^{19} Suppose you had an electric charge q_{e} and a magnetic monopole q_{m}. The field of the electric charge is
E =\frac{1}{4 \pi \epsilon_{0}} \frac{q_{e}}{ᴫ^{2}} \hat{ ᴫ}(of course), and the field of the magnetic monopole is
B =\frac{\mu_{0}}{4 \pi} \frac{q_{m}}{ᴫ^{2}} \hat{ᴫ } .
Find the total angular momentum stored in the fields, if the two charges are separated by a distance d. \text { [Answer: ( } \left.\left.\mu_{0} / 4 \pi\right) q_{e} q_{m} .\right]^{20}
E =\frac{q_{e}}{4 \pi \epsilon_{0}} \frac{ r }{r^{3}};
B =\frac{\mu_{0} q_{m}}{4 \pi} \frac{ r ^{\prime}}{r^{\prime 3}}=\frac{\mu_{0} q_{m}}{4 \pi} \frac{( r -d \hat{ z })}{\left(r^{2}+d^{2}-2 r d \cos \theta\right)^{3 / 2}} .
Momentum density (Eq. 8.32):
g =\epsilon_{0}( E \times B ) (8.32)
g =\epsilon_{0}( E \times B )=\frac{\mu_{0} q_{e} q_{m}}{(4 \pi)^{2}} \frac{(-d)( r \times \hat{ z })}{r^{3}\left(r^{2}+d^{2}-2 r d \cos \theta\right)^{3 / 2}} .
Angular momentum density (Eq. 8.33):
\ell = r \times g =\epsilon_{0}[ r \times( E \times B )] (8.33)
\ell =( r \times g )=-\frac{\mu_{0} q_{e} q_{m} d}{(4 \pi)^{2}} \frac{ r \times( r \times \hat{ z })}{r^{3}\left(r^{2}+d^{2}-2 r d \cos \theta\right)^{3 / 2}} . \quad \text { But } r \times( r \times \hat{ z })= r ( r \cdot \hat{ z })-r^{2} \hat{ z }=r^{2} \cos \theta \hat{ r }-r^{2} \hat{ z } .
The x and y components will integrate to zero; using (\hat{ r })_{z}=\cos \theta , we have:
L =-\frac{\mu_{0} q_{e} q_{m} d}{(4 \pi)^{2}} \hat{ z } \int \frac{r^{2}\left(\cos ^{2} \theta-1\right)}{r^{3}\left(r^{2}+d^{2}-2 r d \cos \theta\right)^{3 / 2}} r^{2} \sin \theta d r d \theta d \phi . \quad \text { Let } u \equiv \cos \theta:
=\frac{\mu_{0} q_{e} q_{m} d}{(4 \pi)^{2}} \hat{ z }(2 \pi) \int_{-1}^{1} \int_{0}^{\infty} \frac{r\left(1-u^{2}\right)}{\left(r^{2}+d^{2}-2 r d u\right)^{3 / 2}} d u d r .
Do the r integral first:
\int_{0}^{\infty} \frac{r d r}{\left(r^{2}+d^{2}-2 r d u\right)^{3 / 2}}=\left.\frac{(r u-d)}{d\left(1-u^{2}\right) \sqrt{r^{2}+d^{2}-2 r d u}}\right|_{0} ^{\infty}=\frac{u}{d\left(1-u^{2}\right)}+\frac{d}{d\left(1-u^{2}\right) d}=\frac{u+1}{d\left(1-u^{2}\right)}=\frac{1}{d(1-u)} .
Then
L =\frac{\mu_{0} q_{e} q_{m} d}{8 \pi} \hat{ z } \frac{1}{d} \int_{-1}^{1} \frac{\left(1-u^{2}\right)}{(1-u)} d u=\frac{\mu_{0} q_{e} q_{m}}{8 \pi} \hat{ z } \int_{-1}^{1}(1+u) d u=\left.\frac{\mu_{0} q_{e} q_{m}}{8 \pi} \hat{ z }\left(u+\frac{u^{2}}{2}\right)\right|_{-1} ^{1}=\frac{\mu_{0} q_{e} q_{m}}{4 \pi} \hat{ z } .
