Answer:
292.3254055 W/m²
469.26267 V/m
[tex]1.56421\times 10^{-6}\ T[/tex]
Explanation:
P = Power of bulb = 90 W
d = Diameter of bulb = 7 cm
r = Radius = [tex]\frac{d}{2}=\frac{7}{2}=3.5\ cm[/tex]
[tex]\epsilon_0[/tex] = Permittivity of free space = [tex]8.85\times 10^{-12}\ F/m[/tex]
c = Speed of light = [tex]3\times 10^8\ m/s[/tex]
The intensity is given by
[tex]I=\frac{P}{A}\\\Rightarrow I=\frac{90}{4\pi 0.035^2}\\\Rightarrow I=5846.50811\ W/m^2[/tex]
5% of this energy goes to the visible light so the intensity is
[tex]I=0.05\times 5846.50811\\\Rightarrow I=292.3254055\ W/m^2[/tex]
The visible light intensity at the surface of the bulb is 292.3254055 W/m²
Energy density of the wave is
[tex]u=\frac{1}{2}\epsilon_0E^2[/tex]
Energy density is also given by
[tex]\frac{I}{c}=\frac{1}{2}\epsilon_0E^2\\\Rightarrow E=\sqrt{\frac{2I}{c\epsilon_0}}\\\Rightarrow E=\sqrt{\frac{2\times 292.3254055}{3\times 10^8\times 8.85\times 10^{-12}}}\\\Rightarrow E=469.26267\ V/m[/tex]
The amplitude of the electric field at this surface is 469.26267 V/m
Amplitude of a magnetic field is given by
[tex]B=\frac{E}{c}\\\Rightarrow B=\frac{469.26267}{3\times 10^8}\\\Rightarrow B=1.56421\times 10^{-6}\ T[/tex]
The amplitude of the magnetic field at this surface is [tex]1.56421\times 10^{-6}\ T[/tex]
