Question

What is the amplitude of electric field produced by radiation coming from a `100` W bulb at a distance of `4` m? Assume efficiency of bulb to be `3.14%` and assume it to the point source:
A. `2.42 V//m`
B. `3.43 V//m`
C. `4.2xx10^(4) V//m`
D. `14xx10^(4) V.m`

Answer 1

Correct Answer - b
Intensity at a distance `r=4` m is
`I=(Ixx3.14%)/(4pi r^(2))=(100xx(3.14//100))/(4xx3.14xx(4)^(2))`
`(1)/(64) W//m^(2)`
Half of the intensity is provide by electric field and remaining half by magnetic field
`:. (1)/(2)I=(1)/(2)(epsilon_(0)E_(rms)^(2)C)`
`implies E_(rms)=sqrt((I)/(epsilon_(0)c))=sqrt({(1//64)/(8.85xx10^(-12)xx3xx10^(8))})`
`sqrt(5.88)=2.42 V//m =sqrt(2)xx2.42=3.43 V//m`