Problem 21.43

21.43.
We need to find the drift velocity. This is given by
I = n q v_d A
$I = n q v_{d} \pi r^{2}$
solving for v_d
$v_{d} = \frac{I}{nq\pi r^{2}}$
$v_{d} = \frac{1000 A}{(8 \times 10^{28} m^{-3})(1,6 \times 10^{-19} C)(\pi)(10^{-2} m)^{2}}$
$v_{d} = 2.49 \times 10^{-4} m/s$
Now, to find the time that it takes for the electrons to get to the end of the cable,
$t = \frac{x}{v}$
$t = \frac{200 \times 10^{3}m}{2.49 \times 10^{-4} m/s}$
$t = 8.04 \times 10^{8} s$
t = 25.5 years.