Question

The electron mobility in copper is $\mathbf{4}\mathbf{.}\mathbf{5}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{3}}\mathbf{}\left(m/s\right)\mathbf{/}\left(N/C\right)$, and the density of mobile electrons is$\mathbf{8}\mathbf{.}\mathbf{4}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{28}}\mathbf{}{\mathbf{m}}^{\mathbf{-}\mathbf{3}}$ . In a copper wire $\mathbf{1}\mathbf{}\mathbf{mm}$ in diameter there is an electric field whose magnitude is $\mathbf{0}\mathbf{.}\mathbf{05}\mathbf{}\mathbf{N}\mathbf{/}\mathbf{C}$. What is the conventional current in this wire? (Note that a small electric field drives a sizable current in copper, which is a good conductor.)

Short answer

The conventional current in the wire is $2.38\mathrm{A}$.

Step-by-step solution

Identification of the given data

The given data can be listed below as,

• The electron mobility in copper wire is, $\mathrm{u}=4.5\times {10}^{-3}\left(\mathrm{m}/\mathrm{s}\right)/\left(\mathrm{N}/\mathrm{C}\right)$.
• The electron density is, $\mathrm{n}=8.4\times {10}^{28}{\mathrm{m}}^3$.
• The diameter of copper wire is, $\mathrm{d}=\left[1\mathrm{mm}\times \left(\frac{{10}^{-3}\mathrm{m}}{1\mathrm{mm}}\right)\right]={10}^{-3}\mathrm{m}$.
• The magnitude of the electric field is, $\mathrm{E}=0.05\mathrm{N}/\mathrm{C}$.

Significance of conventional current in a conducting wire

In this question, the conventional current in a copper wire will be determined by using the values of the electric field, electron mobility, and cross-sectional area of the wire. The relation between the conventional current and the electron mobility is direct linear.

Determination the conventional current in the copper wire

The relation of magnitude of the conventional current in the copper wire is expressed as,

$\begin{array}{rcl}\mathrm{I}& =& \left|\mathrm{q}\right|\mathrm{nA}\left(\mathrm{uE}\right)\\ & =& \left|\mathrm{q}\right|\mathrm{n}\left(\frac{\mathrm{\pi }}{4}\right){\mathrm{d}}^2\left(\mathrm{uE}\right)\end{array}$

Here, $\mathrm{I}$ is the conventional current in the copper wire, $\mathrm{A}$ is the cross-sectional area of the wire and $\mathrm{q}$ is the charge on an electron whose value is $-1.6\times {10}^{-19}\mathrm{C}$.

Substitute all the known values in the above equation.

$\begin{array}{rcl}\mathrm{I}& =& \left|-1.6\times {10}^{-19}\mathrm{C}\right|\left(8.4\times {10}^{28}{\mathrm{m}}^{-3}\right)\left[\frac{\mathrm{\pi }}{4}{\left({10}^{-3}\mathrm{m}\right)}^2\right]\left[\left(4.5\times {10}^{-3}\right)\left(\mathrm{m}/\mathrm{s}\right)\left(\mathrm{N}/\mathrm{C}\right)\left(0.05\mathrm{N}/\mathrm{C}\right)\right]\\ & \approx & 2.38\mathrm{C}/\mathrm{s}\\ & \approx & 2.38\mathrm{C}/\mathrm{s}\times \left(\frac{1\mathrm{A}}{1\mathrm{C}/\mathrm{s}}\right)\\ & \approx & 2.38\mathrm{A}\end{array}$

Thus, the conventional current in the copper wire is $2.38\mathrm{A}$.