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(D) The current $I$ is given as $10^{-6} \ A$,which means $10^{-6} \ C$ of charge passes through the cross-section per second.The charge of a single e

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$6 	imes 10^{12}$

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(D) The current $I$ is given as $10^{-6} \ A$,which means $10^{-6} \ C$ of charge passes through the cross-section per second.The charge of a single electron is $e = 1.602 	imes 10^{-19} \ C$.The number of electrons $n$ passing per second is calculated using the formula $n = \frac{I}{e}$.$n = \frac{10^{-6} \ C/s}{1.602 	imes 10^{-19} \ C} \approx 6.24 	imes 10^{12} \ 	ext{electrons/s}$.Rounding to the nearest provided option,the correct answer is $6 	imes 10^{12}$.

The number of electrons passing per second through a cross-section of copper wire carrying $10^{-6} \ A$ of current is found to be:

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