The number density of free electrons in a copper conductor estimated in textual example 6.1 is `8.5 xx 10^(28) m^(-3)`. How long does an electrons take to drift from one end of a wire 3.0 m long to its other end? The area of cross-section of the wire is `2.0 xx 10^(-6) m^(2)` and it is carrying a current of 3.0 A

Explanation

Here `n = 8.5 xx 10^(28) m^(-3), L = 3.0 m: A = 2.0 xx 10^(-6) m^(2), I = 3.0A, t = ?`
As I = n A eVd
`:. V_(d) = (1)/(Ae)`
Now, `t = (1)/(V_(d)) = (LnAe)/(I)`
`= (3.08 xx 8.5 xx 10^(28) xx 2.0 xx 10^(-6) xx 1.6 xx 10^(-19))/(3.0)`
`= 2.72 xx 10^(4) S`
= 7 hour 33 minutes.