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--- 184_notes:examples:week6_drift_speed [2017/09/26 15:57] – [Solution] tallpaul
+++ 184_notes:examples:week6_drift_speed [2017/09/26 16:05] – [Solution] tallpaul
@@ Line 28: / Line 28: @@
 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: $i = I/q$.
+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*}