Question

A current of $1\,mA$ is flowing through a copper wire. How many electrons will pass a given point in one second. Take $\left[ {e = 1.6 \times {{10}^{ - 19}}\,C} \right]$?
(A) $6.25 \times {10^{19}}$
(B) $6.25 \times {10^9}$
(C) $6.25 \times {10^{31}}$
(D) $6.25 \times {10^{15}}$

Answer 1

Hint: the number of electrons will pass a given point in one second can be determined by using the current formula. In the current formula, the charge is written as the product of the number of electrons and the charge of the electrons, so that the number of electrons passed can be determined.

Useful formula
The flow of charge is given by,
$q = ne$
Where, $q$ is the charge in the copper wire, $n$ is the number of electrons and $e$ is the charge of the electrons.
The current flow in the copper wire is given by,
$I = \dfrac{q}{t}$
Where, $I$ is the current in the copper wire, $q$ is the charge in the copper wire and $t$ is the time.

Complete step by step solution
Given that,
The current flow sin the copper wire is, $I = 1\,mA \Rightarrow 1 \times {10^{ - 3}}\,A$
The charge of the electron is, $e = 1.6 \times {10^{ - 19}}\,C$
The time to pass the given point, $t = 1\,s$
Now,
The flow of charge is given by,
$q = ne\,..................\left( 1 \right)$
 The current flow in the copper wire is given by,
$I = \dfrac{q}{t}\,....................\left( 2 \right)$
By substituting the equation (1) in the equation (2), then the equation (2) is written as,
$I = \dfrac{{ne}}{t}$
By rearranging the terms, then the above equation is written as,
$ne = It$
By keeping the term $n$ in one side and the other terms in other side, then the above equation is written as,
$n = \dfrac{{It}}{e}$
By substituting the current flow, time and charge of the electron in the above equation, then the above equation is written as,
$n = \dfrac{{1 \times {{10}^{ - 3}} \times 1}}{{1.6 \times {{10}^{ - 19}}}}$
By rearranging the terms, then the above equation is written as,
$n = \dfrac{{1 \times {{10}^{ - 3}} \times 1 \times {{10}^{19}}}}{{1.6}}$
On multiplying the terms in the numerator, then
$n = \dfrac{{1 \times {{10}^{16}}}}{{1.6}}$
On dividing the terms in the above equation, then
$n = 0.625 \times {10^{16}}$
Then the above equation is written as,
$n = 6.25 \times {10^{15}}$

Hence, the option (D) is the correct answer

Note: When the current in the circuit increases, the greater the number of electrons flowing in the circuit, and also the time increases the number of the electrons will also increase. The charge of electrons is inversely proportional to the number of electrons but the charge of electrons is a constant value, it will never change.