Chapter 14, Problem 14.13P

To determine

(a)

The $E_{\theta s}\text{ at }P(r=100,\theta ={90}^0,\varphi ={30}^0)$.

Answer to Problem 14.13P

The value of electric field $E_{\theta s}=0.2e^{-j1000\pi }\text{V/m}$.

Explanation of Solution

Given:

The following information is given:

   $r=100,\theta ={90}^0,\varphi ={30}^0$.

The radiation field $E_{\theta s}$ is given by,

   $E_{\theta s}=(\raisebox{1ex}{$20$}\!\left/ \!\raisebox{-1ex}{$r$}\right.)\sin \theta e^{-j10\pi r}\text{ V/m}$

Calculation:

Using the condition $r=100,\theta =90°,\varphi =30°$ in the formula of $E_{\theta s}$ ,

   $\begin{array}{l}{E}_{\theta s}=(\raisebox{1ex}{$20$}\!\left/ \!\raisebox{-1ex}{$100$}\right.)\sin ({90}^0)e^{-j1000\pi }\text{ V/m}\\ E_{\theta s}=0.2e^{-j1000\pi }\text{ V/m}\end{array}$

Conclusion:

The value of electric field $E_{\theta s}=0.2e^{-j1000\pi }\text{V/m}$.

To determine

(b)

The $E_{\theta s}\text{ at }P(100,90°,30°)$.

Answer to Problem 14.13P

The value of electric field $E_{\theta s}=0.2e^{-j1000\pi }{e}^{j0.5\pi }\text{ V/m}$.

Explanation of Solution

Given:

The vertical current element location is $A(0.1,90°,90°)$ . The radiation field $E_{\theta s}$ is given by,

   $E_{\theta s}=(\raisebox{1ex}{$20$}\!\left/ \!\raisebox{-1ex}{$r$}\right.)\sin \theta e^{-j10\pi r}\text{ V/m}$

Concept Used:

Due to the change in the position of the vertical current element, there will be practically no change in the distance of P. However, this will enforce a change in the phase term which needs to be included in the overall electric field expression.

Calculation:

Consider two lines are drawn. One line is drawn from the origin to the point $P$ and the other line is being drawn from $y=0.1$ to $P$ . Therefore, the difference in their length will be $g=0.1\sin (30°)=0.05$

The overall electric field will be obtained by using the modified result obtained in part (a) with the phase factor due to the path difference. The electric field will be,

   $\begin{array}{l}{E}_{\theta s}=0.2e^{-j1000\pi }{e}^{-j10\pi (-g)}\text{ V/m}\\ E_{\theta s}=0.2e^{-j1000\pi }{e}^{j0.5\pi }\text{ V/m}\end{array}$

Conclusion:

The value of electric field $E_{\theta s}=0.2e^{-j1000\pi }{e}^{j0.5\pi }\text{ V/m}$.

To determine

(c)

The $E_{\theta s}\text{ at }P(100,{90}^0,{30}^0)$.

Answer to Problem 14.13P

The value of electric field $E_{\theta s}=0\text{ V/m}$.

Explanation of Solution

Given:

The radiation field $E_{\theta s}$ is given by,

   $E_{\theta s}=(\raisebox{1ex}{$20$}\!\left/ \!\raisebox{-1ex}{$r$}\right.)\sin \theta e^{-j10\pi r}\text{ V/m}$

Identical vertical elements are located at $A(0.1,90°,90°)$ and $B(0.1,90°,270°)$.

Concept Used:

Due to the presence of the element at $A(0.1,{90}^0,{90}^0)$ , the electric field obtained in part (b) will be present. In addition to this, due to the presence of the element at $B(0.1,{90}^0,{270}^0)$ , there will be additional path difference which will give rise to some additional phase difference.

Overall electric field can be found by adding the contribution of $A(0.1,{90}^0,{90}^0)$ and $B(0.1,{90}^0,{270}^0)$.

Calculation:

Since the phase angle of $90°$ and $270°$ are $180°$ apart, the element at $B(0.1,90°,270°)$ is essentially located at $y=-0.1$ . This will give rise to a path difference of $g\text{'}=0.1\sin ({30}^0)=0.05$ . Since this path is longer, the electric field due to the element at $B(0.1,90°,270°)$ will be,

   $E_{\theta s}=0.2e^{-j1000\pi }{e}^{-j0.5\pi }\text{ V/m}$

Therefore, overall electric field will be,

   $\begin{array}{l}{E}_{\theta s}=0.2e^{-j1000\pi }(e^{-j0.5\pi }+e^{+j0.5\pi })\text{ V/m}\\ E_{\theta s}=0.2e^{-j1000\pi }2\cos (0.5\pi )\text{ V/m}\\ E_{\theta s}=0\text{ V/m}\text{.}\end{array}$

Conclusion:

The value of electric field $E_{\theta s}=0\text{ V/m}$.