Question

In the United States, household electric power is provided at a frequency of 60 HzHz, so electromagnetic radiation at that frequency is of particular interest. On the basis of the ICNIRP guidelines, what is the maximum intensity of an electromagnetic wave at this frequency to which the general public should be exposed

Answer 1

Answer:

the maximum intensity of an electromagnetic wave at the given frequency is 45 kW/m²

Explanation:

Given the data in the question;

To determine the maximum intensity of an electromagnetic wave, we use the formula;

$I$ = $\frac{1}{2}$ε₀cE$_{max$²

where ε₀ is permittivity of free space ( 8.85 × 10⁻¹² C²/N.m² )

c is the speed of light ( 3 × 10⁸ m/s )

E$_{max$ is the maximum magnitude of the electric field

first we calculate the maximum magnitude of the electric field ( E$_{max$  )

E$_{max$ = 350/f kV/m

given that frequency of 60 Hz, we substitute

E$_{max$ = 350/60 kV/m

E$_{max$ = 5.83333 kV/m

E$_{max$ = 5.83333 kV/m × ( $\frac{1000 V/m}{1 kV/m}$ )

E$_{max$ = 5833.33 N/C

so we substitute all our values into the formula for  intensity of an electromagnetic wave;

$I$ = $\frac{1}{2}$ε₀cE$_{max$²

$I$ = $\frac{1}{2}$ × ( 8.85 × 10⁻¹² C²/N.m² ) × ( 3 × 10⁸ m/s ) × ( 5833.33 N/C )²

$I$ = 45 × 10³ W/m²

$I$ = 45 × 10³ W/m² × ( $\frac{1 kW/m^2}{10^3W/m^2}$ )

$I$ = 45 kW/m²

Therefore, the maximum intensity of an electromagnetic wave at the given frequency is 45 kW/m²