Question 26.74: The current density in a wire is uniform and has magnitude 2......
The current density in a wire is uniform and has magnitude 2.0 × 10^{{6}} A/m², the wire’s length is 5.0 m, and the density of conduction electrons is 8.49 × 10^{28}\,\mathrm{m}^{-3}. How long does an electron take (on the average) to travel the length of the wire?
We find the drift speed from Eq. 26-7:
\vec{J}=(n e) \vec{\nu}_d (26-7)
\nu_d=\frac{|\vec{J}|}{n e}=\frac{2.0 \times 10^6 \mathrm{~A} / \mathrm{m}^2}{\left(8.49 \times 10^{28} / \mathrm{m}^3\right)\left(1.6 \times 10^{-19} \mathrm{C}\right)}=1.47 \times 10^{-4} \mathrm{~m} / \mathrm{s}
At this (average) rate, the time required to travel L = 5.0 m is
t=\frac{L}{\nu_d}=\frac{5.0 \mathrm{~m}}{1.47 \times 10^{-4} \mathrm{~m} / \mathrm{s}}=3.4 \times 10^4 \mathrm{~s} .