62.5 times 10^{18} electrons per second are f | vedclass

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(C) The current $I$ is defined as the rate of flow of charge,given by the formula $I = \frac{q}{t}$.Since $q = ne$,where $n$ is the number of electron

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$10$

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(C) The current $I$ is defined as the rate of flow of charge,given by the formula $I = \frac{q}{t}$.Since $q = ne$,where $n$ is the number of electrons and $e$ is the elementary charge $(1.6 	imes 10^{-19} \, C)$,we have $I = \frac{ne}{t}$.Given $n/t = 62.5 	imes 10^{18} \, s^{-1}$ and $e = 1.6 	imes 10^{-19} \, C$:$I = (62.5 	imes 10^{18}) 	imes (1.6 	imes 10^{-19}) \, A$$I = 62.5 	imes 1.6 	imes 10^{-1} \, A$$I = 100 	imes 10^{-1} \, A = 10 \, A$.Therefore,the current flowing through the wire is $10 \, A$.

$62.5 \times 10^{18}$ electrons per second are flowing through a wire of area of cross-section $0.1 \, m^2$. The value of the current flowing will be ............ $A$.

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