Chapter 24: Q34PE (page 889)

Suppose the maximum safe intensity of microwaves for human exposure is taken to be $1.00 W / m^{2}$ . (a) If a radar unit leaks 10.0 Wof microwaves (other than those sent by its antenna) uniformly in all directions, how far away must you be to be exposed to an intensity considered to be safe? Assume that the power spreads uniformly over the area of a sphere with no complications from absorption or reflection. (b) What is the maximum electric field strength at the safe intensity? (Note that early radar units leaked more than modern ones do. This caused identifiable health problems, such as cataracts, for people who worked near them.)

Short Answer

Expert verified

(a) The distance is 0.892m .

(b) The electric field strength is 27.44 V/m .

Step by step solution

Define electric field strength

The intensity of an electric field at a certain area is quantified as electric field strength. The volt per meter is the standard unit.

Evaluating the distance

(a)

The intensity of a wave is defined as the power emitted per unit area, and is calculated using the following formula,

$I = \frac{P}{A} I = \frac{P}{4 {ττr}^{2}} r = \sqrt{\frac{P}{4 ττI}}$

Where P is the power radiated,r is the distance from the source, and I is the intensity.

Substitute the values in the above equation,

$r = \sqrt{\frac{10 W}{4 ττ \times 1 W / m^{2}}} = 0.892 m Therefore, the distance is 0.892 m .$

Evaluating the electric field strength

(b)

The formula also gives intensity in terms of maximum electric field and light speed,

$I = \frac{{cε}_{0} E_{0}^{2}}{2} E_{0} = \sqrt{\frac{2 I}{{cε}_{0}}} Substitute the values in the above equation, E_{0} = \sqrt{\frac{2 \times 1.00 W / m^{2}}{(3 \times 10^{8} m / s) \times (8.85 \times 10^{- 12} C^{2} / N - m^{2})}} = 27.44 V / m Therefore, the electric field strength is 27.44 V / m .$