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• Notes

$\begin{array}{l}{\text { Copper has } 8.5 \times 10^{28} \text { free electrons per cubic meter. } A} \\ {71.0-\mathrm{cm} \text { length of } 12 \text { -gauge copper wire that is } 2.05 \mathrm{mm} \text { in diame- }} \\ {\text { ter carries } 4.85 \mathrm{A} \text { of current. (a) How much time does it take for an }} \\ {\text { electron to travel the length of the wire? (b) Repeat part (a) for }} \\ {6 \text { -gauge copper wire (diameter } 4.12 \mathrm{mm} \text { ) of the same length that }} \\ {\text { carries the same current. (c) Generally speaking, how does changing }} \\ {\text { the diameter of a wire that carries a given amount of current affect }} \\ {\text { the drift velocity of the electrons in the wire? }}\end{array}$

$J = \frac{1}{A}$

$J=n|q| V_{d}$

$V_{d}=\frac{L}{t}$

(a) $J = \frac{1}{A} = \frac{1}{\pi r^{2}}=\frac{1}{\pi\left(1 \cdot 025 * 10^{-3}\right)^{2}}$

$=1.469 * 10^{6} \mathrm{A} / \mathrm{m}^{2}$

$V_{d}= \frac{J}{n|q|}$$=\frac{1.469 * 10^{6}}{8.5 * 10^{28} * 1.602*10^{-19}}$

$=1.074 * 10^{-4} \mathrm{m/s}$

$t=\frac{L}{V_{d}}=\frac{0.71}{1.074 * 10^{-4}}$

$=6.58*10^{3} 5=110 \mathrm{min}$

(b) $V_{d}=\frac{I}{\pi r_{1} n|q|}$

$t=\frac{L}{V_{d}} \longrightarrow t=$$\frac{\pi r^2 n |q| L}{I}=$

$=\frac{\pi\left(2 \cdot 06*10^{-3}\right)^{2} 8.5*10^{28}*1.602*10^{-19}*0.71}{4.85 }=2.66*10^{4}5÷60=443 min$

$V_{d}=\frac{I}{\pi r^{2} n|q|}$

$\rightarrow \quad t=\frac{L}{V_ d}$$= \frac{\pi r^{2} n |q| L}{I}$

$t=\left(6.58 * 10^{3}\right)\left(\frac{4.12}{2.05}\right)^{2}$

$=2.66*10^{4}$