Chapter 28, Problem 81PQ

To determine

The time taken for one electron to travel from the generating station to the switching station in the wire.

Answer to Problem 81PQ

The time taken for one electron to travel from the generating station to the switching station in the wire is $\underset{\_}{1.19\times {10}^9\text{s}}$.

Explanation of Solution

Given that the distance between the generating station and the switching station is $140.0\text{km}$, the transmitted current is $850.0\text{A}$, the cross-sectional area of the wire is $750.0{\text{mm}}^2$, and the density of free electrons in the wire is $6.022\times {10}^{28}{\text{m}}^{-3}$.

Write the expression for the drift velocity of electron in terms of the current density.

  $v_d=\frac{J}{ne}$                                                                                                                 (I)

Here, $v_d$ is the drift velocity, $J$ is the current density, $n$ is the number density of electrons, and $e$ is the charge of electron.

Write the expression for the current density in the wire.

  $J=\frac{I}{A}$                                                                                                                   (II)

Here, $I$ is the current, and $A$ is the cross-sectional area of the wire.

Use equation (II) in (I).

  $v_d=\frac{I}{neA}$                                                                                                             (III)

Write the expression for the time taken for the electron to travel a particular distance through the wire.

  $\Delta t=\frac{d}{v_d}$                                                                                                                (IV)

Here, $d$ is the distance, and $\Delta t$ is the time taken to cover the distance.

Use equation (III) in (IV).

  $\Delta t=\frac{neAd}{I}$                                                                                                            (V)

Conclusion:

Substitute $6.022\times {10}^{28}{\text{m}}^{-3}$ for $n$, $1.60\times {10}^{-19}\text{C}$ for $e$, $750.0{\text{mm}}^2$ for $A$, $140.0\text{km}$ for $d$, and $850.0\text{A}$ for $I$ in equation (V) to find $\Delta t$.

  $\begin{array}{c}\Delta t=\frac{\left(6.022\times {10}^{28}{\text{m}}^{-3}\right)\left(1.60\times {10}^{-19}\text{C}\right)\left(750.0{\text{mm}}^2\right)\left(140.0\text{km}\right)}{850.0\text{A}}\\ =\frac{\left(6.022\times {10}^{28}{\text{m}}^{-3}\right)\left(1.60\times {10}^{-19}\text{C}\right)\left(750.0{\text{mm}}^2\times \frac{1{\text{m}}^2}{1\times {10}^6{\text{mm}}^2}\right)\left(140.0\text{km}\times \frac{1000\text{m}}{1\text{km}}\right)}{850.0\text{A}}\\ =1.19\times {10}^9\text{s}\end{array}$

Therefore, the time taken for one electron to travel from the generating station to the switching station in the wire is $\underset{\_}{1.19\times {10}^9\text{s}}$.