Answer:
14.7 A
Explanation:
The magnitude current flowing in a conductor is given by
[tex]I=\frac{Q}{t}[/tex]
where
Q is the total charge
t is the time interval
The total charge passing through a point in the conductor can be written as
[tex]Q=neAd[/tex]
where
n is the density of free electrons
A is the cross-sectional area of the conductor
[tex]e=1.6\cdot 10^{-19}C[/tex] is the electron charge
d is the length of the conductor
The time interval can be written as
[tex]t=\frac{d}{v_d}[/tex]
where
d is the length of the conductor
[tex]v_d[/tex] is the drift velocity of the electrons
Re-arranging the three equations, we get:
[tex]I=\frac{neAd}{d/v_d}=neAv_d[/tex]
For the copper wire here we have:
[tex]v_d = 0.500 mm/s = 0.5\cdot 10^{-3} m/s[/tex]
[tex]n=8.34\cdot 10^{28} m^{-3}[/tex]
The diameter is 1.676 mm, so the area is
[tex]A=\pi (\frac{d}{2})^2=\pi (\frac{1.676\cdot 10^{-3}}{2})^2=2.2\cdot 10^{-6}m^2[/tex]
So, the current in the wire is
[tex]I=(8.34\cdot 10^{28})(1.6\cdot 10^{-19})(2.2\cdot 10^{-6})(0.5\cdot 10^{-3})=14.7 A[/tex]
