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, \[\begin{align*} J&=\frac{I}{A}\\ &=\frac{1.67}{\pi r^2}\\ &=\frac{1.67}{\pi \times (0.51 \times 10^{-3}) ^2}\\ &=2043742.829\\ & \approx 2.04 \times 10^{6} \hspace{0.1cm} Am^{-2}\\ \end{align*}\] Also, \[\begin{align*} J&=v_d e n\\ v_d&=\frac{J}{ne}\\ &=\frac{2.04 \times 10^6}{1.6 \times 10^{-19} \times 8.5 \times 10^{28}}\\ \therefore v_d&=1.5 \times 10^{-4}\hspace{0.1cm} m/s\\ \end{align*}\] Hence, the current density is \(2.04\times 10^6\) A/m² and drift velocity is \(1.5 \times 10^{-4}\) m/s.