Question

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?

Answer 1

maximum value of the magnetic field B = 1 ×
`10^(-8)` T

average intensity of the light = 0.011937 W/m²

power of source = 14.40 J

Step-by-step 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 =
`(E)/(C)` .........1

maximum value of the magnetic field =
`(3)/(3 * 10^8)`

maximum value of the magnetic field B = 1 ×
`10^(-8)` T

and

now we get average intensity of the light that is

average intensity of the light =
`(EB)/(2\mu _o)` .........2

average intensity of the light =
`(3 * 1 * 10^(-8))/(2 * 4\pi * 10^(-7))`

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