Page:ThomsonMagnetic1889.djvu/9

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

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

From (5) we have

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

From these equations we have

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

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

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

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

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

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

or

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