184_notes:examples:week6_drift_speed

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184_notes:examples:week6_drift_speed [2017/09/26 15:51] – [Solution] tallpaul184_notes:examples:week6_drift_speed [2021/06/08 00:49] (current) schram45
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 +[[184_notes:current|Return to current in wires]]
 +
 =====Example: Drift Speed in Different Types of 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=== ===Facts===
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   * The zinc wire has $I=5 \text{ A}$, $r = 0.1 \text{ mm}$.   * The zinc wire has $I=5 \text{ A}$, $r = 0.1 \text{ mm}$.
   * The charge of an electron is $q=-1.6\cdot 10^{-19} \text{ C}$.   * 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=== ===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=== ===Representations===
   * We represent electron current as $i=nAv_{avg}$.   * We represent electron current as $i=nAv_{avg}$.
-  * 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.+  * 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==== ====Solution====
-We can look up electron+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. 
 + 
 +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. 
 + 
 +$$v_{avg} = \frac{I}{\pi r^2 n |q|}$$ 
 + 
 +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*} 
 + 
 +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.
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