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(A) The conducting current density is given by $j_c = \sigma E$. For an electromagnetic wave, $E = E_0 \sin(\omega t - kx)$, so the maximum conducting

Vedclass · Accepted Answer

$1800$

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(A) The conducting current density is given by $j_c = \sigma E$. For an electromagnetic wave, $E = E_0 \sin(\omega t - kx)$, so the maximum conducting current density is $(j_c)_{\max} = \sigma E_0$. The displacement current density is given by $j_d = \epsilon_0 \frac{\partial E}{\partial t}$. Substituting $E$, we get $j_d = \epsilon_0 E_0 \omega \cos(\omega t - kx)$, so the maximum displacement current density is $(j_d)_{\max} = \epsilon_0 E_0 \omega$. The ratio is $\frac{(j_c)_{\max}}{(j_d)_{\max}} = \frac{\sigma E_0}{\epsilon_0 \omega E_0} = \frac{\sigma}{\epsilon_0 \omega}$. Given $\sigma = 10 	ext{ mho/m}$, $f = 100 	imes 10^6 	ext{ Hz}$, and $\omega = 2 \pi f = 2 \pi 	imes 10^8 	ext{ rad/s}$. Also, $\frac{1}{4 \pi \epsilon_0} = 9 	imes 10^9$, so $\frac{1}{\epsilon_0} = 4 \pi 	imes 9 	imes 10^9 = 36 \pi 	imes 10^9$. Ratio $= \frac{10 	imes 36 \pi 	imes 10^9}{2 \pi 	imes 10^8} = \frac{360 \pi 	imes 10^9}{2 \pi 	imes 10^8} = 180 	imes 10 = 1800$.

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