A wire of diameter $0.02\,meter$ contains $10^{28}$ free electrons per cubic meter. For an electrical current of $100\, A$, the drift velocity of the free electrons in the wire is nearly
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(c) By using ${v_d} = \frac{i}{{neA}} = \frac{{100}}{{{{10}^{28}} \times 1.6 \times {{10}^{ - 19}} \times \frac{\pi }{4} \times {{(0.02)}^2}}}$
$ = 2 \times {10^{ - 4}}\,m/\sec $
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