A copper wire has a diameter of 1.02 mm and carries a constant current of 1.67 A. If the density of free currents in copper is 8.5 × 1028 m3, calculate the current density and the drift velocity of the electrons.

A copper wire has a diameter of 1.02 mm and carries a constant current of 1.67 A. If the density of free currents in copper is 8.5 $\times$ 10²⁸ m³, calculate the current density and the drift velocity of the electrons.
Solution:
Given,
diameter , $d$=1.02 mm
radius, $r$ = $\frac{1.02}{2}$ = 0.51 $\times$ 10${}^{-3}$ m
current, $I$ = 1.67 A
number density of electrons, $n$ = 8.5 $\times$ 10${}^{28}$ / m${}^3$
current density, $J$ = ?
drift velocity, $v_d$ = ?

Now, $$\begin{array}{rl}J& =\frac{I}{A}\\ & =\frac{1.67}{\pi r^2}\\ & =\frac{1.67}{\pi \times (0.51\times {10}^{-3}{)}^2}\\ & =2043742.829\\ & \approx 2.04\times {10}^{6}A{m}^{-2}\end{array}$$ Also, $$\begin{array}{rl}J& =v_{d}en\\ v_d& =\frac{J}{ne}\\ & =\frac{2.04\times {10}^6}{1.6\times {10}^{-19}\times 8.5\times {10}^{28}}\\ \therefore v_d& =1.5\times {10}^{-4}m/s\end{array}$$ Hence, the current density is $2.04\times {10}^6$ A/m² and drift velocity is $1.5\times {10}^{-4}$ m/s.