Seawater at a frequency f = 9 times 10^{2} , | vedclass

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(D) The conduction current density is $J_{c} = \sigma E = \frac{E}{\rho} = \frac{V(t)}{\rho d}$.The displacement current density is $J_{d} = \varepsil

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$6$

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(D) The conduction current density is $J_{c} = \sigma E = \frac{E}{ho} = \frac{V(t)}{ho d}$.The displacement current density is $J_{d} = \varepsilon \frac{\partial E}{\partial t} = \varepsilon \frac{1}{d} \frac{dV}{dt} = \frac{\varepsilon}{d} V_{0} (2 \pi f) \cos(2 \pi ft)$.Given $J_{c} = 10^{x} J_{d}$,we have $\frac{V_{0} \sin(2 \pi ft)}{ho d} = 10^{x} \frac{\varepsilon}{d} V_{0} (2 \pi f) \cos(2 \pi ft)$.This simplifies to $	an(2 \pi ft) = 10^{x} \cdot ho \cdot \varepsilon \cdot 2 \pi f$.Given $f = 900 \, Hz$,$t = \frac{1}{800} \, s$,$ho = 0.25 \, \Omega m$,and $\varepsilon = 80 \varepsilon_{0}$.$2 \pi ft = 2 \pi 	imes 900 	imes \frac{1}{800} = \frac{9 \pi}{4} = 2 \pi + \frac{\pi}{4}$.So,$	an(2 \pi ft) = 	an(\pi/4) = 1$.Now,$1 = 10^{x} 	imes 0.25 	imes 80 \varepsilon_{0} 	imes 2 \pi 	imes 900$.Using $\frac{1}{4 \pi \varepsilon_{0}} = 9 	imes 10^{9}$,we have $\varepsilon_{0} = \frac{1}{36 \pi 	imes 10^{9}}$.$1 = 10^{x} 	imes 20 	imes \frac{1}{36 \pi 	imes 10^{9}} 	imes 1800 \pi = 10^{x} 	imes \frac{20 	imes 1800}{36 	imes 10^{9}} = 10^{x} 	imes \frac{36000}{36 	imes 10^{9}} = 10^{x} 	imes 10^{-6}$.$10^{x} = 10^{6}$,which gives $x = 6$.

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