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(B) The current $I$ is given by $I = \frac{q}{t}$,where $q = ne$.Given $I = 1\, mA = 1 	imes 10^{-3}\, A$ and $t = 1\, s$.The charge of an electron i

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$6.25 	imes 10^{15}$

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(B) The current $I$ is given by $I = \frac{q}{t}$,where $q = ne$.Given $I = 1\, mA = 1 	imes 10^{-3}\, A$ and $t = 1\, s$.The charge of an electron is $e = 1.6 	imes 10^{-19}\, C$.Using the formula $I = \frac{ne}{t}$,we can solve for the number of electrons $n$:$n = \frac{I 	imes t}{e} = \frac{1 	imes 10^{-3} 	imes 1}{1.6 	imes 10^{-19}}$$n = \frac{1}{1.6} 	imes 10^{16} = 0.625 	imes 10^{16} = 6.25 	imes 10^{15}$.Therefore,$6.25 	imes 10^{15}$ electrons will pass the point in one second.

$A$ current of $1\, mA$ is flowing through a copper wire. How many electrons will pass a given point in one second?

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