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The number of electrons per second that pass through a section of wire carrying a current of 0.7 A is

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4.4 \times 10^{17}

3.4 \times 10^{18}

4.4 \times 10^{16}

4.4 \times 10^{18}

answer is D.

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Detailed Solution

t = 1 s, i = 0.7 A, n = ?

$i = \frac{- ne}{t}$

$0.7 = \frac{(- n) (- 1.6 \times 10^{- 19})}{(1)}$

$n = \frac{0.7}{1.6 \times 10^{- 19}} = 4.4 \times 10^{18}$

$n = 4.4 \times 10^{18} electrons$

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