A silver wire 2.6 mm in diameter transfers a charge of 420 C in 80

min. Silver containsfree electrons per cubic meter. (a) What is the

current in the wire? (b) What is the magnitude of thedrift velocity of the

electrons in the wire?

Consider the expression for the current is:

$\mathbf{I}\mathbf{=}\frac{\mathbf{△}\mathbf{Q}}{\mathbf{△}\mathbf{t}}$ ….. (1)

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

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

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

(b)

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

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

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

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

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

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