A radiation is emitted by $1000 \mathrm{~W}$ bulb and it generates an electric field and magnetic field at $\mathrm{P}$, 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)
$\left[\right.$ Take $\varepsilon_{0}=8.85 \times 10^{-12} \mathrm{C}^{2} \mathrm{~N}^{-1} \mathrm{~m}^{-2}, \mathrm{c}=3 \times 10^{8}$ $\left.\mathrm{ms}^{-1}\right]$
$\mathrm{I}_{\mathrm{avg}}=\frac{1}{2} \varepsilon_{0} \mathrm{E}_{0}^{2} \mathrm{C}$
$\frac{1.25}{100} \times \frac{1000}{4 \pi(2)^{2}}=\frac{1}{2} \times 8.85 \times 10^{-12} \times 3$
$\times 10^{8} \times \mathrm{E}_{0}^{2}$
$\mathrm{E}_{0}^{2}=187.4$
$\therefore \mathrm{E}_{0}=13.689 \mathrm{~V} / \mathrm{m}$
$=136.89 \times 10^{-1} \mathrm{~V} / \mathrm{m}$
$\therefore x=136.89$
Rounding off to nearest integer
$x=137$
Click here to get exam-ready with eSaral
For making your preparation journey smoother of JEE, NEET and Class 8 to 10, grab our app now.