \sin \theta \hat{\phi} .

But this is exactly the surface current produced by a spinning sphere:  K =\sigma v =\sigma \omega R \sin \theta \hat{\phi}, \text { with }(\sigma \omega R) \leftrightarrow \chi_{m} m\left(\frac{A}{\mu}-\frac{1}{4 \pi R^{3}}\right) . So the field it produces (for points inside) is (Eq. 5.70):

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)

\underset{\text { current }}{ B _{\text {surface }}}=\frac{2}{3} \mu_{0}(\sigma \omega R)=\frac{2}{3} \mu_{0} \chi_{m} m \left(\frac{A}{\mu}-\frac{1}{4 \pi R^{3}}\right) .

Everything is consistent, therefore, provided  A=\frac{2}{3} \mu_{0} \chi_{m}\left(\frac{A}{\mu}-\frac{1}{4 \pi R^{3}}\right), \text { or } A\left(1-\frac{2 \mu_{0}}{3 \mu} \chi_{m}\right)=-\frac{2}{3} \frac{\mu_{0} \chi_{m}}{4 \pi R^{3}} .

But  \chi_{m}=\left(\frac{\mu}{\mu_{0}}\right)-1, \text { so } A\left(1-\frac{2}{3}+\frac{2}{3} \frac{\mu_{0}}{\mu}\right)=-\frac{2}{3} \frac{\left(\mu-\mu_{0}\right)}{4 \pi R^{3}}, \text { or } A\left(1+\frac{2 \mu_{0}}{\mu}\right)=2 \frac{\left(\mu_{0}-\mu\right)}{4 \pi R^{3}} ; A=\frac{\mu}{4 \pi} \frac{2\left(\mu_{0}-\mu\right)}{R^{3}\left(2 \mu_{0}+\mu\right)} , and hence

B =\frac{\mu}{4 \pi}\left\{\frac{1}{r^{3}}[3( m \cdot \hat{ r }) \hat{ r }- m ]+\frac{2\left(\mu_{0}-\mu\right) m }{R^{3}\left(2 \mu_{0}+\mu\right)}\right\} . qed

The exterior field is that of the central dipole plus that of the surface current, which, according to Prob. 5.37, is also a perfect dipole field, of dipole moment

\underset{\text { current }}{ m _{\text {surface }}}=\frac{4}{3} \pi R^{3}(\sigma \omega R)=\frac{4}{3} \pi R^{3}\left(\frac{3}{2 \mu_{0}} B _{\text {current  surface }}\right)=\frac{2 \pi R^{3}}{\mu_{0}} \frac{\mu}{4 \pi} \frac{2\left(\mu_{0}-\mu\right) m }{R^{3}\left(2 \mu_{0}+\mu\right)}=\frac{\mu\left(\mu_{0}-\mu\right) m }{\mu_{0}\left(2 \mu_{0}+\mu\right)} .

So the total dipole moment is:

and hence the field (for r > R) is

B =\frac{\mu_{0}}{4 \pi}\left(\frac{3 \mu}{2 \mu_{0}+\mu}\right) \frac{1}{r^{3}}[3( m \cdot \hat{ r }) \hat{ r }- m ] .