The peak electric field produced by the radia | vedclass

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(D) The power of the bulb is $P = 80\, W$. The efficiency is $10\, \%$,so the radiated power is $P_{rad} = 80 	imes 0.10 = 8\, W$.Since it is a point

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$2$

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(D) The power of the bulb is $P = 80\, W$. The efficiency is $10\, \%$,so the radiated power is $P_{rad} = 80 	imes 0.10 = 8\, W$.Since it is a point source,the intensity $I$ at a distance $r = 10\, m$ is given by $I = \frac{P_{rad}}{4 \pi r^{2}} = \frac{8}{4 \pi (10)^{2}} = \frac{8}{400 \pi} = \frac{1}{50 \pi} \, W/m^{2}$.The intensity of an electromagnetic wave is related to the peak electric field $E_{0}$ by $I = \frac{1}{2} c \epsilon_{0} E_{0}^{2}$.Using $\epsilon_{0} = \frac{1}{\mu_{0} c^{2}}$,we get $I = \frac{1}{2} c \left( \frac{1}{\mu_{0} c^{2}} ight) E_{0}^{2} = \frac{E_{0}^{2}}{2 \mu_{0} c}$.Equating the two expressions for $I$: $\frac{E_{0}^{2}}{2 \mu_{0} c} = \frac{1}{50 \pi}$.$E_{0}^{2} = \frac{2 \mu_{0} c}{50 \pi} = \frac{\mu_{0} c}{25 \pi}$.$E_{0} = \sqrt{\frac{\mu_{0} c}{25 \pi}} = \frac{1}{5} \sqrt{\frac{\mu_{0} c}{\pi}} = \frac{2}{10} \sqrt{\frac{\mu_{0} c}{\pi}}$.Comparing this with $\frac{x}{10} \sqrt{\frac{\mu_{0} c}{\pi}}$,we find $x = 2$.

The peak electric field produced by the radiation coming from the $80\, W$ 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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