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) = 9×109 Nm2/C2 and speed of light in vacuum = 3 × 108 ms-1)
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) = 9×109 Nm2/C2 and speed of light in vacuum = 3 × 108 ms-1)
Option 1 -
2
Option 2 -
4
Option 3 -
20
Option 4 -
40
-
1 Answer
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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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d_lm = Distance covered by transmitting antenna + Distance covered by receiving antenna
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% increment = (8000/16000) × 100 = 50
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