A radiation is emitted by $1000$ W bulb and it generates an electric field and magnetic field at $P$, placed at a distance of $2$ m. The efficiency of the bulb is $1.25\%$. The value of peak electric field at $P$ is $x\times10^{-1}$ V m$^{-1}$. Value of $x$ is ______ (Rounded-off to the nearest integer) [Take $\varepsilon_0 = 8.85\times10^{-12}$ C$^2$ N$^{-1}$ m$^{-2}$, $c = 3\times10^8$ m s$^{-1}$]
Numerical value type. Enter your answer.
Answer: 137
Radiated power $= 1.25\%$ of $1000$ W $= 12.5$ W.
Intensity at $2$ m: $I = \dfrac{12.5}{4\pi(2)^2} \approx 0.249$ W m$^{-2}$.
$I = \frac{1}{2}c\varepsilon_0E_0^2$:
$$E_0 = \sqrt{\frac{2\times0.249}{3\times10^8\times8.85\times10^{-12}}} = \sqrt{187.4} \approx 13.7\ \text{V m}^{-1} = 137\times10^{-1}\ \text{V m}^{-1}$$
Solution by Sreeraj P, M.Sc Physics