Chapter 14, Problem 14.3P

To determine

(a)

The $E_s$ at the distant point $\text{(}0,1000,0\text{)}$.

Answer to Problem 14.3P

The value of electric field is $E_s=-j(1.5\times {10}^{-2})e^{-j1000}{a}_z\text{ V/m}$.

Explanation of Solution

Given:

The following parameters are given:

$\lambda =2\pi \text{ m, }d=0.1\text{ m}$ and $(x=0,y=1000,z=0)$.

Concept Used:

This point lies along the axis of the antenna in the ${\text{a}}_y$ direction and thus its contribution is zero. The contribution will be coming from the antenna in the ${\text{a}}_z$ direction. The observation has been made from the far field zone. Since $\theta ={90}^0$ , only $E_{\theta s}$ component will be present.

$E_s=E_{\theta s}{\text{a}}_{\theta }=j\frac{I_{0}d{\eta }_0}{2\lambda r}\sin \theta e^{-j\raisebox{1ex}{$2\pi r$}\!\left/ \!\raisebox{-1ex}{$\lambda $}\right.}{\text{a}}_{\theta }$ . Now, intrinsic impedance of free space ${\eta }_0=377\text{ Ω}$.

Calculation:

Using the given values, the electric field $E_s$ becomes.

   $\begin{array}{l}{E}_s\text{=}j\frac{5(0.1)(120\pi )}{4\pi (1000)}(1)(e^{-j1000}){\text{ a}}_{\theta }\\ E_s\text{=}j(1.5\times {10}^{-2})e^{-j1000}{\text{ a}}_{\theta }\\ E_s\text{=}-j(1.5\times {10}^{-2})e^{-j1000}{\text{ a}}_z\text{V/m}\end{array}$

Conclusion:

The value of electric field is $E_s=-j(1.5\times {10}^{-2})e^{-j1000}{a}_z\text{ V/m}$.

To determine

(b)

The $E_s$ at the distant point $\text{(}0,0,1000\text{)}$.

Answer to Problem 14.3P

The value of electric field is $E_s=-j(1.5\times {10}^{-2})e^{-j1000}{\text{a}}_y\text{ V/m}$.

Explanation of Solution

Given:

The following parameters are given:

$\lambda =2\pi \text{ m and }d=0.1\text{ m}$ and $(x=0,y=0,z=1000)$.

Concept Used:

This point lies along the axis of the antenna in the ${\text{a}}_z$ direction and thus its contribution is zero. The contribution will be coming from the antenna in the ${\text{a}}_y$ direction. The distance is same and thus the result obtained in part a can be used. However, the direction will be different (along $-a_y$ ) in this case.

Calculation:

Using the given values, the electric field $E_s$ becomes.

   $E_s=-j(1.5\times {10}^{-2})e^{-j1000}{\text{ a}}_y\text{ V/m}$

Conclusion:

The value of electric field is $E_s=-j(1.5\times {10}^{-2})e^{-j1000}{\text{ a}}_y\text{ V/m}$.

To determine

(c)

The value of $E_s$ at the distant point where $\text{(}x=\text{100}0,y=0,z=0\text{)}$.

Answer to Problem 14.3P

The value of electric field is, $E_s=-j(1.5\times {10}^{-2})e^{-j1000}{\text{(a}}_z{\text{+ a}}_y\text{) V/m}$.

Explanation of Solution

Given:

The following parameters are given:

$\lambda =2\pi \text{ m and }d=0.1\text{ m}$ and $\text{(}x=\text{100}0,y=0,z=0\text{)}$.

Concept Used:

This point lies along the x direction, contribution from both the antenna will be counted. Since the distance is same, the result obtained from part (a) and (b) needs to be added.

Calculation:

Under this given condition, the electric field $E_s$ becomes.

   $E_s=-j(1.5\times {10}^{-2})e^{-j1000}{\text{ (a}}_z{\text{+ a}}_y\text{) V/m}$

Conclusion:

The value of electric field is $E_s=-j(1.5\times {10}^{-2})e^{-j1000}{\text{ (a}}_z{\text{+ a}}_y\text{) V/m}$.

To determine

(d)

The value of $E$ at the distant point where $\text{(}x=\text{100}0,y=0,z=0\text{)}$ at $t=0$.

Answer to Problem 14.3P

The value of electric field is $E(0)=-(1.24\times {10}^{-2})({\text{a}}_y+{\text{a}}_z\text{) V/m}$.

Explanation of Solution

Given:

The following parameters are given:

$\lambda =2\pi \text{ m and }d=0.1\text{ m}$ and $\text{(}x=\text{100}0,y=0,z=0\text{)}$ at $t=0$.

Concept Used:

At any time t, the electric field $\text{E}(t)$ can be obtained from the relation $\text{E(t)}={\text{Re{E}}_{\text{s}}{e}^{jwt}\text{}}$.

Calculation:

Using the result obtained in the previous section,

   $\begin{array}{l}\text{E(}t\text{)}{\text{= Re{E}}_s{\text{e}}^{jwt}\text{}}\\ \text{E}(t)\text{ = Re }\{-j(1.5\times {10}^{-2})e^{-j(1000-wt)}\}{\text{ (a}}_{\text{y}}{\text{+a}}_{\text{z}}\text{) V/m}\\ \text{E}(t)\text{ = }(1.5\times {10}^{-2})\sin (wt-1000){\text{ (a}}_{\text{y}}+{\text{a}}_{\text{z}}\text{) V/m}\end{array}$

For $t=0$ , the value becomes,

   $\begin{array}{l}\text{E}(0)\text{ = }(1.5\times {10}^{-2})\sin (-1000){\text{ (a}}_{\text{y}}+{\text{a}}_{\text{z}}\text{) V/m}\\ \text{ = }-(1.24\times {10}^{-2}){\text{ (a}}_{\text{y}}+{\text{a}}_{\text{z}}\text{) V/m}\end{array}$

Conclusion:

The value of electric field is, $\text{E}(0)\text{ = }-(1.24\times {10}^{-2}){\text{(a}}_{\text{y}}+{\text{a}}_{\text{z}}\text{) V/m}$

To determine

(e)

The value of $\left|\text{E}\right|$ at the distant point where $\text{(}x=\text{100}0,y=0,z=0\text{)}$ at $t=0$.

Answer to Problem 14.3P

The value of magnitude of electric field is $\left|\text{E}\right|\text{= }(1.75\times {10}^{-2})\text{ V/m}$.

Explanation of Solution

Given:

The following parameters are given:

$\lambda =2\pi \text{ m and }d=0.1\text{ m}$ and $\text{(}x=\text{100}0,y=0,z=0\text{)}$ at $t=0$.

Concept Used:

At any time t, the electric field $\text{E}(t)$ can be obtained from the relation $\text{E(}t\text{) }=\text{ Re{}{E}_{\text{s}}{e}^{jwt}\text{}}$.

Calculation:

The magnitude of part (d) result is,

   $\left|\text{E}\right|\text{= }(1.75\times {10}^{-2})\text{ V/m}$

Conclusion:

The value of magnitude of electric field is, $\left|\text{E}\right|\text{= }(1.75\times {10}^{-2})\text{ V/m}$.