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

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

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

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

(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) $$
V_{d}=\frac{I}{\pi r_{1} n|q|}
$$

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

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

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

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

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

$$
=2.66*10^{4}
$$

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