The maximum electric field 9.8 m from a point light source is 3.0 V/m. What are (a) the maximum value of the magnetic field and (b) the average intensity of the light there? (c) What is the power of the source?

maximum value of the magnetic field  B  = 1 × T

average intensity of the light = 0.011937 W/m²

power of source =  14.40 J

Explanation:

given data

maximum electric field E = 3.0 V/m

distance from a point source r = 9.8 m

solution

first we get here maximum value of the magnetic field  

maximum value of the magnetic field   =   .........1

maximum value of the magnetic field   =

maximum value of the magnetic field  B  = 1 × T

and

now we get average intensity of the light that is

average intensity of the light =     .........2

average intensity of the light =  

average intensity of the light = 0.011937 W/m²

and

now we get power of source that is express as

power of source = average intensity × 4×π×r²   ..........3

power of source = 0.011937 × 4×π×9.8²

power of source =  14.40 J

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explanation: