Question

A sinusoidal electromagnetic wave from a radio station passes perpendicularly through an open window that has area of 0.500 m2 . At the window, the electric field of the wave has an rms value 0.0600 V/m .

How much energy does this wave carry through the window during a 30.0-s commercial? Express your answer with the appropriate units.

Answer 1

The energy of the wave is 1.435 x 10⁻⁴ J

Step-by-step explanation:

Given;

area of the window, A = 0.5 m²

the rms value of the field, E = 0.06 V/m

The peak value of electric field is given by;

`E_o = √(2) *E_(rms)\\\\E_o = √(2)*0.06\\\\E_o = 0.0849 \ V/m`

The average intensity of the wave is given by;

`I_(avg) = (c \epsilon_o E_o^2 )/(2)\\\\I_(avg) = ((3*10^8)( 8.85*10^(-12)) (0.0849)^2 )/(2)\\\\I_(avg) = 9.569*10^(-6) \ W/m^2`

The average power of the wave is given by;

P = I x A

P = (9.569 x 10⁻⁶ W/m²) (0.5 m²)

P = 4.784 x 10⁻⁶ W

The energy of the wave is given by;

E = P x t

E = (4.784 x 10⁻⁶ W)(30 s)

E = 1.435 x 10⁻⁴ J

Therefore, the energy of the wave is 1.435 x 10⁻⁴ J