Numerical check of the good-conductor skin-depth bounds and the $\sigma =0$ exclusion

[Substitution and solution witness, clause (i)] By [F1], $a=\mu \epsilon {\omega }^2=1\cdot 4\cdot 9=36$ and $b=\mu \sigma \omega =1\cdot 100\cdot 3=300$; hence $r=\sqrt{a^2+b^2}=\sqrt{{36}^2+{300}^2}=\sqrt{91296}$ and $t=a\mathrm{/}b=36\mathrm{/}300=3\mathrm{/}25$. To meet [A1]'s field hypotheses, write $u_1=(1,0,0)$, $u_2=(0,1,0)$, $u_3=(0,0,1)$ for the Cartesian unit directions, choose $E_{\ast }=1\mathrm{V}{\mathrm{m}}^{-1}$, and set $\lambda =p+iq$, $n=u_3$, $k=\lambda u_3$, $e=E_{\ast }{u}_1\mathrm{\ne }0$ and $h=\lambda E_{\ast }{u}_2\mathrm{/}(\mu \omega )$, which has units $\mathrm{A}{\mathrm{m}}^{-1}$. By [F1], ${\lambda }^2=\mu \epsilon {\omega }^2+i\mu \sigma \omega$, so $k\cdot e=k\cdot h=0$, $k\times e=\lambda E_{\ast }{u}_2=\mu \omega h$ and $k\times h=-{\lambda }^2{E}_{\ast }{u}_1\mathrm{/}(\mu \omega )=-(\epsilon \omega +i\sigma )e$. Thus the $C^{\mathrm{\infty }}$ maps $E,H:{\mathbb{R}}^3\times \mathbb{R}\to {\mathbb{R}}^3$ of [F2], with these amplitudes and the stated $D,B,J_f$, satisfy the equations and are transverse; restrict them to $\mathrm{\Omega }\times I$ with $\mathrm{\Omega }=\{x_3>0\}$ and $I=\mathbb{R}$, where the decaying branch is bounded. This is an auxiliary ideal solution, with no interface or empirical assertion. Since $\mathrm{\mid }n\mathrm{\mid }=1$ and $3\mathrm{/}25\le 1\mathrm{/}4$ (equivalently $12\le 25$), all hypotheses of A1,(c) now hold and its good-conductor bounds apply.

[Exact identities, clause (ii)] By [F1] and A1, $p=\sqrt{(r+a)\mathrm{/}2}=\sqrt{(\sqrt{91296}+36)\mathrm{/}2}$, $q=\sqrt{(r-a)\mathrm{/}2}=\sqrt{(\sqrt{91296}-36)\mathrm{/}2}$ and $pq=b\mathrm{/}2=150$ exactly. Since $91296<{336}^2=112896$, positivity and the square-root order give $r=\sqrt{91296}<336$, hence $q^2=(r-36)\mathrm{/}2<300\mathrm{/}2=b\mathrm{/}2$, so $q<\sqrt{150}=\sqrt{b\mathrm{/}2}$; and the exact bounds of A1 read $\mathrm{\mid }q-\sqrt{150}\mathrm{\mid }\le (6\mathrm{/}25)\sqrt{150}=\frac{2\epsilon \omega }{\sigma }\sqrt{\frac{\mu \sigma \omega }{2}}$ and $\mathrm{\mid }\delta -\sqrt{2\mathrm{/}b}\mathrm{\mid }\le (12\mathrm{/}25)\sqrt{2\mathrm{/}b}=\frac{4\epsilon \omega }{\sigma }\sqrt{\frac{2}{\mu \sigma \omega }}$.

[The $\sigma =0$ exclusion, clause (iii)] For the same $\epsilon ,\mu ,\omega$ and $\sigma =0$, [F1] gives $b=0$, $r=a$, $q=\sqrt{(r-a)\mathrm{/}2}=0$ and $\delta =+\mathrm{\infty }$ by the convention of [F1]; the identities and bounds of A1,(c) carry the hypothesis $b>0$, equivalently $\sigma >0$, and are not evaluated at $b=0$, which is the exclusion recorded in A1.

[Decimal displays and the bound comparisons] The decimal numbers displayed in clause (ii) of the Example are the stated roundings of the exact quantities $r=\sqrt{91296}$, $p=\sqrt{(r+36)\mathrm{/}2}$, $q=\sqrt{(r-36)\mathrm{/}2}$, $\delta =1\mathrm{/}q$, $\sqrt{150}$ and $\sqrt{2\mathrm{/}300}$; reading them as roundings, the comparisons $\mathrm{\mid }q-\sqrt{150}\mathrm{\mid }\le (6\mathrm{/}25)\sqrt{150}$ and $\mathrm{\mid }\delta -\sqrt{2\mathrm{/}b}\mathrm{\mid }\le (12\mathrm{/}25)\sqrt{2\mathrm{/}b}$ displayed there are instances of the exact inequalities of step 1.2, and the numerical magnitudes quoted in the Example are not independent assertions of exact decimal arithmetic. In particular the inequalities hold with the margin predicted by A1 for the declared dimensionless ratio $t=3\mathrm{/}25$.

[Scope, clause (iv)] The computation used only the declared parameters, the branch assumption [A1] and the exact definitions [F1]; it reports no measurement, no material identification and no empirical or statistical claim, and the decimal displays of the Example are approximate illustrations of the exact expressions, as stated. No choice principle is used.