Question

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?

Answer 1

Answer:

See answer

Explanation:

Given quantities:

$\eta = 0.05\\ W=90[W]\\r=0.0285[m]$

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

Intensity is power divided by area:

$I =\frac{P}{A}$

a) Now the effective power is $\eta*W$, therefore:

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

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

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

now $E$ and $B$ are related:

$E=cB\\ B=\frac{E}{c}$ and $c=\frac{1}{\sqrt{\epsilon_{0} \mu_{0}}}$

replace in $S_{average}$

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

we replace the values and we get:

$E= \sqrt{\frac{2I}{\epsilon_{0}c}}$

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

therefore

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