Copper has $\mathbf{8}\mathbf{.}\mathbf{5}\mathbf{\times }{\mathbf{10}}^{\mathbf{22}}$free electrons per cubic meter. A 71.0-cm

length of 12-gauge copper wire that is 2.05 mm in diameter carries 4.85 A of

current. (a) How much time does it take for an electron to travel the length

of the wire? (b) Repeat part (a) for 6-gauge copper wire (diameter 4.12 mm)

of the same length that carries the same current. (c) Generally speaking,

how does changing the diameter of a wire that carries a given amount of

current affect the drift velocity of the electrons in the wire?

Consider the formula for the drift velocity as:

${\mathbf{V}}_{\mathbf{d}}\mathbf{=}\frac{\mathbf{I}}{\mathbf{neA}}$

Here,n is the free electron density, e is the charge of the electron and A is the

cross sectional area.

Consider the formula for the time taken as:

$\mathbf{t}\mathbf{=}\frac{\mathbf{L}}{{\mathbf{V}}_{\mathbf{d}}}$

Here, L is the length.

(a)

Given with the number of electrons as$8.5\times {10}^{22}$.

The current as 4.85 A.

Solve for the area of cross section as:

$$\begin{gathered} A=\frac{\mathrm{\pi }}{4}{\left(0.00205\right)}^2 \\ =3.301\times {10}^{-6}{\mathrm{m}}^2 \end{gathered}$$

Solve for the drift velocity as,

role="math" localid="1655719544982" $$\begin{gathered} {\mathrm{v}}_{\mathrm{d}}=\frac{\mathrm{I}}{\mathrm{neA}} \\ {\mathrm{v}}_{\mathrm{d}}=\frac{4.85\mathrm{A}}{8.5\times {10}^{28}\times 1.6\times {10}^{-19}\left(3.301\times {10}^{-6}{\mathrm{m}}^2\right)} \\ {\mathrm{v}}_{\mathrm{d}}=1.08\times {10}^{-4}\frac{\mathrm{m}}{\mathrm{s}} \end{gathered}$$

Solve for the time taken as,

$$\begin{gathered} \mathrm{t}=\frac{0.71\mathrm{m}}{1.08\times {10}^{-4}{\displaystyle \frac{\mathrm{m}}{\mathrm{s}}}} \\ \mathrm{t}=109.5\min \end{gathered}$$

(b)

The cross sectional area will be,

$$\begin{gathered} A=\frac{\mathrm{\pi }}{4}{\left(0.00412\right)}^2 \\ =1.333\times {10}^{-5}{\mathrm{m}}^2 \end{gathered}$$

Solve for the drift velocity as,

$$\begin{gathered} {\mathrm{v}}_{\mathrm{d}}=\frac{\mathrm{I}}{\mathrm{neA}} \\ {\mathrm{v}}_{\mathrm{d}}=\frac{4.85\mathrm{A}}{(8.5\times {10}^{28})(1.6\times {10}^{-19}C)\left(1.333\times {10}^{-6}{\mathrm{m}}^2\right)} \\ {\mathrm{v}}_{\mathrm{d}}=2.675\times {10}^{-5}\frac{\mathrm{m}}{\mathrm{s}} \end{gathered}$$

Therefore the time taken will be,

$$\begin{gathered} \mathrm{t}=\frac{0.71\mathrm{m}}{2.675\times {10}^{-5}{\displaystyle \frac{\mathrm{m}}{\mathrm{s}}}} \\ =442.32\min \end{gathered}$$

(c)

From part (a) and (b)

The diameter of a wire is inversely proportional to the current and the velocity

drift.This means an increase in diameter will decrease the current and drift

velocity or vice-versa.

Therefore, the diameter is inversely proportional with current and drift velocity so

it effects the drift velocity inversely.