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JEE MAIN 2021`
25-02-2021 S2
Question
The peak electric field produced by the radiation coming from the 8 W bulb at a distance of 10 m is $\frac{x}{10} \sqrt{\frac{\mu_0 c}{\pi}} \frac{V}{m}$. The efficiency of the bulb is $10 \%$ and it is a point source. The value of $x$ is $\_\_\_\_$ .
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Solution
$\begin{aligned} & I=\frac{1}{2} c \in_0 E_0^2 \\ & \frac{8}{4 \pi \times 10^2} \times \frac{1}{2}=\frac{1}{4} \times c \times \frac{1}{\mu_0 c^2} \times E_0^2 \\ & E_0=\frac{2}{10} \times \sqrt{\frac{\mu_0 c}{\pi}} \Rightarrow x=2\end{aligned}$
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