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?

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DeniceSandidge DeniceSandidge · Answer 1 · 2020-03-04T10:06:51+01:00

Answer:

maximum value of the magnetic field B = 1 ×[tex]10^{-8}[/tex] 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 = [tex]\frac{E}{C}[/tex] .........1

maximum value of the magnetic field = [tex]\frac{3}{3 \times 10^8}[/tex]

maximum value of the magnetic field B = 1 ×[tex]10^{-8}[/tex] T

and

now we get average intensity of the light that is

average intensity of the light = [tex]\frac{EB}{2\mu _o}[/tex] .........2

average intensity of the light = [tex]\frac{3 \times 1 \times 10^{-8}}{2 \times 4\pi \times 10^{-7}}[/tex]

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