Answer
a) The current in amps is: $I =8.75\times10^{-2}A$
b) The drift velocity is $v_{d}$ $= 1.77\times10^{-6} m/s^{2}$
Work Step by Step
a) $I=\frac{Q}{t}$
$I = \frac{420 C}{4800 s}$
$I =8.75\times10^{-2}A$
b) $I = qnv_{d}A$
Rearranging for the drift velocity, yields the equation: $v_{d} = \frac{I}{qnA}$.
Since the area is : $A =πr^{2}$, this can be substituted for the area in the equation.
Therefore, the drift velocity is: $v_{d} =\frac{8.75\times10^{-2}A}{5.8\times10^{28}\timesπ\times(0.0013m)^{2}\times(1.6\times10^{-19}C)}$
$v_{d}$ $= 1.77\times10^{-6} m/s^{2}$
