First slide

Intensity of electromagnetic wave

Question

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

Moderate

A

64.7 V/m
B

57.8 V/m
C

56.72 V/m
D

54.77 V/m

Solution

Intensity of EM wave is given by
$I = \frac{P}{4 {πR}^{2}} = \frac{1}{2} ε_{0} E_{0}^{2} \times c$
$\Rightarrow E_{0} = \sqrt{\frac{P}{2 {πR}^{2} ε_{0} c}}$
$= \sqrt{\frac{800}{2 \times 3 .14 \times (4)^{2} \times 8 .85 \times 10^{- 12} \times 3 \times 10^{8}}}$
$= 54.77 \frac{V}{m}$

Get Instant Solutions

When in doubt download our app. Now available Google Play Store- Doubts App

Recieve an sms with download link

Doubts App

OR

SCAN ME

Doubts App

Doubts App