diff --git a/_psets/5.md b/_psets/5.md index d3a8df09558b13c973bc1569e5a3d2cd981c6dad..cf048e84386da0c0c0830bacf52682b529acca6e 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