Question 5.44: Calculate the magnetic force of attraction between the north...

From Eq. 5.24, F =\int\left( K \times B _{\text {ave }}\right) d a . \text { Here } K =\sigma v , v =\omega R \sin \theta \hat{\phi}, d a=R^{2} \sin \theta d \theta d \phi , and  B _{\text {ave }}=\frac{1}{2}\left( B _{\text {in }}+ B _{\text {out }}\right). From Eq. 5.70,

F _{ mag }=\int( v \times B ) \sigma d a=\int( K \times B ) d a                                  (5.24)

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)

B _{\text {in }}=\frac{2}{3} \mu_{0} \sigma R \omega \hat{ z }=\frac{2}{3} \mu_{0} \sigma R \omega(\cos \theta \hat{ r }-\sin \theta \hat{ \theta })  . From Eq. 5.69,

A (r, \theta, \phi)= \begin{cases}\frac{\mu_{0} R \omega \sigma}{3} r \sin \theta \hat{\phi}, & (r \leq R), \\ \frac{\mu_{0} R^{4} \omega \sigma}{3} \frac{\sin \theta}{r^{2}} \hat{\phi}, & (r \geq R).\end{cases}                             (5.69)

=\frac{\mu_{0} R^{4} \omega \sigma}{3 r^{3}}(2 \cos \theta \hat{ r }+\sin \theta \hat{ \theta })=\frac{\mu_{0} R \omega \sigma}{3}(2 \cos \theta \hat{ r }+\sin \theta \hat{ \theta })(\text { since } r=R) .

B _{\text {ave }}=\frac{\mu_{0} R \omega \sigma}{6}(4 \cos \theta \hat{ r }-\sin \theta \hat{ \theta }) .

K \times B _{\text {ave }}=(\sigma \omega R \sin \theta)\left(\frac{\mu_{0} R \omega \sigma}{6}\right)[\hat{\phi} \times(4 \cos \theta \hat{ r }-\sin \theta \hat{ \theta })]=\frac{\mu_{0}}{6}(\sigma \omega R)^{2}(4 \cos \theta \hat{ \theta }+\sin \theta \hat{ r }) \sin \theta .

Picking out the z component of \hat{ \theta }(\text { namely, }-\sin \theta) \text { and of } \hat{ r } \text { (namely, } \cos \theta) , we have \left( K \times B _{\text {ave }}\right)_{z}=-\frac{\mu_{0}}{2}(\sigma \omega R)^{2} \sin ^{2} \theta \cos \theta , so

F_{z}=-\frac{\mu_{0}}{2}(\sigma \omega R)^{2} R^{2} \int \sin ^{3} \theta \cos \theta d \theta d \phi=-\left.\frac{\mu_{0}}{2}\left(\sigma \omega R^{2}\right)^{2} 2 \pi\left(\frac{\sin ^{4} \theta}{4}\right)\right|_{0} ^{\pi / 2}, \text { or } \quad F =-\frac{\mu_{0} \pi}{4}\left(\sigma \omega R^{2}\right)^{2} \hat{ z } .