Chapter 26, Problem 51A

To determine

To find: The speed of wave’s in a coaxial cable.

Answer to Problem 51A

The speed of the wave in the coaxial cable is $1.98\times {10}^{8}m/s$ .

Explanation of Solution

Given:

The transmitted frequency of coaxial cable $100\text{MHz}$

Dielectric constant is $2.30$

Formula:

The velocity factor of a coaxial cable is given by

  $VF=\frac{1}{\sqrt{K}}$  ...... (1)

Here, $VF$ is the velocity factor, $K$ is the dielectric constant.

Velocity of the wave in the cable is given by

  $v=VF\times c$  ...... (2)

Here, $v$ is the velocity, $c$ is the velocity of light.

Calculation:

Substitute $2.3$ for $K$ in equation (1)

  $\begin{array}{c}VF=\frac{1}{\sqrt{K}}\\ =\frac{1}{\sqrt{2.3}}\\ =0.66\end{array}$

Substitute $0.66$ for $VF$ in equation (2)

  $\begin{array}{c}v=0.66\times 3\times {10}^8\\ =1.98\times {10}^{8}m/s\end{array}$

Conclusion:

Hence, the required speed of the wave in the coaxial cable is $1.98\times {10}^{8}m/s$ .