A point source of electromagnetic radiation h | vedclass

Name: PDF Generator App | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(d)Intensity of $EM$ wave is given by
$I = \frac{P}{{4\pi {R^2}}} = {v_{av}}.c = \frac{1}{2}{\varepsilon _0}E_0^2 	imes c$
==> ${E_0} = \sqrt {\

Vedclass · Accepted Answer

$54.77$

Vedclass · Answer

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

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....$V/m$

Similar Questions

If a source is transmitting electromagnetic wave of frequency $8.2 \times {10^6}Hz$, then wavelength of the electromagnetic waves transmitted from the source will be.....$m$

A radio transmitter transmits at $830\, kHz$. At a certain distance from the transmitter magnetic field has amplitude $4.82\times10^{-11}\,T$. The electric field and the wavelength are respectively

$TV$ waves have a wavelength range of $1-10 \,meter$. Their frequency range in $MHz$ is

A $1.5 \,kW$ laser beam of wavelength $6400 \,\mathring A$ is used to levitate a thin aluminium disc of same area as the cross-section of the beam. The laser light is reflected by the aluminium disc without any absorption. The mass of the foil is close to ......... $kg$