91Ó°ÊÓ

Consider the expression for the current is:

$\mathbf{I}\mathbf{=}\frac{\mathbf{△}\mathbf{Q}}{\mathbf{△}\mathbf{t}}$ ….. (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 drift velocity as:

${\mathbf{V}}_{\mathbf{d}}\mathbf{=}\frac{\mathbf{I}}{\mathbf{neA}}$ ….. (2)

Here, n is the free electron density, e is the charge of the electron and A is the

cross sectional area.

(a)

Substitute the values in the equation (1) and solve.

$$\begin{gathered} \mathrm{I}=\frac{420\mathrm{C}}{4800\mathrm{s}} \\ =8.75×{10}^{-2}\mathrm{A} \end{gathered}$$

Therefore, the value for the current is $8.75×{10}^{-2}\mathrm{A}$.

(b)

The charge of an electron is $1.6×{10}^{-19}\mathrm{C}$

The number of electrons is given as$5.8×{10}^{28}$

The radius of the wire is$0.0013\mathrm{m}$.

Substitute the values in the equation (2) and solve.

$$\begin{gathered} {\mathrm{V}}_{\mathrm{d}}=\frac{8.75×{10}^{-2}\mathrm{A}}{\mathrm{π}\left(5.8×{10}^{28}\right)\left(1.6×{10}^{-19}\mathrm{C}\right){\left(0.0013\mathrm{m}\right)}^2} \\ =1.77×{10}^{-6}\frac{\mathrm{m}}{\mathrm{s}} \end{gathered}$$

Therefore, the drift velocity is $1.77×{10}^{-6}\frac{\mathrm{m}}{\mathrm{s}}$.