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.

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;

where;

root-mean-square electric field,

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;

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

568.98

explanation:

its math

the correct choice is a .

do not post any more test or exam questions on