Calculate the electric and magnetic fields produced by the radiation coming from a 100 W bulb at a distance of 2 m. Assume that efficiency of the bulb is 2.5% and it is a point source. [Intensity provided by electric field `=(1)/(2)(epsi_(0)E_(rms)^(2)xxc)`.

Explanation

Surface area of 2 m radius is
`A=4pir^(2)" "` [radius, r = 2m]
`=4pir^(2)`
`=4pi(2)^(2)`
`=16pi=50.24m^(2)`
Now, intensity `l=("Power")/("Area")`
`=(2.5%" of "100)/(50.24)`
`=(2.5)/(50.24)`
`=0.0498Wm^(-2)`
Given, Intensity of electric field
i.e., `l=(1)/(2)epsi_(0)E_(rms)^(2)xxc`
`impliesE_(rms)^(2)=(2l)/(epsi_(0)c)`
RMS value of electric field,
`E_(rms)=sqrt(2(1)/(epsi_(0)c))`
`=sqrt((2xx0.0498)/(8.55xx10^(-12)xx3xx10^(8)))`
`=6.12Vm^(-1)`
Now, peak value `E_(0)=sqrt2E_(rms)`
`=sqrt2xx6.12`
`=8.65Vm^(-1)`
We know that, `(E_(0))/(B_(0))=c`
`implies(8.65)/(B_(0))=3xx10^(8)`
`implies` Magnetic field,
`B_(0)=2.88xx10^(-8)T`