What is the drift speed of the conduction electrons in a copper wire with radius `r = 900 mum` when it has a uniform current `I = 17mA`? Assume that each copper atom contributes one conduction electron to the current and the current density is uniform across the wire's cross section. The density of copper is `9.0xx10^3kgm^(−3)` and its atomic mass is `63.5u`.

Explanation

(1) The drift speed `v_(d)` is related to the current density `vecJ` and the number n of conduction electrons per unit volume according to which we can write as `J ne v_(d)`.
(2) Because the current density is uniform, its magnitude J is related to the given current i and wire size by (J=i/A), where A is the cross-sectional area of the wire).
(3) Because we assume one conduction electron per atom, the number n of conduction electrons per nit volume is the same as the number of atoms per unit volume.
Calculations : Let us start with the third idea by writting
n=(atoms per unit volume) =(atoms per mole) (moles per unit mass) (mass per unit volume)
The number of atoms per mole is just Avogadro.s number `N_(A) (=6.02 xx 10^(23)"mol"^(-1))`. Moles per unit mass is the inverse of the mass per mole, which here is the molar mass M of copper. The mass per unit volume is the (mass) density `p_("mass")` of copper. Thus, `n=N_(A) (1/M) p_("mass")=(N_(A) p_("mass"))/(M)`
Taking copper.s molar mass M and density `p_("mass.)` from Appendix F, we then have (with sme conversions of units)
`n=((6.02 xx 10^(23) "mol"^(-1)) (8.96 xx 10^(3) kg//m^(3)))/(63.54 xx 10^(-3)"kg/mol")`
`=8.49 xx 10^(28)" electrons/m"^(3)`
`or n=8.49 xx 10^(28) m^(-3)`
Next let us combine the first two key ideas by writting `i/A=nev_(d)`
Substituting for A with `pir^(2) (=2.54 xx 10^(-6) m^(2))` and solving for `v_(d)`, we then find
`v_(d)=i/(ne(pi r^(2))`
`=(17 xx 10^(-3)A)/((8.49 xx 10^(28) m^(-3)) (1.6 xx 10^(-19)C) (2.54 xx 10^(-6) m^(2))`
`=4.9 xx 10^(-7)m//s`.
which is only 1.8mm/h, slower than a sluggish snail.