Calculate the force of magnetic attraction between the northern and southern hemispheres of a uniformly charged spinning spherical shell, with radius R, angular velocity ω, and surface charge density σ. [This is the same as Prob. 5.44, but this time use the Maxwell stress tensor and Eq. 8.21.]
F =\oint_{ S } \overleftrightarrow{ T } \cdot d a (8.21)
The force is clearly in the z direction, so we need
(\overleftrightarrow{T} \cdot d a )_{z}=T_{z x} d a_{x}+T_{z y} d a_{y}+T_{z z} d a_{z}=\frac{1}{\mu_{0}}\left(B_{z} B_{x} d a_{x}+B_{z} B_{y} d a_{y}+B_{z} B_{z} d a_{z}-\frac{1}{2} B^{2} d a_{z}\right)
=\frac{1}{\mu_{0}}\left[B_{z}( B \cdot d a )-\frac{1}{2} B^{2} d a_{z}\right] .
Now B =\frac{2}{3} \mu_{0} \sigma R \omega \hat{ z } \text { (inside) and } B =\frac{\mu_{0} m}{4 \pi r^{3}}(2 \cos \theta \hat{ r }+\sin \theta \hat{ \theta }) \text { (outside), where } m=\frac{4}{3} \pi R^{3}(\sigma \omega R) .
(From Eq. 5.70, Prob. 5.37, and Eq. 5.88.) We want a surface that encloses the entire upper hemisphere—say a hemispherical cap just outside r = R plus the equatorial circular disk.
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 {dip }}( r )=\nabla \times A =\frac{\mu_{0} m}{4 \pi r^{3}}(2 \cos \theta \hat{ r }+\sin \theta \hat{ \theta }) (5.88)
Hemisphere:
B_{z}=\frac{\mu_{0} m}{4 \pi R^{3}}\left[2 \cos \theta(\hat{ r })_{z}+\sin \theta(\hat{ \theta })_{z}\right]=\frac{\mu_{0} m}{4 \pi R^{3}}\left[2 \cos ^{2} \theta-\sin ^{2} \theta\right]=\frac{\mu_{0} m}{4 \pi R^{3}}\left(3 \cos ^{2} \theta-1\right).
d a =R^{2} \sin \theta d \theta d \phi \hat{ r } ; B \cdot d a =\frac{\mu_{0} m}{4 \pi R^{3}}(2 \cos \theta) R^{2} \sin \theta d \theta d \phi ; d a_{z}=R^{2} \sin \theta d \theta d \phi \cos \theta;
B^{2}=\left(\frac{\mu_{0} m}{4 \pi R^{3}}\right)^{2}\left(4 \cos ^{2} \theta+\sin ^{2} \theta\right)=\left(\frac{\mu_{0} m}{4 \pi R^{3}}\right)^{2}\left(3 \cos ^{2} \theta+1\right).
(\overleftrightarrow{T} \cdot d a )_{z}=\frac{1}{\mu_{0}}\left(\frac{\mu_{0} m}{4 \pi R^{3}}\right)^{2}\left[\left(3 \cos ^{2} \theta-1\right) 2 \cos \theta R^{2} \sin \theta d \theta d \phi-\frac{1}{2}\left(3 \cos ^{2} \theta+1\right) R^{2} \sin \theta \cos \theta d \theta d \phi\right]=\mu_{0}\left(\frac{\sigma \omega R}{3}\right)^{2}\left[\frac{1}{2} R^{2} \sin \theta \cos \theta d \theta d \phi\right]\left(12 \cos ^{2} \theta-4-3 \cos ^{2} \theta-1\right)
=\frac{\mu_{0}}{2}\left(\frac{\sigma \omega R^{2}}{3}\right)^{2}\left(9 \cos ^{2} \theta-5\right) \sin \theta \cos \theta d \theta d \phi
\left(F_{\text {hemi }}\right)_{z}=\frac{\mu_{0}}{2}\left(\frac{\sigma \omega R^{2}}{3}\right)^{2} 2 \pi \int_{0}^{\pi / 2}\left(9 \cos ^{3} \theta-5 \cos \theta\right) \sin \theta d \theta=\left.\mu_{0} \pi\left(\frac{\sigma \omega R^{2}}{3}\right)^{2}\left[-\frac{9}{4} \cos ^{4} \theta+\frac{5}{2} \cos ^{2} \theta\right]\right|_{0} ^{\pi / 2}
=\mu_{0} \pi\left(\frac{\sigma \omega R^{2}}{3}\right)^{2}\left(0+\frac{9}{4}-\frac{5}{2}\right)=-\frac{\mu_{0} \pi}{4}\left(\frac{\sigma w R^{2}}{3}\right)^{2} .
Disk:
B_{z}=\frac{2}{3} \mu_{0} \sigma R \omega ; \quad d a =r d r d \phi \hat{ \phi }=-r d r d \phi \hat{ z };
B \cdot d a =-\frac{2}{3} \mu_{0} \sigma R \omega r d r d \phi ; \quad B^{2}=\left(\frac{2}{3} \mu_{0} \sigma R \omega\right)^{2} ; \quad d a_{z}=-r d r d \phi.
(\overleftrightarrow{T} \cdot d a )_{z}=\frac{1}{\mu_{0}}\left(\frac{2}{3} \mu_{0} \sigma R \omega\right)^{2}\left[-r d r d \phi+\frac{1}{2} r d r d \phi\right]=-\frac{1}{2 \mu_{0}}\left(\frac{2}{3} \mu_{0} \sigma R \omega\right)^{2} r d r d \phi.
\left(F_{ disk }\right)_{z}=-2 \mu_{0}\left(\frac{\sigma \omega R}{3}\right)^{2} 2 \pi \int_{0}^{R} r d r=-2 \pi \mu_{0}\left(\frac{\sigma \omega R^{2}}{3}\right)^{2}.
Total:
F =-\pi \mu_{0}\left(\frac{\sigma \omega R^{2}}{3}\right)^{2}\left(2+\frac{1}{4}\right) \hat{ z }=-\pi \mu_{0}\left(\frac{\sigma \omega R^{2}}{2}\right)^{2} \hat{ z } (agrees with Prob. 5.44).
Alternatively, we could use a surface consisting of the entire equatorial plane, closing it with a hemispherical surface “at infinity,” where (since the field is zero out there) the contribution is zero. We have already done the integral over the disk; it remains to do rest of the integral over the plane, from R to ∞. On the plane,
\theta=0 \text {, and }(\text { for } r>R) B =\frac{\mu_{0} m}{4 \pi r^{3}} \hat{ \theta }=-\frac{\mu_{0} m}{4 \pi r^{3}} \hat{ z } , so
\left(F_{\text {rest }}\right)_{z}=-\frac{1}{2 \mu_{0}}\left(\frac{\mu_{0} m}{4 \pi}\right)^{2} 2 \pi \int_{R}^{\infty} \frac{1}{r^{5}} d r=-\left.\frac{\mu_{0}}{16 \pi}\left(\frac{4}{3} \pi R^{4} \sigma \omega\right)^{2}\left[-\frac{1}{4 r^{4}}\right]\right|_{R} ^{\infty}=-\mu_{0} \pi\left(\frac{R^{4} \sigma \omega}{3}\right)^{2}\left[\frac{1}{4 R^{4}}\right]=-\frac{\mu_{0} \pi}{4}\left(\frac{\sigma \omega R^{2}}{3}\right)^{2}
This is the same as \left(F_{\text {hemi }}\right)_{z}, \text { so-when added to }\left(F_{\text {disk }}\right)_{z} —it will yield the same total force as before.