91Ó°ÊÓ

Consider the area of cross section is

$A=\frac{{\mathrm{Ï€»å}}^2}{4}$

Here, dis the diameter.

Consider the expression for the current is:

${\mathbf{V}}_{\mathbf{d}}\mathbf{=}\frac{\mathbf{J}}{\mathbf{nq}{\displaystyle \frac{{\mathbf{Ï€»å}}^{\mathbf{2}}}{\mathbf{4}}}}$ ….. (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:

$\mathbf{I}\mathbf{=}\mathbf{J}\frac{{\mathbf{Ï€»å}}^{\mathbf{2}}}{\mathbf{4}}$ …… (2)

(a)

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

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

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

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

(b)

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

$1.6×{10}^{-19}$coulombs.

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

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