An electromagnetic wave of frequency 100 MHz propagates through a medium of conductivity, $\sigma=10

Question

An electromagnetic wave of frequency 100 MHz propagates through a medium of conductivity, $\sigma=10 \mathrm{mho} / \mathrm{m}$. The ratio of maximum conduction current density to maximum displacement current density is $\_\_\_\_$ . $$ \left[	ext { Take } \frac{1}{4 \pi \epsilon_{\mathrm{o}}}=9 	imes 10^9 \mathrm{Nm}^2 / \mathrm{C}^2ight] $$

Professor PC Thomas Classes · Answer

Conduction current density $J_c = \sigma E$. Displacement current density $J_d = \epsilon_0 \frac{dE}{dt}$. If $E = E_0 \sin(\omega t)$, then $J_d = \epsilon_0 \omega E_0 \cos(\omega t)$. The ratio of their maximum values is $\frac{J_{c,max}}{J_{d,max}} = \frac{\sigma E_0}{\epsilon_0 \omega E_0} = \frac{\sigma}{\epsilon_0 \omega}$. Here, $\omega = 2\pi f = 2\pi 	imes 10^8	ext{ rad/s}$. Since $\frac{1}{4\pi\epsilon_0} = 9 	imes 10^9$, we have $\frac{1}{\epsilon_0} = 36\pi 	imes 10^9$. The ratio is $\frac{10 	imes 36\pi 	imes 10^9}{2\pi 	imes 10^8} = 180 	imes 10 = 1800$.