Problem Set 6

(8.1)

Find the electric field for an infinitesimal dipole radiator.

This problem is most naturally expressed in spherical coordinates. We know that the vector potential is

$\begin{align*} A_r &= \mu_0 \frac{I_0 d e^{-i k r}}{4 \pi r} \cos \theta \\ A_\theta &= -\mu_0 \frac{I_0 d e^{-i k r}}{4 \pi r} \sin \theta \\ A_\phi &= 0 \end{align*}$

So the electric field is “simply”

$E = \frac{1}{i \omega \mu_0 \epsilon_0} \nabla ( \nabla \cdot A) - i \omega A$

First let’s find $\nabla \cdot A$ .

$\begin{align*} \nabla \cdot A &= \frac{1}{r^2} \frac{\partial}{\partial r} (r^2 A_r) + \frac{1}{r \sin \theta} \frac{\partial}{\partial \theta} ( A_\theta \sin \theta ) + \frac{1}{r \sin \theta} \frac{\partial A_\phi}{\partial \phi} \\ &= \frac{\mu_0 I_0 d}{4 \pi} \left( \frac{\cos \theta}{r^2} \frac{\partial}{\partial r} \left( r e^{-ikr} \right) - \frac{e^{-ikr}}{r^2 \sin \theta} \frac{\partial}{\partial \theta} \sin^2 \theta \right) \\ &= \frac{\mu_0 I_0 d}{4 \pi} \left( \frac{\cos \theta}{r^2} \left( e^{-ikr} - r i k e^{-ikr} \right) - \frac{e^{-ikr}}{r^2 \sin \theta} \left( 2 \sin \theta \cos \theta \right) \right) \\ &= \frac{\mu_0 I_0 d}{4 \pi} \cos \theta e^{-ikr} \left( \frac{1}{r^2} - \frac{ik}{r} - \frac{2}{r^2} \right) \\ &= - \frac{\mu_0 I_0 d}{4 \pi} \cos \theta e^{-ikr} \left( \frac{1}{r^2} + \frac{ik}{r} \right) \end{align*}$

For any function $f(r, \theta, \phi)$ in spherical coordinates,

$\nabla f = \frac{\partial f}{\partial r} \hat{r} + \frac{1}{r} \frac{\partial f}{\partial \theta} \hat{\theta} + \frac{1}{r \sin \theta} \frac{\partial f}{\partial \phi} \hat{\phi}$

So the three components of $\nabla (\nabla \cdot A)$ are

$\begin{align*} [\nabla (\nabla \cdot A)]_r &= \frac{\partial (\nabla \cdot A)}{\partial r} \\ &= - \frac{\mu_0 I_0 d}{4 \pi} \cos \theta \left( \left( \frac{-2}{r^3} - \frac{ik}{r^2} \right) e^{-ikr} + \left( \frac{1}{r^2} + \frac{ik}{r} \right) (-ik) e^{-ikr} \right) \\ &= - \frac{\mu_0 I_0 d}{4 \pi} \cos \theta e^{-ikr} \left( \frac{-2}{r^3} - \frac{ik}{r^2} - \frac{ik}{r^2} + \frac{k^2}{r} \right) \\ &= \frac{\mu_0 I_0 d}{4 \pi} \cos \theta e^{-ikr} \left( \frac{2}{r^3} + \frac{2ik}{r^2} - \frac{k^2}{r} \right) \\ [\nabla (\nabla \cdot A)]_\theta &= \frac{1}{r} \frac{\partial (\nabla \cdot A)}{\partial \theta} \\ &= - \frac{\mu_0 I_0 d}{4 \pi} e^{-ikr} \left( \frac{1}{r^3} + \frac{ik}{r^2} \right) \frac{\partial}{\partial \theta} \cos \theta\\ &= \frac{\mu_0 I_0 d}{4 \pi} e^{-ikr} \left( \frac{1}{r^3} + \frac{ik}{r^2} \right) \sin \theta \\ [\nabla (\nabla \cdot A)]_\phi &= 0 \end{align*}$

Putting these together (and recalling that $k = \omega \sqrt{\mu_0 \epsilon_0}$ ) we find that

$\begin{align*} E_r &= \frac{1}{i \omega \mu_0 \epsilon_0} \frac{\mu_0 I_0 d}{4 \pi} \cos \theta e^{-ikr} \left( \frac{2}{r^3} + \frac{2ik}{r^2} - \frac{k^2}{r} \right) - i \omega \frac{\mu_0 I_0 d}{4 \pi} \frac{\cos \theta}{r} e^{-ikr} \\ &= \frac{\mu_0 I_0 d}{4 \pi} \cos \theta e^{-ikr} \left( \frac{1}{i \omega \mu_0 \epsilon_0} \left( \frac{2}{r^3} + \frac{2ik}{r^2} - \frac{k^2}{r} \right) - \frac{i \omega}{r} \right) \\ &= \frac{I_0 d}{4 \pi} \cos \theta e^{-ikr} \left( \frac{2}{i \omega \epsilon_0 r^3} + \frac{2k}{\omega \epsilon_0 r^2} - \frac{k^2}{i \omega \epsilon_0 r} - \frac{i \omega \mu_o}{r} \right) \\ &= \frac{I_0 d}{4 \pi} \cos \theta e^{-ikr} \left( \frac{2}{i \omega \epsilon_0 r^3} + \frac{2 \omega \sqrt{\epsilon_0 \mu_0}}{\omega \epsilon_0 r^2} - \frac{\omega^2 \epsilon_0 \mu_0}{i \omega \epsilon_0 r} - \frac{i \omega \mu_o}{r} \right) \\ &= \frac{I_0 d}{4 \pi} \cos \theta e^{-ikr} \left( \frac{2}{i \omega \epsilon_0 r^3} + \frac{2}{r^2} \sqrt{\frac{\mu_0}{\epsilon_0}} + \frac{i \omega \mu_0}{r} - \frac{i \omega \mu_o}{r} \right) \\ &= \frac{I_0 d}{4 \pi} \cos \theta e^{-ikr} \left( \frac{2}{i \omega \epsilon_0 r^3} + \frac{2}{r^2} \sqrt{\frac{\mu_0}{\epsilon_0}} \right) \\ E_\theta &= \frac{1}{i \omega \mu_0 \epsilon_0} \frac{\mu_0 I_0 d}{4 \pi} \sin \theta e^{-ikr} \left( \frac{1}{r^3} + \frac{ik}{r^2} \right) + i \omega \frac{\mu_0 I_0 d}{4 \pi} \frac{\sin \theta}{r} e^{-ikr} \\ &= \frac{I_0 d}{4 \pi} \sin \theta e^{-ikr} \left( \frac{1}{i \omega \epsilon_0 r^3} + \frac{k}{\omega \epsilon_0 r^2} + \frac{i \omega \mu_0}{r} \right) \\ &= \frac{I_0 d}{4 \pi} \sin \theta e^{-ikr} \left( \frac{1}{i \omega \epsilon_0 r^3} + \frac{1}{r^2} \sqrt{\frac{\mu_0}{\epsilon_0}} + \frac{i \omega \mu_0}{r} \right) \\ E_\phi &= 0 \end{align*}$

(8.2)

What is the magnitude of the Poynting vector at a distance of 1 km from an antenna radiating 1 kW of power, assuming that it is an isotropic radiator with a wavelength much less than 1 km? What is the peak electric field strength at that distance?

Since it’s an isotropic radiator, the Poynting vector must point along $\hat{r}$ . Similarly its magnitude must depend only on $r$ , so we can compute its time average value by dividing the total power by the total area.

$\begin{align*} \langle |P| \rangle &= \frac{10^3 \si{W}}{4 \pi \cdot 10^6 \si{m^2}} \\ &= \num{8e-5} \si{W/m^2} \end{align*}$

Since $\lambda << 1\si{km}$ , locally the radiation will look like a plane wave. Thus E and H are perpendicular, and $|H| = \sqrt{\epsilon_0/\mu_0} |E|$ .

$\begin{align*} \langle |P| \rangle &= \langle |E| |H| \rangle \\ &= \sqrt{\frac{\epsilon_0}{\mu_0}} \langle |E|^2 \rangle \\ &= \frac{1}{2} \sqrt{\frac{\epsilon_0}{\mu_0}} E^2_\text{max} \end{align*}$

Solving for $E_\text{max}$ ,

$\begin{align*} E_\text{max} &= \left( 2 \sqrt{\frac{\mu_0}{\epsilon_0}} \langle P \rangle \right)^\frac{1}{2} \\ &= \sqrt{2 \cdot 377 \si{\ohm} \cdot \num{8e-5} \si{W/m^2}} \\ &= 0.24 \si{V/m} \end{align*}$

(8.3)

For what value of $R_\text{load}$ is the maximum power delivered to the load in Figure 8.3?

By Ohm’s Law $V = I (R_r + R_l)$ (where $R_r$ is the radiation resistance, and $R_l$ the load resistance). Thus $I = V / (R_r + R_l)$ and the power dissipated in the load resistance is

$\begin{align*} P &= I^2 R \\ &= \frac{V^2 R_l}{(R_r + R_l)^2} \end{align*}$

To maximize this we take the derivative

$\frac{d P}{d R_l} = V^2 ( (R_r + R_l)^{-2} - 2 R_l (R_r + R_l)^{-3} )$

Setting this to zero implies $R_r + R_l = 2 R_l$ , or

$R_l = R_r$

(8.4)

For an infinitesimal dipole antenna, what are the gain and the area, and what is their ratio?

From the text we know that

$\langle |P| \rangle = \frac{I_0^2 k^2 d^2}{32 \pi^2 r^2} \sqrt{\frac{\mu_0}{\epsilon_0}} \sin^2 \theta$

and

$W = \frac{I_0^2 \pi}{3} \sqrt{\frac{\mu_0}{\epsilon_0}} \left( \frac{d}{\lambda} \right)^2$

Thus

$\begin{align*} G &= \max_{\theta, \phi} \frac{\langle |P(r = 1, \theta, \phi)| \rangle}{W/4 \pi} \\ &= 4 \pi \frac{I_0^2 k^2 d^2}{32 \pi^2} \sqrt{\frac{\mu_0}{\epsilon_0}} \frac{3}{I_0^2 \pi} \sqrt{\frac{\epsilon_0}{\mu_0}} \left( \frac{\lambda}{d} \right)^2 \\ &= \frac{(2 \pi \lambda^{-1})^2}{8 \pi^2} 3 \lambda^2 \\ &= \frac{3}{2} \end{align*}$

since $k = 2 \pi / \lambda$ .

To find the area, recall that the maximum power received is $V^2 / 8 R$ . This will be equal to the area times the Poynting vector. Thus

$\begin{align*} \frac{V^2}{8 R} &= \langle | P | \rangle A \\ &= \frac{1}{2} \sqrt{\frac{\epsilon_0}{\mu_0}} E^2_\text{max} A \end{align*}$

Solving for $A$ ,

$\begin{align*} A &= \frac{V^2}{4 R} \sqrt{\frac{\mu_0}{\epsilon_0}} \frac{1}{E^2_\text{max}} \\ &= \frac{V^2}{4} \sqrt{\frac{\mu_0}{\epsilon_0}} \frac{1}{E^2_\text{max}} \frac{3}{2 \pi} \sqrt{\frac{\epsilon_0}{\mu_0}} \left( \frac{\lambda}{d} \right)^2 \\ &= \frac{3}{8 \pi} \frac{V^2}{E^2_\text{max}} \left( \frac{\lambda}{d} \right)^2 \end{align*}$

since for an infinitesimal dipole radiator

$R = \frac{2 \pi}{3} \sqrt{\frac{\mu_0}{\epsilon_0}} \left( \frac{d}{\lambda} \right)^2$

Now we need to relate the voltage and the maximum magnitude of the electric field. The whole point of the antenna is to pick up oscillating electric and magnetic fields, and the presence of the latter makes voltage path dependent. So let’s integrate along our (infinitely thin) antenna. Our antenna has an infinitesimal length $d$ , so the field will be constant across it. Thus the integral is just $d$ times the electric field, i.e. $V = E_\text{max} d$ . As such

$A = \frac{3}{8 \pi} \lambda^2$

Finally we see that the area gain ratio is

$\begin{align*} \frac{A}{G} &= \frac{3}{8 \pi} \lambda^2 \frac{2}{3} \\ &= \frac{1}{4 \pi} \lambda^2 \end{align*}$