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184_notes:examples:week6_drift_speed [2017/09/26 15:50] – [Example: Drift Speed in Different Types of Wires] tallpaul | 184_notes:examples:week6_drift_speed [2018/02/03 22:24] – [Solution] tallpaul | ||
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=====Example: | =====Example: | ||
- | 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 |
- | {{ 184_notes: | + | {{ 184_notes: |
===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 | + | |
===Approximations & Assumptions=== | ===Approximations & Assumptions=== | ||
- | * The table is accurate for our wires. | ||
* The wires have circular cross-sections. | * The wires have circular cross-sections. | ||
- | * The wires do not experience any external | + | * Using the [[184_notes: |
===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 | + | * 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==== | ||
- | 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' | + | We can use the [[184_notes: |
- | {{ 184_notes: | + | |
- | 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 | + | There is a lot going on in this problem, so let's make a plan. |
- | We could write this as a comparison between | + | <WRAP TIP> |
- | $$\Phi_{left}=-\Phi_{right}$$ | + | === Plan === |
+ | We will do the following steps for each wire. | ||
+ | * Find the electron density of each material | ||
+ | * 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. | ||
+ | </ | ||
- | Putting it together, we tentatively write: | + | 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: $A=\pi r^2$. |
- | $$\Phi_{\text{cylinder}}=\Phi_{left}+\Phi_{right}=0$$ | + | |
- | We gain more confidence when we read the [[184_notes: | + | We are given current, and we can solve for electron current using the charge of an electron: $i = \frac{I}{|q|}$. |
- | $$\Phi_{\text{total}}=\int \vec{E} \cdot \text{d}\vec{A}=\frac{Q_{\text{enclosed}}}{\epsilon_0}$$ | + | |
- | Since the total charge of the dipole | + | 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 | ||
+ | \begin{align*} | ||
+ | v_{\text{avg, Cu}} = 0.47 \text{ mm/s} &,& v_{\text{avg, Zn}} = 7.5 \text{ | ||
+ | \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. |