Chapter 26, Problem 20SP

To determine

The number of electrons per second passing through a section of the current carrying wire having a current of 0.70 A.

Answer to Problem 20SP

Solution:

$4.4\times {10}^{18}\text{ electrons/s}$

Explanation of Solution

Given data:

The current passing through the wire is $0.70\text{ A}$.

Formula used:

Write the expression for current passing through the wire:

$I=\frac{q}{t}$

Here, $I$ is the current, $q$ is the magnitude of charge, and $t$ is the time.

Write the expression for number of electrons:

$n=\frac{q}{q_e}$

Here, $n$ is the number of electrons and $q_e$ is the magnitude of charge of one electron, thatis, $1.6\times {10}^{-19}\text{ C}$.

Explanation:

Recall the expression for currentpassing through the wire:

$I=\frac{q}{t}$

Substitute $0.70\text{ A}$ for $I$ and $1\text{ s}$ for $t$

$\begin{array}{c}0.70\text{ A}=\frac{q}{1\text{ s}}\\ q=0.70\text{ C}\end{array}$

Recall the expression for number of electrons:

$n=\frac{q}{q_e}$

Substitute $0.7\text{ C}$ for $q$ and $1.6\times {10}^{-19}\text{ C}$ for $q_e$

$\begin{array}{c}n=\frac{0.7\text{ C}}{1.6\times {10}^{-19}\text{ C}}\\ =4.375\times {10}^{18}\text{ electrons/s}\\ \approx 4.4\times {10}^{18}\text{ electrons/s}\end{array}$

Conclusion:

Therefore, the number of electrons passing through the given section is $4.4\times {10}^{18}\text{ electrons/s}$.