Gujarati
Hindi
8.Electromagnetic waves
medium

Assume a bulb of efficiency $2.5\%$ as a point source. The peak values of electric field produced by the radiation coming from a $100\, W$ bulb at a distance of $3\, m$ is respectively.....$V\,{m^{ - 1}}$

A

$2.5$

B

$4.2$

C

$4.08$

D

$3.6$

Solution

Here intensity, $I$ $=\frac{\text { power }}{\text { area }}$

$=\frac{100 \times 2.5}{4 \pi(3)^{2} \times 100}=\frac{2.5}{36 \pi} \mathrm{\,W} \mathrm{m}^{-2}$

We know, $I=\frac{1}{2} \varepsilon_{0} E_{0}^{2} C$

or $\quad \mathrm{E}_{0}=\sqrt{\frac{2 \mathrm{I}}{\varepsilon_{0} \mathrm{c}}}=\sqrt{\frac{2 \times \frac{2.5}{36 \pi}}{\frac{1}{4 \pi \times 9 \times 10^{9}} \times 3 \times 10^{2}}}=4.08 \mathrm{\,Vm}^{-1}$

Standard 12
Physics

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