Answer: 7.47 milliamperes (mA).
Explanation:The drift velocity of electrons in a wire is related to the current flowing through the wire by the equation:
I = nAevd
where I is the current, n is the number density of electrons, A is the cross-sectional area of the wire, e is the charge of an electron, and vd is the drift velocity.
The number density of electrons in copper is approximately 8.5 x 10^28 electrons per cubic meter.
The cross-sectional area of the wire is given by the formula for the area of a circle:
A = πr^2
where r is the radius of the wire. The diameter of the wire is given as 1.628 mm, so the radius is half of this value:
r = 0.814 mm = 0.814 x 10^-3 m
Therefore, the cross-sectional area of the wire is:
A = π(0.814 x 10^-3 m)^2 = 5.21 x 10^-7 m^2
The charge of an electron is -1.6 x 10^-19 C.
The drift velocity is given as 1.00 mm/s = 1.00 x 10^-3 m/s.
Substituting these values into the equation for the current, we get:
I = (8.5 x 10^28 electrons/m^3) x (5.21 x 10^-7 m^2) x (-1.6 x 10^-19 C) x (1.00 x 10^-3 m/s)
I = -7.47 x 10^-3 A
The magnitude of the current is:
|I| = 7.47 x 10^-3 A
Therefore, the magnitude of the current flowing through the wire is 7.47 milliamperes (mA).