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<math>Δq=ρv⋅nΔAΔt</math>. | <math>Δq=ρv⋅nΔAΔt</math>. | ||
<math>ρ = Nq</math> | <math>ρ = Nq</math> | ||
The charge per unit time is then <math>ρv⋅nΔS</math>, from which we get the current density to be<math>Nqv</math> | |||
The charge per unit time is then <math>ρv⋅nΔS</math>, from which we get the current density to be <math>Nqv</math> | |||
The current I through the surface is <math>I=∫Nqv⋅dA</math> | The current I through the surface is <math>I=∫Nqv⋅dA</math> | ||
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<math>I = 0.5A</math> | <math>I = 0.5A</math> | ||
===Middling=== | ===Middling=== | ||
=== | Current will not always be what you solve for, though. In this problem, let's use the equations we've discussed to find the drift velocity <math>v</math>. | ||
Given the current is <math>1A</math>, the electron density is <math>8 ⋅ 10^{27}m^{-3}</math>, and the diameter of our wire is <math>1mm</math>. | |||
First, let's find the area <math>{0.001\over 2}^{2}⋅π=7.85⋅10^{-7}</math> | |||
Then, plug in the rest: <math>1=|-1.6⋅10^{-19}|⋅5⋅10^{28}⋅7.85⋅10^{-7}⋅v, v = 9.9477.85⋅10^{-4}{m\over s}</math> | |||
==References== | ==References== | ||
https://en.wikipedia.org/wiki/Electric_current | https://en.wikipedia.org/wiki/Electric_current | ||
http://www.feynmanlectures.caltech.edu/II_13.html#Ch13-S2 | http://www.feynmanlectures.caltech.edu/II_13.html#Ch13-S2 | ||
http://hyperphysics.phy-astr.gsu.edu/hbase/electric/elecur.html | http://hyperphysics.phy-astr.gsu.edu/hbase/electric/elecur.html | ||
[[Category:Which Category did you place this in?]] | [[Category:Which Category did you place this in?]] |
Revision as of 19:55, 5 December 2015
Claimed by spencer
Explanation of Current Through a Wire
The Main Idea
The electric current is the continual flow of electric charge through an area. In a wires you will be studying, the current is constant regardless of its width and diameter. It is important to note that the charge in the circuits you will be dealing with are not in equilibrium, but instead, in a steady state. The distinction and the circuits state become clear when you remember that a metal in equilibrium contains no mobile charge in motion. Since there is flow (motion) with a current, the velocity cannot be zero, and equilibrium is not the current state. In a steady state, the particles may be moving however they please, but their net drift velocity stays constant. The most common charge carrier you will encounter is negative. While the positive charge can be the mover, in almost all metals the current consists of drifting electrons. The direction the electrons flow is not the same as the conventional current, however. Before scientists knew that electrons were the common charge carriers, they discovered current. And, with only two choices, they chose to treat the moving charge as positive. This may seem annoying at first, but there are some benefits to using conventional current over the directly opposed "electron current". For one, it flows from the positive end of a battery toward the negative end, and from high to low energy.
A Mathematical Model
Assume uniform density [math]\displaystyle{ n }[/math] of electrons (uniform current) in a wire of area [math]\displaystyle{ A }[/math]
[math]\displaystyle{ i=density⋅volume/seconds =nA v }[/math]
If a charge distribution of density [math]\displaystyle{ ρ }[/math] moves with the velocity [math]\displaystyle{ v }[/math], the charge per unit time through [math]\displaystyle{ ΔA }[/math] is [math]\displaystyle{ ρv⋅nΔA }[/math]
[math]\displaystyle{ Δq=ρv⋅nΔAΔt }[/math].
[math]\displaystyle{ ρ = Nq }[/math]
The charge per unit time is then [math]\displaystyle{ ρv⋅nΔS }[/math], from which we get the current density to be [math]\displaystyle{ Nqv }[/math]
The current I through the surface is [math]\displaystyle{ I=∫Nqv⋅dA }[/math]
[math]\displaystyle{ I = |q|nAv }[/math]
[math]\displaystyle{ i = nAv }[/math]
Examples
Be sure to show all steps in your solution and include diagrams whenever possible
Simple
One type of problem you are sure to encounter involves a battery and resistor. You can read more about that section on the RC page of this wiki, but here's how to solve one now:
Using your formula sheet, notice that [math]\displaystyle{ I = {|ΔV|\over R} }[/math]. where R is the resistance in the circuit and |ΔV| is the voltage.
If you're given that the 50V battery is connected to a 100 [math]\displaystyle{ Ω }[/math] resistor, you simply substitute the values into your equation. [math]\displaystyle{ I = {50V\over 100 Ω} }[/math] [math]\displaystyle{ I = 0.5A }[/math]
Middling
Current will not always be what you solve for, though. In this problem, let's use the equations we've discussed to find the drift velocity [math]\displaystyle{ v }[/math]. Given the current is [math]\displaystyle{ 1A }[/math], the electron density is [math]\displaystyle{ 8 ⋅ 10^{27}m^{-3} }[/math], and the diameter of our wire is [math]\displaystyle{ 1mm }[/math].
First, let's find the area [math]\displaystyle{ {0.001\over 2}^{2}⋅π=7.85⋅10^{-7} }[/math]
Then, plug in the rest: [math]\displaystyle{ 1=|-1.6⋅10^{-19}|⋅5⋅10^{28}⋅7.85⋅10^{-7}⋅v, v = 9.9477.85⋅10^{-4}{m\over s} }[/math]
References
https://en.wikipedia.org/wiki/Electric_current
http://www.feynmanlectures.caltech.edu/II_13.html#Ch13-S2
http://hyperphysics.phy-astr.gsu.edu/hbase/electric/elecur.html