A plane electromagnetic wave varies sinusoidally at 86.0 MHz as it travels through vacuum along the positive x-direction. The peak value of the electric field is 2.30 mV/m, and it is directed along the positive y-direction. Determine the average power per unit area (the intensity) this wave carries through space. (Be careful with your units here and make sure to submit your answer in μW/cm2)

Respuesta :

Answer:

The intensity is  [tex]I = 0.0003053 \mu W/cm^2[/tex]

Explanation:

 From the question we are told

     The  frequency of the electromagnetic wave is  [tex]f = 86.0 Hz[/tex]

     The peak value of the electric field is  [tex]E_o = 2.30 mV/m = \frac{2.30}{1000 } = 2.30 *10^{-3} V/m[/tex]

         

Generally  the intensity of this wave is mathematically represented as

     

                [tex]I = c * \frac{1}{2} * \epsilon_o E^2_o[/tex]

Where c is the speed of light with value  [tex]c = 3 *10^8 m/s[/tex]

           [tex]\epsilon_o[/tex] is the permittivity of free space with value  [tex]\epsilon _o = 8.85 *10^{-12} C^2 /Nm^2[/tex]

Substituting values into equation for intensity

               [tex]I = 3.0 *10^8 * 0.5 * 8.85 *10^{-12} * 2.30*10^{-3}[/tex]

                 [tex]I = 3.053 *10^{-6} W/m^2[/tex]

Converting to [tex]cm^2[/tex] we divide by 10,000

                [tex]I = \frac{3.053 *10^{-6}}{10000} W/cm^2[/tex]

                [tex]= 3.053 *10^{-10} W/cm^2[/tex]

                [tex]= 0.0003053 *10^{-6} W/cm^2[/tex]

                [tex]I = 0.0003053 \mu W/cm^2[/tex]

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