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satela [25.4K]
4 years ago
7

A plane electromagnetic wave varies sinusoidally at 86.0 MHz as it travels through vacuum along the positive x-direction. The pe

ak 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)
Physics
1 answer:
VARVARA [1.3K]4 years ago
6 0

Answer:

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

Explanation:

 From the question we are told

     The  frequency of the electromagnetic wave is  f = 86.0 Hz

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

         

Generally  the intensity of this wave is mathematically represented as

     

                I = c  * \frac{1}{2} * \epsilon_o E^2_o

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

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

Substituting values into equation for intensity

               I = 3.0 *10^8  * 0.5 * 8.85 *10^{-12} * 2.30*10^{-3}

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

Converting to cm^2 we divide by 10,000

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

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

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

                I = 0.0003053 \mu  W/cm^2

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