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

Question

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 [latex]8.80 imes 10^{3} kg / m ^{3}[/latex] .

Strategy
We can calculate the drift velocity using the equation [latex]I=n q A v_{ d }[/latex]. The current I = 20.0 A is given, and [latex]q=-1.60 imes 10^{-19} C[/latex] is the charge of an electron. We can calculate the area of a cross-section of the wire using the formula [latex]A=\pi r^{2}[/latex], where r is one-half the given diameter, 2.053 mm. We are given the density of copper, [latex]8.80 imes 10^{3} kg / m ^{3}[/latex], 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, [latex]6.02 imes 10^{23}[/latex] atoms/mol, to determine n, the number of free electrons per cubic meter.

Accepted Answer

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 [latex]m ^{3}[/latex]. We can now find n as follows:

[latex]n=\frac{1 e^{-}}{ ext {atom }} imes \frac{6.02 imes 10^{23} ext { atoms }}{ mol } imes \frac{1 mol }{63.54 g } imes \frac{1000 g }{ kg } imes \frac{8.80 imes 10^{3} kg }{1 m ^{3}}[/latex] (20.9)

[latex]=8.342 imes 10^{28} e^{-} / m ^{3}[/latex].

The cross-sectional area of the wire is

[latex]A=\pi r^{2}[/latex] (20.10)

[latex]=\pi\left(\frac{2.053 imes 10^{-3} m }{2} ight)^{2}[/latex]

[latex]=3.310 imes 10^{-6} m ^{2}[/latex].

Rearranging [latex]I=n q A v_{ d }[/latex] to isolate drift velocity gives

[latex]v_{ d }=\frac{I}{n q A}[/latex] (20.11)

[latex]=\frac{20.0 A }{\left(8.342 imes 10^{28} / m ^{3} ight)\left(-1.60 imes 10^{-19} C ight)\left(3.310 imes 10^{-6} m ^{2} ight)}[/latex]

[latex]=-4.53 imes 10^{-4} m / s[/latex].

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 [latex]10^{-4} m / s[/latex] ) confirms that the signal moves on the order of [latex]10^{12}[/latex] times faster (about [latex]10^{8} m / s[/latex] ) than the charges that carry it.

Question 20.3: Calculate the drift velocity of electrons in a 12-gauge copp...

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