Problem 6.3.7: A mobile-phone wave on body tissue

$\omega {\epsilon }_0=2\pi \times 1.8\times 10^9\times \frac{10^{-9}}{36\pi }=0.1$ S/m, so $\sigma /\omega {\epsilon }_0=14$ and

since $(7-\mathrm{j}{)}^2=49-14\mathrm{j}-1$. $\tan \delta =0.29$ is neither $\ll 1$ (good dielectric) nor $\gg 1$ (good conductor). Then

and with ${\beta }_0=\omega /c=12\pi =37.7$ rad/m, ${\gamma }_2=\mathrm{j}{\beta }_0(7-\mathrm{j})={\beta }_0+\mathrm{j}7{\beta }_0$:

With ${\eta }_2/{\eta }_0=(7+\mathrm{j})/50$:

In air $|E_1(x)|=20|1+\Gamma {\mathrm{e}}^{2\mathrm{j}{\beta }_{0}x}|$. Its maxima, $20\times 1.754=35.1$ V/m, lie where $2{\beta }_{0}x+{\theta }_{\Gamma }=-2m\pi$:

The minima, $20\times 0.246=4.91$ V/m, lie ${\lambda }_0/4=4.17$ cm from the maxima, where $2{\beta }_{0}x+{\theta }_{\Gamma }=\pm \pi ,\pm 3\pi ,\dots$ The solution with $+\pi$, $x=(\pi -{\theta }_{\Gamma })/(2{\beta }_0)=+0.54$ mm, lies just inside the tissue, where the air expression does not hold. The nearest minimum in air is therefore at

and between $-4.11$ cm and the surface $|E_1|$ falls monotonically to $|E_1(0)|=|\tau |\times 20=4.96$ V/m, only $1\%$ above the minimum value: practically a minimum on the surface.

and the rest, $28/65=43.1\%$, enters the tissue and is absorbed there. In watts: ${\mathcal{P}}^i=20^2/(2\times 120\pi )=0.531$ W/m${}^2$, of which $0.229$ W/m${}^2$ enters.

Just inside, $|E^t(0)|=|\tau |E_0^i=4.96$ V/m, and the Joule loss per unit volume is

a specific absorption rate $p/\rho =0.017$ W/kg for $\rho \approx 1000$ kg/m${}^3$. The field decays as ${\mathrm{e}}^{-{\alpha }_{2}x}$, so $p(x)=p(0){\mathrm{e}}^{-2{\alpha }_{2}x}$, and $p(x)/p(0)=0.01$ at $x=\ln 100/(2{\alpha }_2)$.

From the transmitted wave, Eq. (6.3.17): $\mathrm{Re}\{1/{\eta }_2^{\ast }\}=7/{\eta }_0$, so $T={\eta }_0|\tau {|}^2\mathrm{Re}\{1/{\eta }_2^{\ast }\}=7\times \frac{260}{4225}=0.431$ $✓$. All the power that enters is turned into heat: ${\int }_0^{\infty }pdx=p(0)/(2{\alpha }_2)=17.2/75.4=0.229$ W/m${}^2$ $✓$.