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--- 184_notes:examples:week6_drift_speed [2017/09/26 15:51] – [Example: Drift Speed in Different Types of Wires] tallpaul
+++ 184_notes:examples:week6_drift_speed [2017/09/26 16:05] – [Example: Drift Speed in Different Types of Wires] tallpaul
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 =====Example: Drift Speed in Different Types of Wires=====
-Suppose you have a two wires. Each has a current of $5 \text{ A}$. One is made of copper (Cu) and has radius $0.5 \text{ mm}$. The other is made of zinc (Zn) and has radius $0.1 \text{ mm}$. What is the drift speed of electrons in each wire? You may want to consult the table below.
+Suppose you have a two wires. Each has a current of $5 \text{ A}$. One is made of copper (Cu) and has radius $0.5 \text{ mm}$. The other is made of zinc (Zn) and has radius $0.1 \text{ mm}$. What are the drift speeds of electrons in each wire? You may want to consult the table below.
 {{ 184_notes:6_n_table.jpg?800 |Properties of Metals}}
@@ 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====
-First, notice that we probably do not want to do any calculations here, since the it will not be fun to take a dot-product of the dipole's electric field and the area-vector, and it will get very messy very quickly when we start integrating over the surface of the cylinder. Instead, we evaluate the situation more qualitatively. Consider the electric field vectors of the dipole near the surface of the cylinder:
+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 $n_{\text{Zn}}=13.2\cdot 10^{22} \text{ cm}^{-3}$.
-{{ 184_notes:5_dipole_field_lines.png?300 |Dipole Electric Field Lines}}
-Notice that the vectors near the positive charge are leaving the cylinder, and the vectors near the negative charge are entering. Not only this, but they are mirror images of each other. Wherever an electric field vector points out of the cylinder on the right side, there is another electric field vector on the left that is pointing into the cylinder at the same angle. These mirror image vectors also have the same magnitude, though it is a little tougher to visualize.
+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 could write this as a comparison between the left and right side of the cylinder. The $\text{d}\vec{A}$-vectors are mirrored for left vs. right, and the $\vec{E}$-vectors are mirrored, but opposite $\left(\vec{E}_{left}=-\vec{E}_{right}\right)$. We tentatively write the equality,
+We are given current, and we can solve for electron current using the charge of an electron: $i = \frac{I}{q}$.
-$$\Phi_{left}=-\Phi_{right}$$
-Putting it together, we tentatively write:
+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.
-$$\Phi_{\text{cylinder}}=\Phi_{left}+\Phi_{right}=0$$
-We gain more confidence when we read the [[184_notes:gauss_ex|next section of notes]], where we define "Gauss' Law". This law states that the total flux through a close surface is the amount of charge enclosed divided by $\epsilon_0$, the [[https://en.wikipedia.org/wiki/Vacuum_permittivity|permittivity of free space]].
+$$v_{avg} = \frac{I}{\pi r^2 n q}$$
-$$\Phi_{\text{total}}=\int \vec{E} \cdot \text{d}\vec{A}=\frac{Q_{\text{enclosed}}}{\epsilon_0}$$
-Since the total charge of the dipole is $0$, then indeed the charge enclosed is $0$, and we were correct with our reasoning about the electric field and flux above.
+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*}