Calculate the electric and magnetic fields produced by the radiation coming from a `100` W bulb at a distance of `3 m`. Assume that the efficiency of the bulb is `25%` and it is a point source.

Explanation

The bulb, as a point source, radiates light in all directions uniformly. At a distance of 3m , the surface area of the surrounding sphere is
`A=4pir^(2)=4pi(3)^(2)=113 m^(2)`
The intensity at this distance is
`l=("Power")/("Area")=(100Axx2.5%)/(113 m^(2))=0.22 W//m^(2)`
Half of this intensity is provided by the electric field and half by the magnetic, field
`(1)/(2)l=(1)/(2)(epsi_(0)E_(ms^(c))^(2))=(1)/(2)(0.022W//m^(2))`
`E_(ms)=sqrt((0.200)/((8.85xx10^(-12))(3xx10^(8))))V//m=2.9m`
The value of E found above is teh root mean square value of the electric. field since the electic field in light beam is sinusoidal, the peak electric field `E_(0)` is
`E_(0)=sqrt(2)E_(ms)=sqrt(2)xx2.9V//m=4.07V//m`
Thus, you see that the electric field strength of the light that you use for reading is fairly large. Compare it with electric field strength of TV or FM magnetic field. it is
`B_(ms)=(E_(ms))/(c)=(2.9xxVm^(-1))/(3xx10^(8) ms^(-1))=9.6xx10^(-9)T`
Again, since the filed in the light beam is sinusoidal the peak magnetic fild is `B_(0)=sqrt(2)B_(ms)=1.4xx10^(-8)`. Note that alhtough the energy in the magnetic field in equal to the energy in the electric, field, magnetic field strength is evidently very weak.
Let us consider a cylinder region where E.M. wave is passing through.