Question

A certain electromagnetic wave traveling in seawater was observed to have an amplitude of 98.02 (V/m) at a depth of 10 m, and an amplitude of 81.87 (V/m) at a depth of 100 m. What is the attenuation constant of seawater

Answer 1

The value is
`\alpha =  0.002 Np/m`

Step-by-step explanation:

From the question we are told that

The first amplitude of the wave is
`E_(max)1 =  98.02 \  V/m`

The first depth is
`D_1 =  10 \  m`

The second amplitude is
`E_(max)2 =  81.87 \  (V/m)`

The second depth is
`D_2 = 100 \ m`

Generally from the spatial wave equation we have

`v(x) =  Ae^(-\alpha d)cos(\beta x  + \phi_o)`

=>
`(v(x))/(v(x)) =(  Ae^(-\alpha d)cos(\beta x  + \phi_o))/( Ae^(-\alpha d)cos(\beta x  + \phi_o))`

So considering the ratio of the equation for the two depth

`(A)/(A_S)  =  (e^(-D_1 \alpha ))/(e^(-D_2 \alpha ))`

=>
`(98.02)/(81.87)  =  (e^(-10 \alpha ))/(e^(-100 \alpha ))`

=>
`\alpha  =  (0.18)/(90)`

=>
`\alpha =  0.002 Np/m`