91Ó°ÊÓ

a) Measured current,$i=1.2×{10}^{-10}A$

b) Diameter of copper wire,$d=2.5mmor2.5×{10}^{-3}m$

c) Electron concentration,$n=8.49×{10}^{28}{m}^{-3}$

The term "current density" refers to the quantity of electric current moving across a certain cross-section. We have to use the formula of current density to find the current density and then we have to use the relation between current density and drift velocity, to find the electron drift speed.

Formulae:

The current density due to the current flowing through the area,$J=\frac{i}{A}$ ...(i)

The current density in relation to the drift velocity,$J=ne{V}_d$ ...(ii)

The cross-sectional area of the circle,$A={\mathrm{Ï€°ù}}^2$ ...(iii)

where, n is the particle concentration, i is the current, and $V_d$is the drift velocity, r is the radius of wire and e is the charge of electron

= $1.6×{10}^{-9}C$

Substituting the given data in the formula of the area value from equation (iii) in equation (i), we can get the magnitude value of the current density as follows:

$$\begin{gathered} J=\frac{i}{{\mathrm{Ï€°ù}}^2}\left(âˆ\mu radiusinrelationtodiameter,r=\frac{d}{2}\right) \\ =\frac{4i}{\mathrm{Ï€}{d}^2} \\ =\frac{4\left(1.2×{10}^{-10}A\right)}{\mathrm{Ï€}{\left(2.5×{10}^{-3}\mathrm{m}\right)}^2} \\ =2.44×{10}^{-5}A/m^2 \\ ≈2.44×{10}^{-5}A/m^2 \end{gathered}$$

Hence, the value of the current density is$2.44×{10}^{-5}A/m^2$ .

Using the given data in equation (ii), we can get the required drift velocity of the electron as follows:

$$\begin{gathered} V_d=\frac{J}{ne} \\ =\frac{2.44×{10}^{-5}A/m^2}{\left(8.49×{10}^{28}{m}^{-3}\right)\left(1.6×{10}^{-19}C\right)} \\ =1.796×{10}^{-15}m/s \\ ≈1.796×{10}^{-15}m/s \end{gathered}$$

Hence, the value of the drift velocity is$1.796×{10}^{-15}m/s$ .