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

See answer

Step-by-step 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 =(P)/(A)`

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

`I =(\eta*W)/(\pi r^2)=(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)= (EB)/(2\mu_(0))[W/m]`

now
`E` and
`B` are related:

`E=cB\\ B=(E)/(c)` and
`c=\frac{1}{\sqrt{\epsilon_(0) \mu_(0)}}`

replace in
`S_(average)`

`S_(average)=I= (c \epsilon_(0)E^2)/(2)[W/m]`

we replace the values and we get:

`E= \sqrt{(2I)/(\epsilon_(0)c)}`

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

therefore

`B=(E)/(c)=(576.24)/(3*10^(8))=1.92*10^(-6)[T]`