Chapter 20, Problem 1FC

To determine

The current $I$ in the wire.

Answer to Problem 1FC

The current $I$ in the wire is $\underset{\_}{1.5\text{A}}$.

Explanation of Solution

Given data:

Total number of electrons (N) pass a certain point in a wire is $1.9\times {10}^{19}$

Time $\left(\Delta t\right)$ is 2.0 s.

Formula used:

Consider the expression for number of electrons $\left(N\right)$.

$N=\frac{\Delta q}{e}$ (1)

Here,

$\Delta q$ is the magnitude of the charge, and

e is the electric charge, which is known as $1.60\times {10}^{-19}\text{C}$.

Consider the expression for current, if the rate is constant.

$I=\frac{\Delta q}{\Delta t}$ (2)

Here,

$I$ is the current, and

$\Delta t$ is the time.

Explanation:

Rearrange equation (1).

$\Delta q=eN$

Substitute $1.60\times {10}^{-19}\text{C}$ for $e$ and $1.9\times {10}^{19}$ for $N$,

$\begin{array}{c}\Delta q=\left(1.60\times {10}^{-19}\text{C}\right)\left(1.9\times {10}^{19}\right)\\ =3.04\text{C}\end{array}$

Substitute $3.04\text{C}$ for $\Delta q$ and $2.0\text{s}$ for $\Delta t$ in equation (2),

$\begin{array}{c}I=\frac{3.04\text{C}}{2.0\text{s}}\\ =1.52\text{A}\\ \cong 1.5\text{A}\end{array}$

Conclusion:

Thus, the current $I$ in the wire is $\underset{\_}{1.5\text{A}}$.