There is a current of $1.344\, amp$ in a copper wire whose area of cross-section normal to the length of the wire is $1\,m{m^2}$. If the number of free electrons per $c{m^3}$ is $8.4 \times {10^{22}}$, then the drift velocity would be
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(c) ${v_d} = \frac{i}{{nAe}} = \frac{{1.344}}{{{{10}^{ - 6}} \times 1.6 \times {{10}^{ - 19}} \times 8.4 \times {{10}^{22}}}}$
$ = \frac{{1.344}}{{10 \times 1.6 \times 8.4}} = 0.01\,cm/s = 0.1\,mm/s$
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