Radiation emitted generates |ElectromagneticWavesJEEMain2021

JEE MAIN 2021

26-02-2021 S1

Question

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 $x \times 10^{-1} \mathrm{~V} / \mathrm{m}$. Value of $x$ is_. (Rounded-off to the nearest integer)[Take $\left.\varepsilon_0=8.85 \times 10^{-12} \mathrm{C}^2 \mathrm{~N}^{-1} \mathrm{~m}^{-2}, \mathrm{c}=3 \times 10^8 \mathrm{~ms}^{-1}\right]$

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Solution

$$ \begin{aligned} & \mathrm{I}_{\text {avg }}=\frac{1}{2} \varepsilon_0 \mathrm{E}_0^2 \mathrm{C} \\ & \frac{1.25}{100} \times \frac{1000}{4 \pi(2)^2}=\frac{1}{2} \times 8.85 \times 10^{-12} \times 3 \times 10^8 \times \mathrm{E}_0^2 \\ & \mathrm{E}_0^2=187.4 \\ & \therefore \mathrm{E}_0=13.689 \mathrm{~V} / \mathrm{m} \\ & =136.89 \times 10^{-1} \mathrm{~V} / \mathrm{m} \\ & \therefore \mathrm{x}=136.89 \end{aligned} $$ Rounding off to nearest integer $$ x=137 $$

Question Tags

JEE Main

Physics

Medium

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