Question

A traveling electromagnetic wave in a vacuum has an electric field amplitude of 93.3 V/m. Calculate the intensity S of this wave. Then, determine the amount of energy U that flows through area of 0.0287 m 2 over an interval of 13.7 s, assuming that the area is perpendicular to the direction of wave propagation. S

Answer 1

Intensity = 11.56W/m²

The energy flowing through the given area is 4.55 J

Step-by-step explanation:

The expression for the intensity of the electromagnetic wave is,

`I = (1)/(2) C{ {\varepsilon _0}E_m^2`

Here,
`\varepsilon _0` is the permittivity of the free space,

`E_m` is the electric field amplitude and

c is the speed of the light.

substitute

⁸m/s for c

8.85×10 −12 C² /N⋅m² for
`{\varepsilon _0}`

and 93.3 V/m for
`{E_{\rm{m}`

`I = (1)/(2) * (3*10^8)*(8.85*10^-^1^2)(93.3)\\\\I = 11.56W/m^2`

The expression for the energy is,

E = I×A×t

Here, I is the intensity of the electromagnetic wave,

A is the area, and

t is the time.

Substitute

11.56W/m² for I

0.0287m ² for A

13.7s for t

`E = (11.56)*(0,0287)*(13.7)\\E = 4.55J`

The energy flowing through the given area is 4.55 J