Answer:
The current in the wire is 0.0875 A and the drift velocity of the electrons in the wire is 1.77x10^{-6} m/s.
Explanation:
Facts that you know from the problem statement:
Using the diameter of the wire the radius is r=d/2. r=2.6/2= 1.3 mm. It is useful to have the time in seconds so t=80 min = 4800 s. You know the charge Q= 420 C. The concentration of electrons in the wire is n = 5.8*10^{28}.
The current in the wire is the net charge that flows through the area in a second, the units are amperes (A), and it is given by:
[tex]I=\frac{Q}{t}[/tex]
Replace the values:
[tex]I=\frac{420C}{4800s}=0.0875 A[/tex]
The drift velocity in the electrons in the wire (V) is given by the equation:
[tex]V=\frac{I}{n|q|(pi)r^2}[/tex]
Replace the values that you have and the charge of an electron taht is equal to 1.6*10^{-19}
[tex]V=\frac{0.0875}{(5.8*10^{28})(1.6*10^{-19})0.0013^2(pi)}[/tex]
V=1.77x10^{-6} m/s
