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  • The peak electric field produced by the radiation coming from the 8 W bulb at a distance of 10 m is <img alt="\sqrt{\frac{\mu _{0}C}{\pi }}\frac{V}{m}

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 ______.
Option: 1 1
Option: 2 2
Option: 3 3
Option: 4 4

Answers (1)

best_answer

\begin{array}{l} \mathrm{I}=\frac{1}{2} \mathrm{c} \in_{0} \mathrm{E}_{0}^{2} \\ and \ I= \frac{8}{4 \pi \times 10^{2}} \\ So \ \frac{8}{4 \pi \times 10^{2}} \times \frac{1}{2}=\frac{1}{4} \times \mathrm{c} \times \frac{1}{\mu_{0} \mathrm{c}^{2}} \times \mathrm{E}_{0}^{2} \\ \\ \Rightarrow \mathrm{E}_{0}=\frac{2}{10} \times \sqrt{\frac{\mu_{0} \mathrm{c}}{\pi}} \\ \\ \Rightarrow \mathrm{x}=2 \end{array}

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Deependra Verma

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