Exercise 6.2.11: How much does a phone heat the head?

$E_0=\sqrt{\frac{2|{\eta }_c|S_0}{\cos {\theta }_{\eta }}}=\sqrt{\frac{2\times 57.0\times 5}{\cos 8.86^{\circ }}}=24.0V/m$.

$\mathrm{SAR}(0)=\frac{\sigma E_0^2}{2\rho }=\frac{1.405\times 24.0^2}{2000}=0.405W/kg$. Power balance: $S_{av}=S_0{e}^{-2\alpha z}$ gives $-d{S}_{av}/dz=2\alpha S_0=405W/{\mathrm{m}}^3$ at $z=0$, that is $0.405W/kg$.

The SAR is halved at $\ln 2/2\alpha =8.55\mathrm{mm}$. Over $1\mathrm{cm}$ the power factor is $e^{-2\alpha \times 0.01}=(1/1.5{)}^2=0.444$ (the amplitude ratio of Problem 6.2.7, squared), so $\mathrm{SAR}(1\mathrm{cm})=0.180W/kg$.

The power absorbed in the first centimetre per square metre is $S_0(1-0.444)=2.78\mathrm{W}$, and the mass is $\rho \times 0.01=10kg/{\mathrm{m}}^2$:

The average is proportional to $S_0$: $S_0=5\times 1.6/0.278=28.8W/{\mathrm{m}}^2$ for $1.6W/kg$ and $36.0W/{\mathrm{m}}^2$ for $2W/kg$.

The two routes in (b) agree because $\gamma /{\eta }_c=\sigma +\mathrm{j}\omega \epsilon$ gives $2\alpha \cos {\theta }_{\eta }/|{\eta }_c|=\sigma$, and $2\times 40.55\times 0.988/57.0=1.405S/m$. The SAR limits are averaged over $1$ g (FCC, a cube of $1\mathrm{cm}$ side) or $10$ g (ICNIRP) of tissue, so (d) is the right kind of number to compare. A phone is not a plane-wave source: its near field concentrates on a few square centimetres, which is why SAR is measured in a head phantom filled with liquid of exactly the parameters of Problem 6.2.7 rather than computed.