Solution:
Solution:
Given, current $$I=20\,A$$,
number of electrons, $$n=10^{29}$$,
area of cross-section,
$$A=1\,mm^{2}=(1\times 10^{-3})^{2}$$
$$=10^{-6}\,m^{2}$$
$$\therefore$$ drift velocity $$V_{d}=\frac{I}{neA}$$
$$=\frac{20}{10^{29}\times1.6\times 10^{-19}\times1\times10^{-6}}$$
$$=\frac{20}{1.6\times 10^{4}}$$
$$=12.5\times10^{-4}$$ m/s
