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184_notes:examples:week6_drift_speed [2017/09/26 16:05] – [Solution] tallpaul | 184_notes:examples:week6_drift_speed [2017/09/27 15:00] – dmcpadden | ||
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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 wires have circular cross-sections. | * The wires have circular cross-sections. | ||
* The wires do not experience any external electric field. | * The wires do not experience any external electric field. | ||
+ | * 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) 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 density $n$ in the table. It is labeled as " | We can look up electron density $n$ in the table. It is labeled as " | ||
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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: $A=\pi r^2$. | 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 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. | 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. | ||
- | $$v_{avg} = \frac{I}{\pi r^2 n q}$$ | + | $$v_{avg} = \frac{I}{\pi r^2 n |q|}$$ |
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: | 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: | ||
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v_{\text{avg, | v_{\text{avg, | ||
\end{align*} | \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. |