Question

An 18-gauge copper wire (diameter 1.02 mm) carries a current

with a current density of $\mathbf{3}\mathbf{.}\mathbf{2}\mathbf{\times }{\mathbf{10}}^{\mathbf{6}}\frac{\mathbf{A}}{{\mathbf{m}}^{\mathbf{2}}}$. The density of free electrons for

copper is$\mathbf{8}\mathbf{.}\mathbf{5}\mathbf{\times }{\mathbf{10}}^{\mathbf{28}}$electrons per cubic meter. Calculate (a) the current in

the wire and (b) the drift velocity of electrons in the wire.

Short answer

a) The current in the wire is $/=2.61\mathrm{A}$.

b) The drift velocity of the electrons in the wire is ${\mathrm{V}}_{\mathrm{d}}=2.353\times {10}^{-4}\frac{\mathrm{m}}{\mathrm{s}}$

Step-by-step solution

Formula of current and velocity drift.

Consider the area of cross section is

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

Here, dis the diameter.

Consider the expression for the current is:

${\mathbf{V}}_{\mathbf{d}}\mathbf{=}\frac{\mathbf{J}}{\mathbf{nq}{\displaystyle \frac{{\mathbf{\pi d}}^{\mathbf{2}}}{\mathbf{4}}}}$ ….. (1)

Here, nis the charge concentration and q is the charge.

Consider the expression for the current density:

$$\begin{gathered} \mathbf{J}\mathbf{=}\frac{\mathbf{I}}{\mathbf{A}} \\ \mathbf{I}\mathbf{=}\mathbf{JA} \end{gathered}$$

Rewrite as:

$\mathbf{I}\mathbf{=}\mathbf{J}\frac{{\mathbf{\pi d}}^{\mathbf{2}}}{\mathbf{4}}$ …… (2)

Determine the value for the current.

(a)

Given is the density of current in the wire as $3.2\times {10}^6\frac{\mathrm{A}}{{\mathrm{m}}^2}$

Diameter of the wire as $1.02\times {10}^{-3}\mathrm{m}$

Substitute the values in the equation (2) and solve as,

$$\begin{gathered} I=\left(3.2\times {10}^6\frac{\mathrm{A}}{{\mathrm{m}}^2}\right)\left(\frac{\mathrm{\pi }{\left(1.02\times {10}^{-3}\mathrm{m}\right)}^2}{4}\right) \\ =2.61\mathrm{A} \end{gathered}$$

Calculation of velocity drift.

(b)

Given the number of electrons as $8.5\times {10}^{28}$and the charge of a electron is

$1.6\times {10}^{-19}$coulombs.

Substitute the values in the equation (1) and solve as,

$$\begin{gathered} {\mathrm{V}}_{\mathrm{d}}=\frac{\left(3.2\times {10}^6{\displaystyle \frac{\mathrm{A}}{{\mathrm{m}}^2}}\right)}{8.5\times {10}^{28}\times 1.6\times {10}^{-19}} \\ =2.353\times {10}^{-4}\frac{\mathrm{m}}{\mathrm{s}} \end{gathered}$$