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--- 184_notes:examples:week6_drift_speed [2017/09/26 15:48] – [Example: Drift Speed in Different Types of Wires] tallpaul
+++ 184_notes:examples:week6_drift_speed [2021/06/08 00:49] (current) – schram45
@@ Line 1: / Line 1: @@
+[[184_notes:current|Return to current in wires]]
 =====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}}
+[{{ 184_notes:6_n_table.jpg?700 |Properties of Metals}}]
 ===Facts===
@@ Line 8: / Line 10: @@
   * The zinc wire has  $I=5 \text{ A}$ ,  $r = 0.1 \text{ mm}$ .
   * The charge of an electron is  $q=-1.6\cdot 10^{-19} \text{ C}$ .
+  * Electron density of copper is  $n_{\text{Cu}}=8.47\cdot 10^{22} \text{ cm}^{-3}$ .
+  * Electron density of zinc is  $n_{\text{Zn}}=13.2\cdot 10^{22} \text{ cm}^{-3}$ .
+  * Electron current as  $i=nAv_{avg}$ .
+  * Current is  $I=|q|i$ .
+    * Units of current is charge per second. Electron current is electrons per second. We multiply by  $q$  (the electron charge) to get charge per second.
-===Lacking===
+===Goal===
-  * Drift speed for both wires.
+  * Find the drift speed for both wires.
-  * Electron charge density for both wires.
-  * Electron current for both wires.
 ===Approximations & Assumptions===
-  * The table is accurate for our wires.
+  * The wires have circular cross-sections. This is typical of real wires and allows us to use the diameter of the wire to calculate the area properly.
-  * The wires have circular cross-sections.
+  * Using the [[184_notes:current|Drude model]] for electrons in the wire - the electrons are accelerated by the electric field, until they run into a positive nucleus, which reduces the speed back to zero.
-  * The wires do not experience any external electric field.
 ===Representations===
-  * We represent the situation with the following diagram.
+  * We represent electron current as  $i=nAv_{avg}$ .
+  * We represent current as  $I=|q|i$ . 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 use the [[184_notes:current|Drude model]] for electrons in the wire - the electrons are accelerated by the electric field, until they run into a positive nucleus, which reduces the speed back to zero. This is where our definition of drift speed comes from, so it is worth including it in our solution for reference.
-{{ 184_notes:5_dipole_field_lines.png?300 |Dipole Electric Field Lines}}
+There are a lot of variables in this problem, so let's make a plan.
+<WRAP TIP>
+=== Plan ===
+We will do the following steps for each wire.
+  * Find the electron density of each material (see listed above, in Facts).
+  * Find the cross-sectional area of the wire.
+  * Find the electron current of each wire, using the given current.
+  * Use all the new information to find the drift speed.
+</WRAP>
+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.
-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.
+ $v_{avg} = \frac{I}{\pi r^2 n |q|}$
-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,
+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:
-$$\Phi_{left}=-\Phi_{right}$$
+\begin{align*}
+v_{\text{avg, Cu}} = 0.47 \text{ mm/s} &,& v_{\text{avg, Zn}} = 7.5 \text{ mm/s}
+\end{align*}
-Putting it together, we tentatively write:
+Notice that this is actually really slow! Depending on the material, the electron only travels somewhere between 1 mm - 1 cm per second on average.
- $\Phi_{\text{cylinder}}=\Phi_{left}+\Phi_{right}=0$
-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]].
- $\Phi_{\text{total}}=\int \vec{E} \cdot \text{d}\vec{A}=\frac{Q_{\text{enclosed}}}{\epsilon_0}$
-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.