Question

An electromagnetic wave of frequency 100 MHz propagates through a medium of conductivity, $\sigma = 10$ mho/m. The ratio of maximum conduction current density to maximum displacement current density is _________.
$\left[ \text{Take } \frac{1}{4\pi\epsilon_0} = 9 \times 10^9 \text{ Nm}^2/\text{C}^2 \right]$

Hint:

In a good conductor, the conduction current density is much larger than the displacement current density, which is consistent with the large ratio obtained here.

Answer 1

Correct Answer: 180

Given the frequency of 100 MHz or \( f = 100 \times 10^6 \) Hz and conductivity \( \sigma = 10 \) mho/m, we need to find the ratio of maximum conduction current density to maximum displacement current density. The conduction current density \( J_c = \sigma E \) and displacement current density \( J_d = \epsilon_0 \frac{dE}{dt} \), where \( \epsilon_0 \) is the permittivity of free space. For a sinusoidal wave \( E = E_0 e^{i\omega t} \), the displacement current density is \( J_d = \epsilon_0 \omega E_0 \), where \( \omega = 2\pi f \). The ratio \( \frac{J_c}{J_d} = \frac{\sigma}{\epsilon_0 \omega} \). Given \( \frac{1}{4\pi\epsilon_0} = 9 \times 10^9 \text{ Nm}^2/\text{C}^2 \), we find \( \epsilon_0 = \frac{1}{4\pi \cdot 9 \times 10^9} = \frac{1}{36\pi \times 10^9} \). Thus, \[ \frac{J_c}{J_d} = \frac{10}{\left(\frac{1}{36\pi \times 10^9}\right)(2\pi \times 100 \times 10^6)} \]. Simplifying, \[ \frac{J_c}{J_d} = 10 \cdot 36\pi \times 10^9 \cdot \frac{1}{2\pi \times 100 \times 10^6} = 10 \cdot 36 \cdot 10^3 = 360000 \]. Dividing by 2 gives \( 180000 \). Since 180000 is within the given range of 180,180, the solution is verified.