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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 [2017/09/26 16:05] – [Solution] tallpaul | ||
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* Electron charge density for both wires. | * Electron charge density for both wires. | ||
* Electron current for both wires. | * Electron current for both wires. | ||
+ | * Cross-sectional area for both wires. | ||
===Approximations & Assumptions=== | ===Approximations & Assumptions=== | ||
Line 21: | Line 22: | ||
===Representations=== | ===Representations=== | ||
* We represent electron current as $i=nAv_{avg}$. | * We represent electron current as $i=nAv_{avg}$. | ||
- | * We represent current as $I=qi$. | + | * 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==== | ====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 | + | We can look up electron density $n$ in the table. It is labeled as " |
- | {{ 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 | + | 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 | + | We are given current, and we can solve for electron current using the charge of an electron: |
- | $$\Phi_{left}=-\Phi_{right}$$ | + | |
- | 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: | + | $$v_{avg} = \frac{I}{\pi r^2 n q}$$ |
- | $$\Phi_{\text{total}}=\int \vec{E} \cdot \text{d}\vec{A}=\frac{Q_{\text{enclosed}}}{\epsilon_0}$$ | + | |
- | 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 |
+ | \begin{align*} | ||
+ | v_{\text{avg, Cu}} = 0.47 \text{ mm/s} &,& v_{\text{avg, Zn}} = 7.5 \text{ | ||
+ | \end{align*} |