Question

A traveling sinusoidal electromagnetic wave in vacuum has an electric field amplitude of 96.9 V/m. Find the intensity of this wave and calculate the energy flowing during 14.9 s through an area of 0.0227 m2 that is perpendicular to the wave\'s direction of propagation.

Answer 1

Answer:

• the intensity of this wave, I = 12.42 W/m²

• the energy of this wave, U = 4.2 J

Explanation:

Given;

peak electric field, E₀ = 96.9 V/m

time of flow, t = 14.9s

area through which the energy flows, A = 0.0227 m²

The intensity of this wave is calculated using the following formula;

$I = \frac{E_{rms}^2}{c \mu_o}$

where;

root-mean-square electric field, $E_{rms} = \frac{E_o}{\sqrt{2}} = \frac{96.9}{\sqrt{2} } = 68.5187 \ V/m$

c is speed of light, c = 3 x 10⁸ m/s

μ₀ is permeability of free space (constant), μ₀ = 1.26 x 10⁻⁶

Substitute these values and calculate the intensity of the wave;

$I = \frac{E_{rms}^2}{c \mu_o} = \frac{(68.5187)^2}{(3*10^8)(1.26*10^{-6})} = 12.42 \ W/m^2$

Thus, intensity of this wave is 12.42 W/m²

The energy of the wave is calculated as follows;

U = IAt

U = 12.42 x 0.0227 x 14.9

U = 4.2 J

Thus, the energy of this wave is 4.2 J