Question 166: EM waves

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 $\_\_\_\_$.
$$ \left[	ext { Take } \frac{1}{4 \pi \epsilon_0}=9 	imes 10^9\, \mathrm{Nm}^2 / \mathrm{C}^2ight] $$

Accepted Answer

According to Ohm's Law in point form, the conduction current density is proportional to the electric field (E) :
$$ \mathrm{J}_{\mathrm{c}}=\sigma \mathrm{E} $$
The maximum value occurs when the electric field is at its peak $\left(\mathrm{E}_0ight)$ :
$$ \left(\mathrm{J}_{\mathrm{c}}ight)_{\max }=\sigma \mathrm{E}_0 $$
Displacement current density arises from a time-varying electric field. For an electromagnetic wave, the electric field can be represented as $E=E_0 \sin (\omega t)$.
Maxwell defined displacement current density as :
$$ \mathrm{J}_{\mathrm{d}}=\epsilon \frac{\partial \mathrm{E}}{\partial \mathrm{t}} $$
$$\Rightarrow $$ $$ \mathrm{J}_{\mathrm{d}}=\epsilon \frac{\partial}{\partial \mathrm{t}}\left(\mathrm{E}_0 \sin (\omega \mathrm{t})ight)=\epsilon \omega \mathrm{E}_0 \cos (\omega \mathrm{t}) $$
The maximum value occurs when the cosine term is 1 :
$$ \left(\mathrm{J}_{\mathrm{d}}ight)_{\max }=\epsilon \omega \mathrm{E}_0 $$
The ratio of the maximum values is:
Ratio $=\frac{\left(\mathrm{J}_{\mathrm{c}}ight)_{	ext {max }}}{\left(\mathrm{J}_{\mathrm{d}}ight)_{	ext {max }}}=\frac{\sigma \mathrm{E}_0}{\epsilon \omega \mathrm{E}_0}=\frac{\sigma}{\epsilon \omega}$
The conductivity $(\sigma)$ of the medium is, $\sigma=10 \mathrm{mho} / \mathrm{m}$
The angular frequency $(\omega)$ is $2 \pi \mathrm{f}=2 \pi 	imes\left(100 	imes 10^6ight)=2 \pi 	imes 10^8 \mathrm{rad} / \mathrm{s}$
Since no specific medium permittivity is given, we use the permittivity of free space ( $\epsilon_0$ ).
Given $\frac{1}{4 \pi \epsilon_0}=9 	imes 10^9$, then $\epsilon_0=\frac{1}{36 \pi 	imes 10^9}$.
Substituting these values into the ratio:
Ratio $=\frac{10}{\frac{1}{36 \pi 	imes 10^9} 	imes 2 \pi 	imes 10^8}=\frac{10 	imes 36 \pi 	imes 10^9}{2 \pi 	imes 10^8}$
Ratio $=180 	imes 10^1=1800$
Therefore, the ratio of maximum conduction current density to maximum displacement current density is 1800 .
Hence, the correct answer is $\mathbf{1 8 0 0}$.

EM waves

Integer / Numeric Answer Type