This problem uses the relationships among current I, current density J, and drift speed vd. We are given the total of electrons that pass through the wire in t = 3s and the area A, so we use the following equation to to find vd, from J and the known electron density n, so:
[tex]v_{d} = \frac{J}{n\left | q \right |}[/tex]
The current I is any motion of charge from one region to another, so this is given by:
[tex]I = \frac{\Delta Q}{\Delta t} = \frac{9.4x1018electrons}{3s} = 3189.73(A)[/tex]
The magnitude of the current density is:
[tex]J = \frac{I}{A} = \frac{3189.73}{2x10^{-6}} = 1594.86(A/m^{2})[/tex]
Being:
[tex]A=2mm^{2} = 2x10^{-6}m^{2}[/tex]
Finally, for the drift velocity magnitude vd, we find:
[tex]v_{d} = \frac{1594.86}{5.8x1028\left |1.60x10^{-19}|\right } = 1.67x10^{18}(m/s)[/tex]
Notice: The current I is very high for this wire. The given values of the variables are a little bit odd
