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 $\_\_\_\_$ .
$\left[\text{Take }\frac{1}{4 \pi \epsilon_{\mathrm{o}}}=9 \times 10^{9} \mathrm{Nm}^{2} / \mathrm{C}^{2}\right]$

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[\text{Take }\frac{1}{4 \pi \epsilon_{\mathrm{o}}}=9 \times 10^{9} \mathrm{Nm}^{2} / \mathrm{C}^{2}\right]

Quantrex Academy · Accepted Answer

Correct answer: 1800

For an electromagnetic wave in a conducting medium, the maximum conduction current density is

J_{c,max} = \sigma E_0

and maximum displacement current density is

J_{d,max} = \epsilon_0 \omega E_0

.
The ratio is:

\frac{J_{c,max}}{J_{d,max}} = \frac{\sigma}{\epsilon_0 \omega}

.
With

\sigma = 10

mho/m,

f = 100

MHz
So

\omega = 2\pi \times 10^8

rad/s, and

\epsilon_0 = \frac{10^{-9}}{36\pi}

F/m:

\frac{J_{c,max}}{J_{d,max}} = \frac{10 \times 36\pi}{10^{-9} \times 2\pi \times 10^8} = \frac{10 \times 36}{2 \times 10^{-1}} = 1800

.

An electromagnetic wave of frequency 100 MHz propagates through a...