Calculate the electric and magnetic fields produced by the radiation coming from a 100 W bulb at a distance of 3m. Assume that the efficiency of the bulb is 2.5% and it is a point source.

Explanation

The bulb, as a point source, radiates light in all directions uniformaly. At a distance of 3m, the surface of the surrounding sphere is, `A=4pi r^(2)=4pi (3)^(2)=113m^(2)`
The intensity at this distance is `I=("Power")/("Area")=(100xx2.5%)/(113)=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)(0.022 W//m^(2))`
`E_(rms)=sqrt((0.022)/((8.85xx10^(-12))(3xx10^(8))))=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=4.07` V//m
Thus, the electric field strength of the light that we use for reading is fairlylarge.
Now, let us calculate the strength of the magnetic field. It is
`B_(rms)=(E_(rms))/(c )=(2.9)/(3xx10^(8))=9.6xx10^(-9)T`
Again, 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 to the energy in the electric field, the magnetic field strength is evidently very weak.