Respuesta :
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
