An electromagnetic wave of frequency 100 MHz propagates through a medium of conductivity σ = 10 mho/m. The ratio of maximum conduction current density to maximum displacement current density is ____.
Q. An electromagnetic wave of frequency \(100\ \text{MHz}\) propagates through a medium of conductivity \( \sigma = 10\ \text{mho/m} \).

The ratio of maximum conduction current density to maximum displacement current density is ____.

[ Take \( \frac{1}{4\pi\varepsilon_0} = 9 \times 10^9 \) ]
Correct Answer: 1800

Step-by-Step Solution

Conduction current density:

\[ J_c = \sigma E \]

Displacement current density:

\[ J_d = \varepsilon_0 \frac{dE}{dt} \]

For sinusoidal wave:

\[ E = E_0 \sin \omega t \]

\[ \frac{dE}{dt} = \omega E_0 \cos \omega t \]

Maximum displacement current density:

\[ J_{d(max)} = \varepsilon_0 \omega E_0 \]

Maximum conduction current density:

\[ J_{c(max)} = \sigma E_0 \]

Required ratio:

\[ \frac{J_c}{J_d} = \frac{\sigma}{\varepsilon_0 \omega} \]

Angular frequency:

\[ f = 100 \text{ MHz} = 10^8 \text{ Hz} \]

\[ \omega = 2\pi f = 2\pi \times 10^8 \]

Given:

\[ \frac{1}{4\pi\varepsilon_0} = 9 \times 10^9 \]

\[ \varepsilon_0 = \frac{1}{36\pi \times 10^9} \]

Now substitute:

\[ \frac{J_c}{J_d} = \frac{10}{\varepsilon_0 \times 2\pi \times 10^8} \]

After simplifying:

\[ \frac{J_c}{J_d} = 1800 \]

Final Answer = 1800

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