Answer:
14.5A
Explanation:
The following data are given
Diameter,D=4.41mm=0.00441m,Hence the area can be calculated as
[tex]Arae=\pi \frac{d^{2}}{4}\\ Arae=\pi \frac{0.00441^{2}}{4}\\Area=1.53*10^{-5}m^{2}[/tex]
also the speed of the electron which we refer to as the drift speed is given as
1.01*10^-4m/s
and lastly the conduction electron density, n =6.00*10^28
From the formula connecting the drift velocity to the current which is
[tex]I=qnAv\\i=current,\\q=charge\\A=area \\v=drift speed[/tex]
if we insert values we arrive at
[tex]I=1.6*10^{-19}*6*10^{28}*1.5*10^{-5}*1.01*10^{-4}\\I=14.5*10^{-19+28-5-4}\\I=14.5A[/tex]
Hence the electron current in the wire is 14.5A
The electron current in the wire is 14.976A in the direction opposite the motion of the charges.
The current (I) flowing through a wire of cross-sectional area (A), made of a material with free charge density (n), is related to the carriers of the current each of charge Q, and the drift velocity v at which the carriers move, as follows;
I = n x Q x A x v -------------------------(i)
From the question;
The wire is made of aluminium and has
diameter d, = 4.41mm = 0.00441m
A = [tex]\frac{\pi d^2}{4}[/tex]
Take [tex]\pi[/tex] = 3.142
=> A = [tex]\frac{3.142 * 0.00441^2}{4}[/tex]
=> A = 1.5 x 10⁻⁵m²
drift speed, v = 1.01 x 10⁻⁴m/s
electron charge density, n = 6.00 x 10²⁸ m⁻³
Q = charge on an electron = -1.6 x 10⁻¹⁹C
Substitute these values into equation (i) to get the current through the wire as follows;
I = 6.00 x 10²⁸ x (-1.6 x 10⁻¹⁹) x 1.5 x 10⁻⁵ x 1.01 x 10⁻⁴
Solve for I;
I = -14.976A
The negative sign indicates that the direction of conventional current is opposite that of the motion of the negative charges.
Therefore, the electron current in the wire is 14.976A in the direction opposite the motion of the charges.
