# A radiation is emitted by

Question:

A radiation is emitted by $1000 \mathrm{~W}$ bulb and it generates an electric field and magnetic field at $\mathrm{P}_{t}$ placed at a distance of $2 \mathrm{~m}$. The efficiency of the bulb is $1.25 \%$. The value of peak electric field at $P$ is $x \times 10^{-1} \mathrm{~V} / \mathrm{m}$. Value of $x$ is (Rounded-off to the nearest integer) [Take $\left.\varepsilon_{0}=8.85 \times 10^{-12} \mathrm{C}^{2} \mathrm{~N}^{-1} \mathrm{~m}^{-2}, \mathrm{c}=3 \times 10^{8} \mathrm{~ms}^{-1}\right\rceil$

Solution:

(137)

Intensity of electro magnetic wave is,

$I=\frac{1}{2} C \varepsilon_{0} E_{0}^{2}=\frac{P}{4 \pi r^{2}}$

$\frac{1}{2} 4 \pi \varepsilon_{0} \times \mathrm{C} \times \mathrm{E}_{0}^{2}=\frac{\mathrm{P}}{\mathrm{r}^{2}}$

$\frac{1}{2} \times \frac{3 \times 10^{5} \times \mathrm{E}_{0}^{2}}{9 \times 10^{9}}=\frac{1000 \times 1.25}{(2)^{2}} \times \frac{1}{100}$

$\mathrm{E}_{0}^{2}=\frac{60 \times 1000 \times 1.25}{4 \times 100}=\frac{125 \times 3}{2}$

$\mathrm{E}_{0}^{2}=\frac{375}{2}=187.5$

$E_{0}=13.69$

$\mathrm{E}_{0} \approx 137 \times 10^{-1} \mathrm{v} / \mathrm{m}$