Chapter 26, Problem 1P

To determine

The number of years does one electron takes to travel the full length of the cable.

Answer to Problem 1P

The number of years that takes one electron to travel the full length of the cable is $27.0\text{yr}$.

Explanation of Solution

Given information: Length of the high voltage transmission line is $200\text{km}$, diameter of the transmission line is $2.00\text{cm}$, steady current is $1000\text{A}$, the number of electrons per unit volume is $8.50\times {10}^{28}\text{electrons}/{\text{m}}^{\text{3}}$.

Write the expression for the area of cross section of a conductor.

$A=\frac{\pi }{4}{d}^2$

Here,

$A$ is the area of cross section of a conductor.

$d$ is the diameter of the transmission line.

Substitute $2.00\text{cm}$ for $d$ in equation (1) to find $A$,

$\begin{array}{c}A=\frac{\pi }{4}{\left(2.00\text{cm}\times \frac{1\text{m}}{100\text{cm}}\right)}^2\\ =3.14\times {10}^{-4}{\text{m}}^{\text{2}}\end{array}$

Thus, the area of cross section of a conductor is $3.14\times {10}^{-4}{\text{m}}^{\text{2}}$.

Formula to calculate the drift velocity of a conductor.

$\begin{array}{c}I=neA{V}_d\\ V_d=\frac{I}{neA}\end{array}$ (2)

Here,

$V_d$ is the drift velocity of a conductor.

$n$ is the number of electrons per unit volume.

$I$ is the steady current.

$e$ is the charge of one electron.

The value of charge of one electron is $1.6\times {10}^{-19}\text{C}$.

Substitute $1000\text{A}$ for $I$, $8.50\times {10}^{28}\text{electrons}/{\text{m}}^{\text{3}}$ for $n$, $1.6\times {10}^{-19}\text{C}$ for $e$, $3.14\times {10}^{-4}{\text{m}}^{\text{2}}$ for $A$ in equation (2) to find $V_d$,

$\begin{array}{c}{V}_d=\frac{1000\text{A}}{\left(8.50\times {10}^{28}\text{electrons}/{\text{m}}^{\text{3}}\right)\times \left(1.6\times {10}^{-19}\text{C}\right)\times \left(3.14\times {10}^{-4}{\text{m}}^{\text{2}}\right)}\\ =2.34\times {10}^{-4}\text{m}/\text{s}\end{array}$

Thus, the drift velocity of a conductor is $2.34\times {10}^{-4}\text{m}/\text{s}$.

Formula to calculate the number of years that takes one electron to travel the full length of the cable.

$T=\frac{D}{V_d}$ (3)

Here,

$T$ is the number of years that takes one electron to travel the full length of the cable.

$D$ is the length of the high voltage transmission line.

Substitute $200\text{km}$ for $D$, $2.34\times {10}^{-4}\text{m}/\text{s}$ for $V_d$ in equation (3) to find $T$,

$\begin{array}{c}T=\frac{200\text{km}\times \frac{1000\text{m}}{1\text{km}}}{2.34\times {10}^{-4}\text{m}/\text{s}}\\ =8.54\times {10}^8\text{s}\times \frac{1\text{yr}}{\left(365\times 24\times 60\times 60\right)\text{s}}\\ =27.0\text{yr}\end{array}$

Thus, the number of years that takes one electron to travel the full length of the cable is $27.0\text{yr}$.

Conclusion:

Therefore, the number of years that takes one electron to travel the full length of the cable is $27.0\text{yr}$.