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$62.5 \times 10^{18}$ electrons per second are flowing through a wire of area of cross-section $0.1 \mathrm{~m}^2$, the value of current flowing will be
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$10 \mathrm{~A}$
$i=\frac{n e}{t}=\frac{62.5 \times 10^{18} \times 1.6 \times 10^{-19}}{1}=10$ ampere
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