Question

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?

Short answer

a) The time taken by an electron to travel the wire with 2.0 5mm as diameter is

109.5 mins

b) The time taken by an electron to travel the wire with 4.12mm as diameter is

442.32 mins

c) The diameter is inversely proportional with current and drift velocity so it

effects the drift velocity inversely.

Step-by-step solution

Formula of velocity drift and time taken.

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.

Calculation of time taken by an electron to travel the wire with 2.05mm as diameter.

(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,

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

Calculation of time taken by an electron to travel the wire with 4.12mm as diameter..

(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,

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

Conclusion of changing the diameter of the wire.

(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.