The peak electric field produced by the radiation coming from

Question:

The peak electric field produced by the radiation coming from the $8 \mathrm{~W}$ bulb at a distance of $10 \mathrm{~m}$ 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

Solution:

(2)

$I=\frac{1}{2} c \in_{0} E_{0}^{2}$

$\frac{8}{4 \pi \times 10^{2}}=\frac{1}{2} \times c \times \frac{1}{\mu_{0} c^{2}} \times E_{0}^{2}$

$E_{0}=\frac{2}{10} \sqrt{\frac{\mu_{0} c}{\pi}}$

$\Rightarrow x=2$

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Sahiba
Aug. 30, 2023, 6:35 a.m.
Why have we not taken efficiency while solving????
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