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

Answer:

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

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   = $\frac{E}{C}$   .........1

maximum value of the magnetic field   = $\frac{3}{3 \times 10^8}$

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 =  $\frac{EB}{2\mu _o}$   .........2

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

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