Question

Calculate the electric and magnetic fields produced by the radiation coming from a $30\pi$W bulb at a distance of $3$m. Assume that the efficiency of the bulb is $10\%$ and it is a point source.

• A. $0.22\times {10}^{-8}$T
• B. $0.88\times {10}^{-6}$T
• C. $0.22\times {10}^{-6}$T
• D. $2.64\times {10}^{-8}$T

Answer 1

The correct option is D $2.64\times {10}^{-8}$T

Let r be the distance between the points S and X.
Intensity at distance r from a point source, $I=\frac{p}{4\pi r^2}$
$\therefore$ Efficiency, $eta=\frac{Output}{Input}=\frac{P}{P^{\prime }}$
$P=\eta P^{\prime }$
$=\left(\frac{10}{100}\right)\left(30\pi \right)=3\pi$W
$I=\frac{P}{4\pi r^2}=\frac{3\pi }{4\pi (3{)}^2}$
$=\frac{1}{12}\frac{W}{m^2}=\frac{1}{2}{\epsilon }_0{E}_0^{2}c$
$\frac{1}{12}=\frac{1}{2}{\epsilon }_0{E}_0^2\times 3\times {10}^8$
$E_0^2=\frac{{10}^{-8}}{18{\epsilon }_0}=\frac{{10}^{-8}}{18\times 8.85\times {10}^{-12}}$
$=\frac{{10}^4}{18\times 8.85}$
$\Rightarrow E_0=\frac{{10}^2}{\sqrt{18\times 8.85}}=7.92$V$/$m
$\therefore B_0=\frac{E_0}{C}=\frac{2.92}{3\times {10}^5}=2.64\times {10}^{-8}$T.