We can reasonably model a 90-W incandescent lightbulb as a sphere 5.7 cm in diameter. Typically, only about 5% of the energy goes to visible light; the rest goes largely to nonvisible infrared radiation.(a) What is the visible light intensity (in W/m^2) at the surface of the bulb?(b) What are the amplitudes of the electric and magnetic fields at this surface, for a sinusoidal wave with this intensity?

where [tex]\eta[/tex] is the efficiency of the lightbulb (visible light is 5% of the total power), [tex]W[/tex] is the total power of the lightbulb, r is the radius of the lightbulb in meters.

Intensity is power divided by area:

[tex]I =\frac{P}{A}[/tex]

a) Now the effective power is [tex]\eta*W[/tex], therefore:

[tex]I =\frac{\eta*W}{\pi r^2}=\frac{0.05*90}{4\pi (0.0285)^2}=440.87[W/m^2][/tex]

b) Now the intensity is the average poynting vector is related to the magnitudes of the maximum electric field and magnetic field amplitudes, following:

[tex]S_{average}= \frac{EB}{2\mu_{0}}[W/m][/tex]

now [tex]E[/tex] and [tex]B[/tex] are related:

[tex]E=cB\\ B=\frac{E}{c}[/tex] and [tex]c=\frac{1}{\sqrt{\epsilon_{0} \mu_{0}}} [/tex]

replace in [tex]S_{average}[/tex]

[tex]S_{average}=I= \frac{c \epsilon_{0}E^2}{2}[W/m][/tex]

we replace the values and we get:

[tex]E= \sqrt{\frac{2I}{\epsilon_{0}c}}[/tex]

[tex]E = \sqrt{\frac{2(440.8)}{8.85*10^{-12}3*10^8}}=576.24[V/m][/tex]

therefore

[tex]B=\frac{E}{c}=\frac{576.24}{3*10^{8}}=1.92*10^{-6}[T][/tex]