Assume a bulb of efficiency `2.5%` as a point source. The peak values of electric and magnetic fields produced by the radiation coming from a 100 W bulb at a distance of 3 m is respectively

Options

A. `2.5"V m"^(-1), 3.6xx10^(-8)T`
B. `4.2"V m"^(-1), 2.8xx10^(-8)T`
C. `4.08"V m"^(-1), 1.36xx10^(-8)T`
D. `3.6"V m"^(-1), 4.2xx10^(-8)T`

Answer

C

Explanation

Here intensity, `I = ("power")/("area")`
`= (100 xx 2.5)/(4pi(3)^(2) xx 100) = (2.5)/(36 pi) Wm^(-2)`
Half of this intensity belongs to electric field and half of that to magnetic field.
`:. (I)/(2) = (1)/(4) epsilon_(0) E_(0)^(2)c`
or `E_(0) = sqrt((2I)/(epsilon_(0)c)) = sqrt((2xx(2.5)/(36 pi))/((1)/(4pi xx 9 xx 10^(9))xx3 xx 10^(8))) = 4.08 Vm^(-1)`
`:. B_(0) = (E_(0))/(c) = (4.08)/(3 xx 10^(8)) = 1.36 xx 10^(-8)T`