Question 8.16: ^17 A sphere of radius R carries a uniform polarization P an...

According to Eqs. 3.104, 4.14, 5.89, and 6.16, the fields are

E _{ dip }( r )=\frac{1}{4 \pi \epsilon_{0}} \frac{1}{r^{3}}[3( p \cdot \hat{ r }) \hat{ r }- p ]                               (3.104)

E =-\nabla V=-\frac{P}{3 \epsilon_{0}} \hat{ z }=-\frac{1}{3 \epsilon_{0}} P , \quad \text { for } \quad r<R                          (4.14)

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

B =\frac{2}{3} \mu_{0} M                                 (6.16)

where p =(4 / 3) \pi R^{3} P , \text { and } m =(4 / 3) \pi R^{3} M . \text { Now } p =\epsilon_{0} \int( E \times B ) d \tau , and there are two contributions, one from inside the sphere and one from outside.

Inside:

p _{ in }=\epsilon_{0} \int\left(-\frac{1}{3 \epsilon_{0}} P \right) \times\left(\frac{2}{3} \mu_{0} M \right) d \tau=-\frac{2}{9} \mu_{0}( P \times M ) \int d \tau=-\frac{2}{9} \mu_{0}( P \times M ) \frac{4}{3} \pi R^{3}=\frac{8}{27} \mu_{0} \pi R^{3}( M \times P ).

Outside:

p _{\text {out }}=\epsilon_{0} \frac{1}{4 \pi \epsilon_{0}} \frac{\mu_{0}}{4 \pi} \int \frac{1}{r^{6}}\{[3( p \cdot \hat{ r }) \hat{ r }- p ] \times[3( m \cdot \hat{ r }) \hat{ r }- m ]\} d \tau.

Now  \hat{ r } \times( p \times m )= p (\hat{ r } \cdot m )- m (\hat{ r } \cdot p ), \text { so } \hat{ r } \times[\hat{ r } \times( p \times m )]=(\hat{ r } \cdot m )(\hat{ r } \times p )-(\hat{ r } \cdot p )(\hat{ r } \times m ) , whereas using the BAC-CAB rule directly gives \hat{ r } \times[\hat{ r } \times( p \times m )]=\hat{ r }[\hat{ r } \cdot( p \times m )]-( p \times m )(\hat{ r } \cdot \hat{ r }) . \text { So }\{[3( p \cdot \hat{ r }) \hat{ r }- p ] \times[3( m \cdot \hat{ r }) \hat{ r }- m ]\}= -3( p \cdot \hat{ r })(\hat{ r } \times m )+3( m \cdot \hat{ r })(\hat{ r } \times p )+( p \times m )=3\{\hat{ r }[\hat{ r } \cdot( p \times m )]-( p \times m )\}+( p \times m )=-2( p \times m )+3 \hat{ r }[\hat{ r } \cdot( p \times m )].

p _{\text {out }}=\frac{\mu_{0}}{16 \pi^{2}} \int \frac{1}{r^{6}}\{-2( p \times m )+3 \hat{ r }[\hat{ r } \cdot( p \times m )]\} r^{2} \sin \theta d r d \theta d \phi .

To evaluate the integral, set the z axis along (p × m); then \hat{ r } \cdot( p \times m )=| p \times m | \cos \theta . Meanwhile, \hat{ r }=\sin \theta \cos \phi \hat{ x }+\sin \theta \sin \phi \hat{ y }+\cos \theta \hat{ z } \text {. But } \sin \phi \text { and } \cos \phi \text { integrate to zero, so the } \hat{ x } \text { and } \hat{ y } terms drop out, leaving 

p _{\text {out }}=\frac{\mu_{0}}{16 \pi^{2}}\left(\int_{0}^{\infty} \frac{1}{r^{4}} d r\right)\left\{-2( p \times m ) \int \sin \theta d \theta d \phi+3| p \times m | \hat{ z } \int \cos ^{2} \theta \sin \theta d \theta d \phi\right\}

=\left.\frac{\mu_{0}}{16 \pi^{2}}\left(-\frac{1}{3 r^{3}}\right)\right|_{R} ^{\infty}\left[-2( p \times m ) 4 \pi+3( p \times m ) \frac{4 \pi}{3}\right]=-\frac{\mu_{0}}{12 \pi R^{3}}( p \times m )

=-\frac{\mu_{0}}{12 \pi R^{3}}\left(\frac{4}{3} \pi R^{3} P \right) \times\left(\frac{4}{3} \pi R^{3} M \right)=\frac{4 \mu_{0}}{27} R^{3}( M \times P ) .

p _{ tot }=\left(\frac{8}{27}+\frac{4}{27}\right) \mu_{0} R^{3}( M \times P )=\frac{4}{9} \mu_{0} R^{3}( M \times P ).