Page:ThomsonMagnetic1889.djvu/9

magnetic force outside and inside $=\omega \epsilon ^{ipt}\sin \theta {\dfrac {e}{a^{2}}}$; hence

$-3\mathrm {C} {\dfrac {d}{da}}\mathrm {E} _{0}(i\lambda a)-3\mathrm {D} {\dfrac {dS_{0}(\lambda _{1}a)}{da}}={\dfrac {\omega e}{a^{2}}}.$

From (5) we have

$\mathrm {C} {\big (}\mathrm {E} _{2}(i\lambda a)+2\mathrm {E} _{0}(i\lambda a){\big )}-\mathrm {D} {\big (}\mathrm {S} _{2}(\lambda _{1}a)+2\mathrm {S} _{0}(\lambda _{1}a){\big )}={\dfrac {\omega e}{p^{2}a^{3}\mathrm {K} }}.$

From these equations we have

$\mathrm {C} ={\frac {1}{\Delta }}\left\{{\dfrac {\omega e}{\mathrm {K} p^{2}a^{3}}}{\dfrac {d}{da}}\mathrm {S} _{0}(\lambda _{1}a)-{\frac {1}{3}}{\dfrac {\omega e}{a^{2}}}{\big (}\mathrm {S} _{2}(\lambda _{1}a)-2\mathrm {S} _{0}(\lambda _{1}a){\big )}\right\}$

(6)

$\qquad \mathrm {D} ={\frac {1}{\Delta }}\left\{-{\frac {1}{3}}{\dfrac {\omega e}{a^{2}}}\left(\mathrm {E} _{2}(i\lambda a)+2E_{0}(i\lambda a)\right)-{\dfrac {\omega e}{\mathrm {K} p^{2}a^{3}}}{\dfrac {d}{da}}\mathrm {E} _{0}(i\lambda a)\right\},$

(7)

where

$\Delta ={\Big (}\mathrm {E} _{2}(i\lambda a)+2\mathrm {E} _{0}(i\lambda a){\dfrac {d}{da}}\mathrm {S} _{0}(\lambda _{1}a)+{\dfrac {d}{da}}\mathrm {E} _{0}(i\lambda a){\big (}\mathrm {S} _{2}(\lambda a)-2\mathrm {S} _{0}(\lambda _{1}a){\big )}{\Big )}.$

Let us first consider the case where $\lambda a$ and $\lambda _{1}$ are both small. In this case $\mathrm {E} _{2}(i\lambda a)$ is large compared with $\mathrm {E} _{0}(i\lambda a)$ and $\mathrm {S} _{2}(\lambda _{1}a)$, very small compared with $\mathrm {S} _{0}(\lambda _{1}a)$; hence we see that

${\begin{array}{l}\mathrm {C} ={\dfrac {\omega e}{\mathrm {K} p^{2}a^{3}}}\mathrm {E} _{2}(i\lambda a),\\\\\mathrm {D} =-{\dfrac {-{\dfrac {\omega e}{a^{3}}}\mathrm {E} _{0}(i\lambda a)}{\mathrm {E} _{2}(i\lambda a){\dfrac {d}{da}}\mathrm {S} _{0}(\lambda _{1}a)}}.\end{array}}$

Since $\mathrm {E} _{2}(i\lambda a)={\frac {3\epsilon ^{-i\lambda a}}{(i\lambda a)^{3}}}$ approximately, and $\mathrm {S} _{0}(\lambda _{1}a)=1$, we have

${\begin{array}{l}\mathrm {C} =-{\frac {1}{3}}i\omega e\lambda \epsilon ^{i\lambda a},\\\\\mathrm {D} ={\dfrac {\omega e}{a}}{\dfrac {ip\sigma \mathrm {K} }{4\pi }}.\end{array}}$

The magnetic force outside the sphere parallel to the axis of $x$ equals

${\dfrac {d\mathrm {H} _{1}}{dy}}-{\dfrac {d\mathrm {G} _{1}}{dz}},$

or

${\begin{array}{c}-3\mathrm {C} {\dfrac {d}{dy}}\mathrm {E} _{0}(i\lambda r)\epsilon ^{ipt},\\\\=i\omega e\lambda \epsilon ^{i\lambda a}{\dfrac {d}{dy}}{\dfrac {\epsilon ^{-i\lambda a}}{i\lambda r}}\cdot \epsilon ^{ipt},\end{array}}$