A point source surrounded by vacuum emits an electromagnetic wave of frequency of 900 kHz. If the power emitted by the source is 15 W, the amplitude of the electric field of the wave (in 10^-3V/m) at a distance 15 km from the source is :
**( (1 / 4πε₀) = 910⁹ Nm²/C² and speed of light in vacuum = 3 10⁸ ms^-1)**

Option 1 - 2

Option 2 - 4

Option 3 - 20

Option 4 - 40

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Answered by

Vishal Baghel | Contributor-Level 10

4 months ago

Correct Option - 1

Detailed Solution:

I = P/ (4πr²) . (i)
and I = (1/2)ε? E? ²c . (ii)
From (i) and (ii)
P/ (4πr²) = (1/2)ε? E? ²c
E? = √ (2P/ (4πε? r²c) ; E? = √ (9*10? *2*15)/ (15*15*10? *3*10? ) = 2*10? ³ V/m

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