Question

A traveling sinusoidal electromagnetic wave in vacuum has an electric field amplitude of 91.1 V/m. Find the intensity of this wave and calculate the energy flowing during 17.1 s through an area of 0.0269 m2 that is perpendicular to the wave\'s direction of propagation.

Answer 1

intensity is 11.0145 W/m^2

Energy is 5.07 J

Step-by-step explanation:

Given data

amplitude E =91.1 V/m

area = 0.0269 m2

time t = 17.1 s

to find out

intensity and energy

solution

we know intensity formula that is given below

intensity = E(rms)² / (cμ0)

here c is speed of light and μo is permeability of free space

we know these constant value c = 2.99 x 10^8

μo = 1.26 x 10^-6

put all these value and we get intensity

intensity = (91.1/√2)² / (2.99 x 10^8 × 1.26 x 10^-6)

intensity = 4149.605 / 3.7674 x 10^2

intensity is 11.0145 W/m^2

and Energy is calculated by this formula

Energy = intensity × time × area

put all value now here

Energy = 11.0145 × 17.1 × 0.0269

Energy is 5.07 J