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Making connections: take-home investigation—filament observations

Find a lightbulb with a filament. Look carefully at the filament and describe its structure. To what points is the filament connected?

We can obtain an expression for the relationship between current and drift velocity by considering the number of free charges in a segment of wire, as illustrated in [link] . The number of free charges per unit volume is given the symbol n size 12{n} {} and depends on the material. The shaded segment has a volume Ax size 12{ ital "Ax"} {} , so that the number of free charges in it is nAx size 12{ ital "nAx"} {} . The charge Δ Q size 12{DQ} {} in this segment is thus qnAx size 12{ ital "qnAx"} {} , where q size 12{q} {} is the amount of charge on each carrier. (Recall that for electrons, q size 12{q} {} is 1 . 60 × 10 19 C size 12{ - 1 "." "60" times "10" rSup { size 8{ - "19"} } "C"} {} .) Current is charge moved per unit time; thus, if all the original charges move out of this segment in time Δ t size 12{Dt} {} , the current is

I = Δ Q Δ t = qnAx Δ t . size 12{I = { {ΔQ} over {Δt} } = { { ital "qnAx"} over {Δt} } "."} {}

Note that x / Δ t size 12{x/Δt} {} is the magnitude of the drift velocity, v d , since the charges move an average distance x size 12{x} {} in a time Δ t size 12{Dt} {} . Rearranging terms gives

I = nqAv d , size 12{I= ital "nqAv" rSub { size 8{"d"} } } {}

where I size 12{I } {} is the current through a wire of cross-sectional area A size 12{A} {} made of a material with a free charge density n size 12{n} {} . The carriers of the current each have charge q size 12{q} {} and move with a drift velocity of magnitude v d size 12{v rSub { size 8{d} } } {} .

Charges are shown moving through a section of a conducting wire. The charges have a drift velocity v sub d along the length of the wire, shown by an arrow pointing to the right. The volume of a segment of the wire is equal to A times x, where x equals the product of the drift velocity, v sub d, and time t. A cross section of the wire is marked as A, and the length of the section is x.
All the charges in the shaded volume of this wire move out in a time t size 12{t} {} , having a drift velocity of magnitude v d = x / t size 12{v rSub { size 8{d} } =x/t} {} . See text for further discussion.

Note that simple drift velocity is not the entire story. The speed of an electron is much greater than its drift velocity. In addition, not all of the electrons in a conductor can move freely, and those that do might move somewhat faster or slower than the drift velocity. So what do we mean by free electrons? Atoms in a metallic conductor are packed in the form of a lattice structure. Some electrons are far enough away from the atomic nuclei that they do not experience the attraction of the nuclei as much as the inner electrons do. These are the free electrons. They are not bound to a single atom but can instead move freely among the atoms in a “sea” of electrons. These free electrons respond by accelerating when an electric field is applied. Of course as they move they collide with the atoms in the lattice and other electrons, generating thermal energy, and the conductor gets warmer. In an insulator, the organization of the atoms and the structure do not allow for such free electrons.

Calculating drift velocity in a common wire

Calculate the drift velocity of electrons in a 12-gauge copper wire (which has a diameter of 2.053 mm) carrying a 20.0-A current, given that there is one free electron per copper atom. (Household wiring often contains 12-gauge copper wire, and the maximum current allowed in such wire is usually 20 A.) The density of copper is 8 . 80 × 10 3 kg/m 3 size 12{8 "." "80" times "10" rSup { size 8{3} } `"kg/m" rSup { size 8{3} } } {} .

Strategy

We can calculate the drift velocity using the equation I = nqAv d . The current I = 20.0 A is given, and q = 1.60 × 10 19 C is the charge of an electron. We can calculate the area of a cross-section of the wire using the formula A = π r 2 , where r is one-half the given diameter, 2.053 mm. We are given the density of copper, 8.80 × 10 3 kg/m 3 , and the periodic table shows that the atomic mass of copper is 63.54 g/mol. We can use these two quantities along with Avogadro’s number, 6.02 × 10 23 atoms/mol , to determine n , the number of free electrons per cubic meter.

Solution

First, calculate the density of free electrons in copper. There is one free electron per copper atom. Therefore, is the same as the number of copper atoms per m 3 . We can now find n as follows:

n = 1 e atom × 6 . 02 × 10 23 atoms mol × 1 mol 63 . 54 g × 1000 g kg × 8.80 × 10 3 kg 1 m 3 = 8 . 342 × 10 28 e /m 3 .

The cross-sectional area of the wire is

A = π r 2 = π 2.053 × 10 −3 m 2 2 = 3.310 × 10 –6 m 2 .

Rearranging I = n q A v d to isolate drift velocity gives

v d = I nqA = 20.0 A ( 8 . 342 × 10 28 /m 3 ) ( –1 . 60 × 10 –19 C ) ( 3 . 310 × 10 –6 m 2 ) = –4 . 53 × 10 –4 m/s.

Discussion

The minus sign indicates that the negative charges are moving in the direction opposite to conventional current. The small value for drift velocity (on the order of 10 4 m/s size 12{"10" rSup { size 8{ - 4} } `"m/s"} {} ) confirms that the signal moves on the order of 10 12 size 12{"10" rSup { size 8{"12"} } } {} times faster (about 10 8 m/s size 12{"10" rSup { size 8{8} } `"m/s"} {} ) than the charges that carry it.

Questions & Answers

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x-2y+3z=-3 2x-y+z=7 -x+3y-z=6
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Need help solving this problem (2/7)^-2
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x+2y-z=7
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-1
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An investment account was opened with an initial deposit of $9,600 and earns 7.4% interest, compounded continuously. How much will the account be worth after 15 years?
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lim x to infinity e^1-e^-1/log(1+x)
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If f(x) = x-2 then, f(3) when 5f(x+1) 5((3-2)+1) 5(1+1) 5(2) 10
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make 5/4 into a mixed number, make that a decimal, and then multiply 32 by the decimal 5/4 turns out to be
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20/(×-6^2)
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A soccer field is a rectangle 130 meters wide and 110 meters long. The coach asks players to run from one corner to the other corner diagonally across. What is that distance, to the nearest tenths place.
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Jeannette has $5 and $10 bills in her wallet. The number of fives is three more than six times the number of tens. Let t represent the number of tens. Write an expression for the number of fives.
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. After 3 months on a diet, Lisa had lost 12% of her original weight. She lost 21 pounds. What was Lisa's original weight?
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Source:  OpenStax, General physics ii phy2202ca. OpenStax CNX. Jul 05, 2013 Download for free at http://legacy.cnx.org/content/col11538/1.2
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