1) Current in the wire: 0.0875 A
The current in the wire is given by:
[tex]I=\frac{Q}{t}[/tex]
where
Q is the charge passing a given point in the conductor
t is the time elapsed
In this problem, we have
Q = 420 C is the total charge passing through a given point in a time of
t = 80 min = 4800 s
So, the current is
[tex]I=\frac{420 C}{4800 s}=0.0875 A[/tex]
2) Drift velocity of the electrons: [tex]1.78\cdot 10^{-6} m/s[/tex]
The drift velocity of the electrons in the wire is given by:
[tex]u = \frac{I}{nAq}[/tex]
where
I = 0.0875 A is the current
[tex]n=5.8\cdot 10^{28}[/tex] is the number of free electrons per cubic meter
A is the cross-sectional area
[tex]q=1.6\cdot 10^{-19} C[/tex] is the charge of one electron
The radius of the wire is
[tex]r=\frac{d}{2}=\frac{2.6 mm}{2}=1.3 mm=0.0013 m[/tex]
So the cross-sectional area is
[tex]A=\pi r^2=\pi (0.0013 m)^2=5.31\cdot 10^{-6} m^2[/tex]
So, the drift velocity is
[tex]u = \frac{(0.0875 A)}{(5.8\cdot 10^{28})(5.31\cdot 10^{-6})(1.6\cdot 10^{-19}C)}=1.78\cdot 10^{-6} m/s[/tex]
