Question

Calculate the propagation constant \( \gamma \) for a conducting medium in which \(\sigma\) = 58 MS/m, \( \mu_r \) = 1 and f = 100 MHz.

• A. $2.14 \times 10^{5} { angle } 90^\circ { m}^{-1}$
• B. $2.14 \times 10^{2} { angle } 60^\circ { m}^{-1}$
• C. $2.14 \times 10^{3} { angle } 15^\circ { m}^{-1}$
• D. $2.14 \times 10^{5} { angle } 45^\circ { m}^{-1}$

Hint:

For a good conductor, the propagation constant simplifies to \( \gamma = \sqrt{j\omega\mu \sigma}\)

Answer 1

The Correct Option is D

Step 1: Given, the conductivity \( \sigma = 58 \times 10^6 S/m \), relative permeability \( \mu_r = 1 \), and frequency \( f = 100 \times 10^6 Hz \). 
Step 2: First calculate angular frequency \(\omega\): \[ \omega = 2\pi f = 2\pi \times 100 \times 10^6 = 2\pi \times 10^8 rad/sec \] 
Step 3: Calculate the propagation constant \( \gamma = \alpha + j\beta \), using the given values of the medium. \[ \gamma = \sqrt{j\omega\mu (\sigma + j\omega \epsilon)} \] Since conductivity is high, we can neglect the \(\omega\epsilon\) term, so the equation becomes: \[ \gamma = \sqrt{j\omega\mu \sigma} = \sqrt{j(2\pi \times 10^8) \times (4\pi \times 10^{-7} ) \times (58 \times 10^6)} = \sqrt{j \times 460224 \times 10^7} \] \[ \gamma = \sqrt{460224 \times 10^7} \sqrt{j} = 214526 \times 10^2 \times e^{j45^\circ} = 2.145 \times 10^5 \angle 45^\circ m^{-1} \] Therefore, \( \gamma = 2.14 \times 10^5 { angle } 45^\circ { m}^{-1} \).