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+---
+title: Problem Set 6
+---
+
+## (8.1)
+
+{:.question}
+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)
+
+{:.question}
+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)
+
+{:.question}
+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)
+
+{:.question}
+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
+
+$$
+\begin{align*}
+\frac{A}{G} &= \frac{3}{8 \pi} \lambda^2 \frac{2}{3} \\
+&= \frac{1}{4 \pi} \lambda^2
+\end{align*}
+$$