Answer 8.4

_psets/6.md

@@ -79,3 +79,67 @@ $$
 
 {:.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 $$A$$,
+
+$$
+\begin{align*}
+A &= \frac{V^2}{4 R} \sqrt{\frac{\mu_0}{\epsilon_0}} \frac{1}{E^2_\text{max}} \\
+&= \frac{V^2}{4} \sqrt{\frac{\mu_0}{\epsilon_0}} \frac{1}{E^2_\text{max}}
+\frac{3}{2 \pi} \sqrt{\frac{\epsilon_0}{\mu_0}} \left( \frac{\lambda}{d} \right)^2 \\
+&= \frac{3}{8} \frac{V^2}{E^2_\text{max}} \left( \frac{\lambda}{d} \right)^2
+\end{align*}
+$$
+
+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 $$d$$, $$V
+= E_\text{max} d$$. As such
+
+$$
+A = \frac{3}{8} \lambda^2
+$$