Conducting Medium Current Density Ratio Calculation

The problem asks for the ratio of the maximum conduction current density ($J_{c,max}$) to the maximum displacement current density ($J_{d,max}$) for an electromagnetic wave in a conducting medium.

Given Information

• Frequency, $f = 100 \text{ MHz} = 100 \times 10^6 \text{ Hz} = 1 \times 10^8 \text{ Hz}$
• Conductivity, $\sigma = 10 \text{ mho/m}$
• Coulomb's constant, $\frac{1}{4\pi\varepsilon_0} = 9 \times 10^9 \text{ Nm}^2/\text{C}^2$

Key Concepts and Formulas

• The conduction current density is given by $J_c = \sigma E$.
• The displacement current density is $J_d = \varepsilon \frac{\partial E}{\partial t}$.
• For a time-harmonic electric field $E(t) = E_{max} \cos(\omega t)$, we have $\frac{\partial E}{\partial t} = -\omega E_{max} \sin(\omega t)$. The maximum magnitude of $\frac{\partial E}{\partial t}$ is $\omega E_{max}$.
• Therefore, the maximum conduction current density is $|J_{c,max}| = \sigma E_{max}$.
• The maximum displacement current density is $|J_{d,max}| = \omega \varepsilon E_{max}$.
• The ratio required is $\frac{|J_{c,max}|}{|J_{d,max}|} = \frac{\sigma E_{max}}{\omega \varepsilon E_{max}} = \frac{\sigma}{\omega \varepsilon}$.

Calculation Steps

1. Calculate the angular frequency ($\omega$):

   $\omega = 2\pi f = 2\pi (1 \times 10^8) \text{ rad/s}$

2. Determine the permittivity ($\varepsilon$):

   Using the given constant $\frac{1}{4\pi\varepsilon_0} = 9 \times 10^9$, we find the permittivity of free space: $\varepsilon_0 = \frac{1}{4\pi (9 \times 10^9)} = \frac{1}{36\pi \times 10^9} \text{ F/m}$. Assuming the relative permittivity $\varepsilon_r = 1$ (as is common when not specified), the medium's permittivity is $\varepsilon = \varepsilon_r \varepsilon_0 = \varepsilon_0$.

3. Compute the ratio:

   Ratio $R = \frac{\sigma}{\omega \varepsilon} = \frac{10 \text{ mho/m}}{(2\pi \times 10^8 \text{ rad/s}) \times (\frac{1}{36\pi \times 10^9} \text{ F/m})}$ $R = \frac{10 \times 36\pi \times 10^9}{2\pi \times 10^8}$ $R = \frac{10 \times 36 \times 10^9}{2 \times 10^8}$ $R = \frac{360 \times 10^9}{2 \times 10^8}$ $R = 180 \times 10^{(9-8)} = 180 \times 10 = 1800$.

Result

The ratio of maximum conduction current density to maximum displacement current density is $1800$.