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

Answer:

Intensity = 11.56W/m²

The energy flowing through the given area is 4.55 J

Explanation:

The expression for the intensity of the electromagnetic wave is,

$I = \frac{1}{2} C{ {\varepsilon _0}E_m^2$

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 ${E_{\rm{m}$

$I = \frac{1}{2} \times (3\times10^8)\times(8.85\times10^-^1^2)(93.3)\\\\I = 11.56W/m^2$

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

$E = (11.56)\times(0,0287)\times(13.7)\\E = 4.55J$

The energy flowing through the given area is 4.55 J