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 3 m, the surface area of the surrounding sphere is
`A=4pir^(2)=4pi(3)^(2)=113m^(2)`
The intensity at this distance is
`I=("Power")/("Area")=(100Wxx2.5%)/(113m^(2))=0.022 W//m^(2)`
Half of this intensity is provided by the electric field and half by the magnetic field.
`(1)/(2)I=(1)/(2)(epsilon_(0)E_("rms")^(2)c)=(1)/(2)(0.222 W//m^(2))`
`E_("rms")=sqrt((0.022)/((8.85xx10^(-12))(3xx10^(8))))V//m=2.9 V//m`
The value of E found above is the root mean square value of the electric field. Since the electric field in a light beam is sinusoidal, the peak electric field `E_(0)` is
`E_(0)=sqrt(2)E_("rms")=sqrt(2)xx2.9 V//m=4.07V//m`
Electric field strength of light is fairly large
`B_("rms")=(E_("rms"))/(c)=(2.9Vm^(-1))/(3xx10^(8)ms^(-1))=9.6xx10^(-9)T`
Agains, since the field in the light beam is sinusoidal the peak magnetic field is `B_(0)=sqrt(2)B_(rms)=1.4xx10^(-8)T`. Note that although the energy in the magnetic field is equal tot he energy in the electric field, the magnetic field strength is evidently very weak.