Question

The current of $0.5$ Amp is flowing in a copper wire whose cross-sectional area is $1m{m}^2$. If the number of free electrons is $8.5\times {10}^{22}perc{m}^3$ in unit volume then find out the drift velocity of electrons?

Answer 1

Area of cross section = $1m{m}^2$
$=1\times ({10}^3{)}^2=1\times {10}^{-6}{m}^2$
Number of free electrons $\left(n\right)=8.5\times {10}^{22}/c{m}^3$
= $8.5\times {10}^{28}/m^3$
Current (I) = 0.5 amp
Drift velocity $\left(v_d\right)=\frac{I}{neA}$
= $\frac{0.5}{8.5\times {10}^{28}\times 1.6\times {10}^{19}\times 1\times {10}^{-6}}$
= $3.7\times {10}^{-5}m/s$