For sea water with σ = 5 mho/m and ε_r = 80, what is the distance for which ratio signal can be transmitted with 90% attenuation at 25 kHz?

• A. 327 m
• B. 3.27 m
• C. 32.7 m
• D. 0.327 m

Answer: (B)

Condition of good conductor:
σ/ϵω >> 1
Equation is given by:
The constant ratio = σ/ϵω
= 44913.6
σ/ϵω >> 1.
hence sea water is a acting as a good conductor Since attenuation is 90%, so transmission is 10%
So,
e^(-αx) = 0.1

Attenuation constant = √(σμϵ/2)
= 0.702
-αx = ln(0.1)
x = 3.27 m