91Ó°ÊÓ

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,$â–³\mathrm{Q}$is the change in charge and$â–³\mathrm{t}$is the change in time

Consider the expression for the current density:

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

Rewrite as:

$\mathbf{J}\mathbf{=}\frac{\mathbf{I}}{{\mathbf{Ï€°ù}}^{\mathbf{2}}}$ …… (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.

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

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

(b)

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

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

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

(c)

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

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

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

(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.