A 5.00-A current runs through a 12-gauge copper wire (diameter

2.05 mm) and through a light bulb. Copper has$\mathbf{8}\mathbf{.}\mathbf{5}\mathbf{\times }{\mathbf{10}}^{\mathbf{8}}$free electrons per

cubic meter. (a) How many electrons pass through the light bulb each

second? (b) What is the current density in the wire? (c) At what speed does

a typical electron pass by any given point in the wire? (d) If you were to use

wire of twice the diameter, which of the above answers would change?

Would they increase or decrease?

Consider the expression for the current is:

$$\begin{gathered} \mathbf{I}\mathbf{=}\frac{\mathbf{△}\mathbf{Q}}{\mathbf{△}\mathbf{t}} \\ \mathbf{I}\mathbf{=}\frac{\mathbf{n}\mathbf{e}}{\mathbf{t}} \end{gathered}$$ ….. (1)

Here,$△\mathrm{Q}$is the change in charge and$△\mathrm{t}$is the change in time

Consider the expression for the current density:

$\mathbf{J}\mathbf{=}\frac{\mathbf{I}}{\mathbf{A}}$

Rewrite as:

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

Here, A is the cross sectional area.

Consider the speed of the electron is:

${\mathbf{V}}_{\mathbf{d}}\mathbf{=}\frac{\mathbf{J}}{\mathbf{ne}}$ …… (3)

Here, J is the density and e is the charge of electron.

(a)

Substitute the values in the equation (1) and determine the total number of

electrons.

$$\begin{gathered} \frac{\mathrm{n}}{\mathrm{t}}=\frac{\mathrm{I}}{\mathrm{e}} \\ \frac{\mathrm{n}}{\mathrm{t}}=\frac{5\mathrm{A}}{\left(1.6\times {10}^{-19}\mathrm{C}\right)} \\ \mathrm{n}=3.125\times {10}^{19}\frac{\mathrm{e}}{\mathrm{s}} \end{gathered}$$

Therefore, the total number of electrons are $3.125\times {10}^{19}\frac{\mathrm{e}}{\mathrm{s}}$.

(b)

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

$$\begin{gathered} \mathrm{J}=\frac{5\mathrm{A}}{3.14{\left(1.025\times {10}^{-3}\mathrm{m}\right)}^2} \\ =\frac{5\mathrm{A}}{3.301\times {10}^{-6}{\mathrm{m}}^2} \\ =1.515\times {10}^6\frac{\mathrm{A}}{{\mathrm{m}}^2} \end{gathered}$$

Therefore, the value of the current density is $1.515\times {10}^6\frac{\mathrm{A}}{{\mathrm{m}}^2}$.

(c)

Substitute the values in equation (3) and solve as,

$$\begin{gathered} {\mathrm{V}}_{\mathrm{d}}=\frac{\left(1.515\times {10}^6\right){\displaystyle \frac{\mathrm{A}}{{\mathrm{m}}^2}}}{\overline{)\left(8.5\times {10}^{28}\right)}\left(-1.6\times {10}^{-19}\right)\overline{)\mathrm{C}}} \\ =1.114\times {10}^{-4}\frac{\mathrm{m}}{\mathrm{s}} \end{gathered}$$

Therefore, the drift velocity is $1.114\times {10}^{-4}\frac{\mathrm{m}}{\mathrm{s}}$.

(d)

Increasing the diameter twice would increase the cross sectional area of the

wire.Since both current density and speed of the electron is inversely

proportional to the area of the wire, therefore both current density and speed of

the electrons will decrease.