Problem 6.1.8: Safety limits for human exposure to electromagnetic radiation

Problem 6.1.8

Safety limits for human exposure to electromagnetic radiation

There exist various effects of electromagnetic fields on humans and many quite different recommendations and standards for the safety limits of human exposure to electromagnetic radiation. The IEEE standards at frequencies between $100 MHz$ and $300 GHz$ are as follows:

(i) $(P_{a v})_{m a x} = 2 W/ m^{2}$ for $100 MHz \leq f \leq 300 MHz$ ,
(ii) $(P_{a v})_{m a x} = (f /150) W/ m^{2}$ ( $f$ in MHz) for $300 MHz \leq f \leq 15 GHz$ , and
(iii) $(P_{a v})_{m a x} = 100 W/ m^{2}$ for $15 GHz \leq f \leq 300 GHz$ .

Based on this standard, compute the maximum permissible levels for the intensities of the electric (and magnetic) fields in air at frequencies of $f_{1} = 150 MHz$ , $f_{2} = 1.5 GHz$ and $f_{3} = 15 GHz$ , respectively.

Adapted from: B. M. Notaroš, Electromagnetics, Problem 9.16.

Try it yourself first.

Strategy

Pick the rule that applies at each frequency, then convert the power density into field amplitudes with the plane-wave relations in air, $P_{a v} = E_{m}^{2} /2 η_{0} = E_{rms}^{2} / η_{0}$ and $H_{m} = E_{m} / η_{0}$ :

E_{m, m a x} = 2 η_{0} (P_{a v})_{m a x} .

1. The limits

$f_{1} = 150 MHz$ , rule (i): $2 W/ m^{2}$ . $f_{2} = 1500 MHz$ , rule (ii): $1500/150 = 10 W/ m^{2}$ . $f_{3} = 15 GHz$ , rules (ii) and (iii) agree: $15 000/150 = 100 W/ m^{2}$ .

2. The fields

E_{1, m a x} E_{2, m a x} E_{3, m a x} = 2 \times 120 π \times 2 = 38.83 V/m, = 2 \times 120 π \times 10 = 86.83 V/m, = 2 \times 120 π \times 100 = 274.6 V/m,

and $H_{m} = E_{m} /377$ . The rms values are $1/ 2$ of the amplitudes:

	$150$ MHz	$1.5$ GHz	$15$ GHz
$E_{m}$ (V/m)	$38.83$	$86.83$	$274.6$
$H_{m}$ (A/m)	$0.103$	$0.230$	$0.728$
$E_{rms}$ (V/m)	$27.46$	$61.40$	$194.2$
$H_{rms}$ (A/m)	$0.0728$	$0.163$	$0.515$

Answer

E_{m, m a x} H_{m, m a x} = 38.83, 86.83, 274.6 V/m = 0.103, 0.230, 0.728 A/m

Check

$E_{rms} H_{rms}$ must return the limit: $27.46 \times 0.0728 = 2.00$ , $61.40 \times 0.163 = 10.0$ and $194.2 \times 0.515 = 100 W/ m^{2}$ . Between $f_{2}$ and $f_{3}$ the power limit grows tenfold and the field limit by $10$ : $86.83 10 = 274.6$ .

Problem 6.1.8. The permitted field amplitude against frequency under the three rules, with the three frequencies of the problem. The limit is lowest around 100 to 300\,\mathrm{MHz}, where a human body, about 1.7\,\mathrm m tall and so comparable with \lambda/2, absorbs most efficiently (whole-body resonance). At higher frequencies the energy is absorbed in a thinner and thinner layer of skin, and the standard lets the power density, and with it E\propto\sqrt{\mathcal P_{av}}, rise. — **Figure 6.1.31.** Problem 6.1.8. The permitted field amplitude against frequency under the three rules, with the three frequencies of the problem. The limit is lowest around $100$ to $300 MHz$ , where a human body, about $1.7 m$ tall and so comparable with $λ /2$ , absorbs most efficiently (whole-body resonance). At higher frequencies the energy is absorbed in a thinner and thinner layer of skin, and the standard lets the power density, and with it $E \propto P_{a v}$ , rise.

Amplitude or rms, and which standard

The statement asks for the "field intensities", and the solution above gives the amplitudes $E_{m}$ . The cited source asks for rms values, $E_{m} / 2$ . Both are in the table. The numbers are those of the IEEE standard for uncontrolled environments used by that source. Later editions and the ICNIRP guidelines differ, for the general public about $10 W/ m^{2}$ above a few GHz, but the method does not change.