8.Electromagnetic waves
medium

A point source of electromagnetic radiation has an average power output of $800\, W.$ The maximum value of electric field at a distance $4.0 \,m$ from the source is....$V/m$

A

$64.7$

B

$57.8 $

C

$56.72 $

D

$54.77$

Solution

(d)Intensity of $EM$ wave is given by
$I = \frac{P}{{4\pi {R^2}}} = {v_{av}}.c = \frac{1}{2}{\varepsilon _0}E_0^2 \times c$
==> ${E_0} = \sqrt {\frac{P}{{2\pi {R^2}{\varepsilon _0}c}}} $
$ = \sqrt {\frac{{800}}{{2 \times 3.14 \times {{(4)}^2} \times 8.85 \times {{10}^{ – 12}} \times 3 \times {{10}^8}}}} $
$= 54.77 \frac{V}{m}$

Standard 12
Physics

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