Respuesta :
Answer:
a) 3.1205*10^19 electron/s
b) 1.51*10^6 A/m^2
c) 1.11*10^-4 m/s
Explanation:
a) to find the number of electrons you use the current in the wire, and the following formula:
[tex]I=5.00A=5.00\frac{C}{s}\\\\1C=6.241*10^{18}e\\\\I=5.00(6.241*10^{18}e)/s=3.1205*10^{19}\frac{e}{s}[/tex]
3.1205*10^19 electrons per second
b) To find the current density you use the formula:
[tex]J=\frac{I}{A}[/tex]
I: current in the wire
A: cross area of the wire
[tex]J=\frac{I}{\pi r^2}=\frac{5.00A}{\pi(\frac{2.05}{2}*10^{-3})^2}=1.51*10^6\frac{A}{m^2}[/tex]
c) To find the speed you use the formula for the drift speed of electrons in the wire:
[tex]I=nqv_dA\\\\v_d=\frac{I}{nqA}[/tex]
n: number of free electrons
q: charge of the electron = 1.6*10^{-19}C
[tex]v_d=\frac{5.00C/s}{(8.5*10^{28}m^{-3})(1.6*10^{-19}C)(3.30*10^{-6}m^2)}=1.11*10^{-4}\frac{m}{s}[/tex]
d) if the diameter of the wire is increased, the number of electron that pass trough the lighy bulb each second is the same.
The current density decreases because J=I/A. If A increases J decreases.
The drift vellocity of the electrons decreases, again, because in the formula for vd the Area is in the denominator.a
The flow of current through the bulb and the copper wire is by the motion
of the electrons in the opposite direction.
- (a) The number of electrons is approximately 3.125 × 10¹⁹
- (b) The current density is approximately 1514859.6606 A/m².
- (c) The speed of a typical electron is approximately 1.114 × 10⁻⁴ m/s.
- (d) Using twice the diameter, decreases; The current density and the speed of a typical electron.
Reasons:
The current that runs through the gauge copper, I = 5.00 A
Diameter of the copper wire = 12-gauge = 2.05 mm = 0.00205 m
Number of free electrons in a cubic feet of copper wire, (electron density) ρ = 8.5 × 10²⁸ electrons
(a) The number of electrons passing through the light bulb per second is
given by the equation;
[tex]\displaystyle n = \frac{I \cdot t}{e}[/tex]
Where;
I = 5.00 A
t = 1 second
e = The charge on one electron
[tex]\mathrm{The \ number \ of \ electrons, }\displaystyle \ n = \frac{5 \times 1}{1.6 \times 10^{-19}} = \underline {3.125 \times 10^{19}}[/tex]
(b) The current density, J, is given by formula;
[tex]\displaystyle J = \frac{I}{A}[/tex]
[tex]\mathrm{The \ area \ of \ the \ wire, } \ A = \displaystyle \frac{\pi \cdot D^2}{4} = \frac{\pi \times 0.00205^2}{4} \approx 3.3006 \times 10^{-6}[/tex]
Therefore;
[tex]\displaystyle J = \frac{5}{ \frac{\pi \times 0.00205^2}{4} } = 1514859.6606[/tex]
The current density, J = 1,514,859.6606 A/m².
(c) Current density, J = Charge density, ρ × Velocity of Charges, v
Charge density, ρ = Electron density × Charge of one electron
Therefore;
[tex]\displaystyle v = \mathbf{\frac{J}{\rho}}[/tex]
Which gives;
[tex]\displaystyle v = \frac{1,514,859.6606}{8.5 \times 10^{28} \times \left(1.6 \times 10^{-19} \right)} \approx \mathbf{1.114 \times 10^{-4}}[/tex]
The speed of a typical electron, v ≈ 1.114 × 10⁻⁴ m/s
(d) Using twice the diameter, increases the area of the wire
[tex]\mathrm{The \ electron \ density \ is,}\displaystyle \ J = \frac{I}{A}[/tex]
- Increasing the area, A, reduces the current density
[tex]\mathrm{Similarly, \, velocity \ of \ an \ electron \ is \ given \ by, }\displaystyle \ v = \frac{J}{\rho}[/tex]
Therefore, when the current density decreases, the velocity decreases.
Therefore;
Using twice the diameter, reduces;
- The current density
- The speed of a typical electron
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