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--- 184_notes:examples:week6_drift_speed [2017/09/26 15:51] – [Solution] tallpaul
+++ 184_notes:examples:week6_drift_speed [2017/09/26 16:05] – [Solution] tallpaul
@@ Line 13: / Line 13: @@
   * Electron charge density for both wires.
   * Electron current for both wires.
+  * Cross-sectional area for both wires.
 ===Approximations & Assumptions===
@@ Line 23: / Line 24: @@
   * We represent current as $I=qi$. Current is charge per second. Electron current is electrons per second. We multiply by $q$ (the electron charge) to get charge per second.
 ====Solution====
-We can look up electron
+We can look up electron density $n$ in the table. It is labeled as "density of valency electrons", which to us just means the density of free electrons, or electrons that are free to drift through the wire. For copper, we get $n_{\text{Cu}}=8.47\cdot 10^{22} \text{ cm}^{-3}$. For zinc, we get $n_{\text{Zn}}=13.2\cdot 10^{22} \text{ cm}^{-3}$.
+To find the cross-sectional area of the wire, we just use the area of a circle. We know the radius, so this should be easy: $A=\pi r^2$.
+We are given current, and we can solve for electron current using the charge of an electron: $i = \frac{I}{q}$.
+We now have enough information to solve for the drift speed of electrons. We use positive numbers below, since we care only about speed for now, not direction.
+$$v_{avg} = \frac{I}{\pi r^2 n q}$$
+Current ($I$), radius ($r$), electron density ($n$), and electron charge ($q$) are all things we know for our two wires. When we plug in the numbers, we get the following:
+\begin{align*}
+v_{\text{avg, Cu}} = 0.47 \text{ mm/s} &,& v_{\text{avg, Zn}} = 7.5 \text{ mm/s}
+\end{align*}