Chapter 13, Problem 1P

To determine

Calculate the electric field intensity $E\left(r,\theta ,\varphi ,t\right)$ and the magnetic field intensity $H\left(r,\theta ,\varphi ,t\right)$ at the far field.

Answer to Problem 1P

The electric field intensity $E\left(r,\theta ,\varphi ,t\right)$ and the magnetic field intensity $H\left(r,\theta ,\varphi ,t\right)$ at the far field are $\frac{50\beta \eta }{\mu r}\sin \left(\omega t-\beta r\right)\left(\sin \varphi a_{\varphi }+\cos \theta \cos \varphi a_{\theta }\right)\text{V}/\text{m}$ and $-\frac{50}{\mu r}\beta \sin \left(\omega t-\beta r\right)\left(\sin \varphi a_{\theta }-\cos \theta \cos \varphi a_{\varphi }\right)\text{A}/\text{m}$ respectively.

Explanation of Solution

Calculation:

Given that,

$A_s=\frac{50e^{-j\beta r}}{r}{a}_x$ where $r^2=x^2+y^2+z^2$.

Using vector transformation (in spherical coordinate system),

$a_x=\sin \theta \cos \varphi a_r+\cos \theta \cos \varphi a_{\theta }-\sin \varphi a_{\varphi }$

Therefore, the given vector function $A_s$ is written as,

$A_s=\frac{50e^{-j\beta r}}{r}\left(\sin \theta \cos \varphi a_r+\cos \theta \cos \varphi a_{\theta }-\sin \varphi a_{\varphi }\right)$        (1)

Write the general expression for magnetic vector potential $A_s$.

$\nabla \times \frac{A_s}{\mu }=H_s$        (2)

Here,

$\mu$ is the permeability of the medium, and

$H_s$ is the magnetic field intensity.

Substitute equation (1) in (2).

$\begin{array}{l}\nabla \times \frac{\left(\frac{50e^{-j\beta r}}{r}\left(\sin \theta \cos \varphi a_r+\cos \theta \cos \varphi a_{\theta }-\sin \varphi a_{\varphi }\right)A_s\right)}{\mu }=H_s\\ \frac{1}{r^2\sin \theta }\left|\begin{array}{ccc}{a}_r& r{a}_{\theta }& r\sin \theta a_{\varphi }\\ \frac{\partial }{\partial r}& \frac{\partial }{\partial \theta }& \frac{\partial }{\partial \varphi }\\ \frac{50e^{-j\beta r}}{\mu r}\sin \theta \cos \varphi & r\left(\frac{50e^{-j\beta r}}{\mu r}\cos \theta \cos \varphi \right)& r\sin \theta \left(-\frac{50e^{-j\beta r}}{\mu r}\sin \varphi \right)\end{array}\right|=H_s\\ \frac{1}{r^2\sin \theta }\left|\begin{array}{ccc}{a}_r& r{a}_{\theta }& r\sin \theta a_{\varphi }\\ \frac{\partial }{\partial r}& \frac{\partial }{\partial \theta }& \frac{\partial }{\partial \varphi }\\ \frac{50e^{-j\beta r}}{\mu r}\sin \theta \cos \varphi & \frac{50e^{-j\beta r}}{\mu }\cos \theta \cos \varphi & -\frac{50e^{-j\beta r}}{\mu }\sin \theta \sin \varphi \end{array}\right|=H_s\\ \frac{1}{r^2\sin \theta }\left[\begin{array}{l}\left(-\frac{50e^{-j\beta r}}{\mu }\cos \theta \sin \varphi -\frac{50e^{-j\beta r}}{\mu }\cos \theta \sin \varphi \right)a_r\\ -\left(\frac{j50\beta e^{-j\beta r}}{\mu }\sin \theta \sin \varphi +\frac{50e^{-j\beta r}}{\mu r}\sin \theta \sin \varphi \right)r{a}_{\theta }\\ +\left(\frac{-j50\beta e^{-j\beta r}}{\mu }\cos \theta \cos \varphi -\frac{50e^{-j\beta r}}{\mu r}\cos \theta \cos \varphi \right)r\sin \theta a_{\varphi }\end{array}\right]=H_s\end{array}$

Reduce the equation as follows,

$-\frac{100\cos \theta \sin \varphi }{\mu r^2\sin \theta }{e}^{-j\beta r}{a}_r-\frac{50}{\mu r^2}\left(1-j\beta r\right)\sin \varphi e^{-j\beta r}{a}_{\theta }-\frac{50}{\mu r^2}\cos \theta \cos \varphi \left(1+j\beta r\right)e^{-j\beta r}{a}_{\varphi }=H_s$

At far field, $\frac{1}{r}$ term only remains. Therefore,

$H_s=\frac{j50}{\mu r}\beta e^{-j\beta r}\left(\sin \varphi a_{\theta }-\cos \theta \cos \varphi a_{\varphi }\right)$        (3)

Consider the general expression to calculate the electric fields intensity.

$E_s=-\eta a_r\times H_s$        (4)

Here,

$\eta$ is the intrinsic impedance.

Substitute equation (3) in (4).

$E_s=-\eta a_r\times \left[\frac{j50}{\mu r}\beta e^{-j\beta r}\left(\sin \varphi a_{\theta }-\cos \theta \cos \varphi a_{\varphi }\right)\right]$

$E_s=-\frac{j50\beta \eta e^{-j\beta r}}{\mu r}\left(\sin \varphi a_{\varphi }+\cos \theta \cos \varphi a_{\theta }\right)$        (5)

The general expression to calculate the electric field intensity is,

$E=\mathrm{Re}\left[E_s{e}^{j\omega t}\right]$        (6)

Substitute equation (5) in (6).

$\begin{array}{c}E=\mathrm{Re}\left[-\frac{j50\beta \eta e^{-j\beta r}}{\mu r}\left(\sin \varphi a_{\varphi }+\cos \theta \cos \varphi a_{\theta }\right)e^{j\omega t}\right]\\ =\mathrm{Re}\left[-\frac{j50\beta \eta e^{j\left(\omega t-\beta r\right)}}{\mu r}\left(\sin \varphi a_{\varphi }+\cos \theta \cos \varphi a_{\theta }\right)e^{j\omega t}\right]\\ =\mathrm{Re}\left\{-\frac{j50\beta \eta \left[\cos \left(\omega t-\beta r\right)+j\sin \left(\omega t-\beta r\right)\right]}{\mu r}\left(\sin \varphi a_{\varphi }+\cos \theta \cos \varphi a_{\theta }\right)e^{j\omega t}\right\}\\ =\frac{50\beta \eta }{\mu r}\sin \left(\omega t-\beta r\right)\left(\sin \varphi a_{\varphi }+\cos \theta \cos \varphi a_{\theta }\right)\text{V}/\text{m}\end{array}$

The general expression to calculate the magnetic field intensity is,

$H=\mathrm{Re}\left[H_s{e}^{j\omega t}\right]$        (7)

Substitute equation (3) in (7).

$\begin{array}{c}H=\mathrm{Re}\left[\frac{j50}{\mu r}\beta e^{-j\beta r}\left(\sin \varphi a_{\theta }-\cos \theta \cos \varphi a_{\varphi }\right)e^{j\omega t}\right]\\ =\mathrm{Re}\left[\frac{j50}{\mu r}\beta e^{j\left(\omega t-\beta r\right)}\left(\sin \varphi a_{\theta }-\cos \theta \cos \varphi a_{\varphi }\right)\right]\\ =\mathrm{Re}\left\{\frac{j50}{\mu r}\beta \left[\cos \left(\omega t-\beta r\right)+j\sin \left(\omega t-\beta r\right)\right]\left(\sin \varphi a_{\theta }-\cos \theta \cos \varphi a_{\varphi }\right)\right\}\\ =-\frac{50}{\mu r}\beta \sin \left(\omega t-\beta r\right)\left(\sin \varphi a_{\theta }-\cos \theta \cos \varphi a_{\varphi }\right)\text{A}/\text{m}\end{array}$

Conclusion:

Thus, the electric field intensity $E\left(r,\theta ,\varphi ,t\right)$ and the magnetic field intensity $H\left(r,\theta ,\varphi ,t\right)$ at the far field are $\frac{50\beta \eta }{\mu r}\sin \left(\omega t-\beta r\right)\left(\sin \varphi a_{\varphi }+\cos \theta \cos \varphi a_{\theta }\right)\text{V}/\text{m}$ and $-\frac{50}{\mu r}\beta \sin \left(\omega t-\beta r\right)\left(\sin \varphi a_{\theta }-\cos \theta \cos \varphi a_{\varphi }\right)\text{A}/\text{m}$ respectively.