From b74081f71285c0a56a86cf949b5ccf8bfedce588 Mon Sep 17 00:00:00 2001
From: Erik Strand <erik.strand@cba.mit.edu>
Date: Wed, 13 Mar 2019 23:01:43 -0400
Subject: [PATCH] Answer 7.3

---
 _psets/5.md | 27 +++++++++++++++++++++++++++
 1 file changed, 27 insertions(+)

diff --git a/_psets/5.md b/_psets/5.md
index d3a8df0..cf048e8 100644
--- a/_psets/5.md
+++ b/_psets/5.md
@@ -41,6 +41,33 @@ $$
 Integrate Poynting’s vector $$P = E \times H$$ to find the power flowing across a cross-sectional
 slice of a coaxial cable, and relate the answer to the current and voltage in the cable.
 
+We saw that the magnetic field has a magnitude of $$I/(2 \pi r)$$ and wraps around the inner
+conductor according to the right hand rule. The electric field has magnitude $$Q/(2 \pi \epsilon
+r)$$ and points radially from the inner conductor toward the outer. So $$P = E \times H$$ points
+along the direction of current in the inner conductor, and has magnitude
+
+$$
+| P | = \frac{I Q}{(2 \pi r)^2 \epsilon}
+$$
+
+Integrating over the area between the conductors, we find that the cross-sectional power transfer is
+
+$$
+\begin{align*}
+\int_0^{2 \pi} \int_{r_i}^{r_o} \frac{I Q}{(2 \pi r)^2 \epsilon} r \mathrm{d} r \mathrm{d} \theta
+&= \frac{I Q}{2 \pi \epsilon} \int_{r_i}^{r_o} r^{-1} \mathrm{d} r \mathrm{d} \theta \\
+&= \frac{I Q}{2 \pi \epsilon} \ln \left( \frac{r_o}{r_i} \right)
+\end{align*}
+$$
+
+But the charge is
+
+$$
+Q = CV = \frac{2 \pi \epsilon}{\ln (r_0 / r_i)}
+$$
+
+so we can write the total power as $$IV$$.
+
 ## (7.4)
 
 Find the characteristic impedance and signal velocity for a transmission line consisting of two
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