Question

1.60 mm diameter tin cylindrical wire carries current of 4A. The drift velocity of the conduction electrons is 0.15 mms^-1. What is the concentration of conduction electrons?

Answer 1

Concentration of conduction electrons $n = \cfrac{I} {e A v}$ ,

where

$\quad I \text{- electrical current, A} \\ \quad e = 1.60 * 10^{-19} \text{ - elementary charge, C} \\ \quad A = \cfrac {\pi d^2}{4} \text{ - cross-sectional area, }m^2\\ \quad d \text{ - diameter of the wire, m} \\ \quad v \text{ - drift velocity, m/s}$

$n = \cfrac{4}{1.60*10^{-19} * \cfrac{3.14 * 1.60^2}{4}*10^{-6} * 0.15 * 10^{-3}} \approx 8.29*10^{28} ( m^{-3} )$