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8.Electromagnetic waves
hard
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 ...... .
A
$1$
B
$3$
C
$4$
D
$2$
(JEE MAIN-2021)
Solution
$I =\frac{1}{2} c \in_{0} E _{0}^{2}$
$\frac{8}{4 \pi \times 10^{2}} \times \frac{1}{2}=\frac{1}{4} \times c \times \frac{1}{\mu_{0} c ^{2}} \times E _{0}^{2}$
$E _{0}=\frac{2}{10} \times \sqrt{\frac{\mu_{0} c }{\pi}} \Rightarrow x =2$
Standard 12
Physics
