Estimate the average drift speed of conduction electrons in a copper wire of cross-sectional area `1.0 xx 10^(-7) m^(2)` carrying a current of 1.5 A.

Question

Estimate the average drift speed of conduction electrons in a copper wire of cross-sectional area `1.0 xx 10^(-7) m^(2)` carrying a current of 1.5 A.

Estimate the average drift speed of conduction electrons in a copper wire of cross-sectional area `1.0 xx 10^(-7) m^(2)` carrying a current of 1.5 A. Assume that each copper atom contributes roughly one conduction electron. The density of copper is `9.0 xx 10^(3) kg//m^(3)` and its atomic mass is 63.5 u.

Monjur Alahi

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2025-01-04 04:42

1 Answers

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Salim Ahmed Kawsar · Accepted Answer · 2023-02-05T15:29:48+06:00

The direction of drift velocity of conduction electrons is opposite to the electric field direction, i.e., electrons drift in the direction of increasing potential. The drift speed `upsilon_(d)` is given by Eq.
`I Delta t = n eA//V_(d)//Delta t`
`V_(d) = (I//n eA)`
Now, `e = 1.6 xx 10^(-19) C,A = 1.0 xx 10^(-7) m^(2), I = 1.5 A`. The density of conduction electrons n is equal to the number of atoms per cubic meter (assuming one conduction electron per Cu atom as is reasonable from its valence electron count of one). A cubic metre of copper has a mass of `9.0 xx 10^(3)` kg. Since `6.0 xx 10^(23)` copper atoms have a mass of 63.5 g
`n = (6.0 xx 10^(23))/(63.5) xx 9.0 xx 10^(6)`
`= 8.5 xx 10^(28) m^(-3)`
Which gives,
`upsilon_(d) = (1.5)/(8.5 xx 10^(28) xx 1.6 xx 10^(-19) xx 10 xx 10^(-7))`
`= 1.1 xx 10^(-3) ms^(-1) = 1.1 m m s^(-1)`