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 $\frac{x}{10} \sqrt{\frac{\mu_{0} \mathrm{c}}{\pi}} \frac{\mathrm{V}}{\mathrm{m}}$. The

efficiency of the bulb is $10 \%$ and it is a point source. The value of $x$ is_________.

Solution:

$I=\frac{1}{2} \mathrm{c} \in_{0} \mathrm{E}_{0}^{2}$

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

$\mathrm{E}_{0}=\frac{2}{10} \times \sqrt{\frac{\mu_{0} \mathrm{c}}{\pi}} \Rightarrow x=2$

The peak electric field produced by the radiation coming from

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