Chapter 14, Problem 14.1P

A short dipole-carrying current I₀ cos $\omega t$ in the a_z direction is located at the origin in free space, (a) If k = 1 rad/m, r = 2 m, $\theta =45°$, $\varphi =0°$, and t = 0, give a unit vector in rectangular components that shows the instantaneous direction of E. (b) What fraction of the total average power is radiated in the belt, $80°<\theta <100°$?

To determine

(a)

A unit vector of electric field that shows the instantaneous direction of E for a short dipole.

Answer to Problem 14.1P

A unit vector of electric field that shows the instantaneous direction of E for a short dipole is $E_N=-0.284a_x-0.959a_z$.

Explanation of Solution

Given:

k =1 rad/m

r = 2 m

   $\begin{array}{l}\theta =45°\\ \varphi =0\\ t=0\end{array}$

Concept Used:

In spherical co-ordinate component of electric field is given by,

   $\begin{array}{l}{E}_{\theta s}=\frac{I_{0}kd\eta }{4\pi r}\sin \theta e^{-jkr}\left(\frac{jk}{r}+\frac{1}{r^2}+\frac{1}{jk{r}^3}\right)\\ E_{rs}=\frac{I_{0}kd\eta }{2\pi r}\cos \theta e^{-jkr}\left(\frac{1}{r^2}+\frac{1}{jk{r}^3}\right)\end{array}$

Calculation:

   $\begin{array}{l}{E}_R=e^{-jkr}\left(\frac{1}{r^2}+\frac{1}{jk{r}^3}\right)\\ E_{\theta }=e^{-jkr}\frac{1}{2}\left(\frac{jk}{r}+\frac{1}{r^2}+\frac{1}{jk{r}^3}\right)\end{array}$

Plugging r =2 in above equations

   $\begin{array}{l}{E}_R=e^{-j2}\left(\frac{1}{4}-\frac{1}{j8}\right)\\ E_R=\frac{1}{4}\left(1.12\right)e^{-j26.6°}{e}^{-j2}\\ E_{\theta }=e^{-jkr}\left(j\frac{1}{4}+\frac{1}{8}-\frac{1}{j16}\right)\\ E_{\theta }=\frac{1}{4}\left(0.90\right)e^{-j56.3°}{e}^{-j2}\end{array}$

Therefore,

Vector $E_N=E_R{a}_R+E_{\theta }{a}_{\theta }$

Normalize above vector by dividing their magnitude.

   $\begin{array}{l}{E}_N=E_R{a}_R+E_{\theta }{a}_{\theta }\\ E_N=0.780e^{-j141.2°}{a}_R+0.627e^{-j58.3°}{a}_{\theta }\end{array}$

Converting above equation in real instantaneous form.

   $E_{N\left(t\right)}=\mathrm{Re}\left(E_{N\left(t\right)}{e}^{j\omega t}\right)=0.780\cos \left(\omega t-141.2°\right)a_R+0.627\cos \left(\omega t-58.3°\right)a_{\theta }$

Hence,

   $\begin{array}{l}{e}_{N\left(0\right)}=\frac{\left(-0.608a_R+0.330a_{\theta }\right)}{\sqrt{{0.608}^2+{0.330}^2}}\\ e_{N\left(0\right)}=\left(-0.879a_R+0.477a_{\theta }\right)\\ e_{N\left(x\right)}=e_{N\left(0\right)}\cdot a_x=\left(-0.879\sin \theta \cos \varphi +0.477\cos \theta \cos \varphi \right)\\ e_{N\left(x\right)}=\frac{1}{\sqrt{2}}\left(-0.879+0.477\right)\\ e_{N\left(x\right)}=-0.284\\ e_{N\left(y\right)}=e_{N\left(0\right)}\cdot a_y=\left(-0.879\sin \theta \sin \varphi +0.477\cos \theta \cos \varphi \right)\\ e_{N\left(y\right)}=0\\ e_{N\left(z\right)}=e_{N\left(0\right)}\cdot a_z=\left(-0.879\cos \theta -0.477\sin \theta \right)\\ e_{N\left(z\right)}=-0.959\end{array}$

at t =0

   $e_{N\left(0\right)}$ in rectangular form is,

   $e_{N\left(0\right)}=-0.284a_x-0.959a_z$

Conclusion:

Therefore, the unit vector of electric field that shows the instantaneous direction of E for a short dipole is $-0.284a_x-0.959a_z$.

To determine

(b)

The fraction of total average power is radiated in the belt $80°\le \theta \le 100°$.

Answer to Problem 14.1P

The fraction of total average power is radiated in the given belt is 0.258.

Explanation of Solution

Given:

   $\begin{array}{l}k=\frac{2\pi }{\lambda }\\ 80°\le \theta \le 100°\end{array}$

Concept Used:

   $P_{avg}=\frac{I_0{}^2{d}^2\eta }{8{\lambda }^2{r}^2}{\sin }^2\theta W/m^2$

Calculation:

   $P_{avg}=\frac{I_0{}^2{d}^2\eta }{8{\lambda }^2{r}^2}{\sin }^2\theta W/m^2$

Integrate above equation over the belt.

   $\begin{array}{l}{P}_{belt}={\displaystyle \underset{0}{\overset{2\pi }{\int }}{\displaystyle \underset{80°}{\overset{100°}{\int }}\frac{I_0{}^2{d}^2\eta }{8{\lambda }^2{r}^2}{\sin }^2\theta r^2\sin \theta d\theta d\varphi }}\\ P_{belt}=\frac{\pi I_0{}^2{d}^2\eta }{4{\lambda }^2}{\displaystyle \underset{80°}{\overset{100°}{\int }}{\sin }^{3}d\theta }\end{array}$

Evaluating above integral.

   $P_{belt}=0.344\frac{\pi I_0{}^2{d}^2\eta }{4{\lambda }^2}$

While total power is found by performing the same integral over $0°\le \theta \le 360°$.

   $P_{total}=1.333\frac{\pi I_0{}^2{d}^2\eta }{4{\lambda }^2}$

Therefore, the fraction of the total power is,

   $\begin{array}{l}f=\frac{0.344}{1.333}\\ f=0.258\end{array}$

Conclusion:

Hence, the fraction of total power radiated to the given belt is 0.258.