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 direction uniformly. At a distance of 3 m. the surface area of the surrounding sphere is
`A = 4 pir^2 = 4 pi (3)^2 = 113m^2`
The intesity at this distance is
`l = ("Power")/("Area") = (100W xx 2.5%)/(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) (epsilon_(0) E_("rms")^(2) c ) = (1)/(2) (0.022 W//m^2)`
`E_("rms") sqrt((0.200)/((8.85 xx 10^(-12)) (3 xx 10^(6) ) )) V//m = 2.9 V//m`
The value of E found above is the root of 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 = sqrt2 E_("rms") = sqrt(2) xx 2.9 V//m = 4.07 V//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 of FM waves, which is of the order of a few microvolts per metre.
Now, let us calculate the strength of the magnetic field, it is
`B_("rms") = (E_("rms) )/(C ) = (2.9 Vm^(-1) )/(3 xx 10^(8)ms^(-1))=9.6 xx 10^(-9) T`
Again, since the field in the light beam is sinusoidal, the peak magnetic field is `B_0 = sqrt(2) B_("rms") = 1.4 xx 10^(-6)`. Note that although the energy in the magnetic field is equal to the energy in the electric field, the magnetic field strength is evidently very weak.
Let us consider a cylindrical region where E. M. wave is passing through.