What is the current flowing through a conductor if $1$ million electrons are crossing in $1$ millisecond through its cross section?
A. $1.6A$
B. $2 \times {10^6}A$
C. $1.6 \times {10^{ - 6}}A$
D. $1.6 \times {10^{ - 10}}A$

Hint: In order to solve this question we need to understand electric current. So when a wire is subjected to potential difference, then an electrical field is set up in the wire in the direction of decreasing potential bonding electrons to flow in the opposite direction as of applied electrical field, hence the electrical current flows in opposite direction of flow of electrons. So electric current is defined as the number of charges crossing per unit cross sectional area in unit interval of time.

Complete step by step answer:
According to question, number of electrons in conductor is,
$n = 1{\kern 1pt} {\kern 1pt} million$
$\Rightarrow n = 1 \times {10^6}$
Since electron charge is given as,
$1{e^ - } = - 1.6 \times {10^{ - 19}}C$
So the total electrons flowing in the conductor is, $q = ne$.
Putting values we get,
$q = (1 \times {10^6}) \times ( - 1.6 \times {10^{ - 19}})$
$\Rightarrow q = - 1.6 \times {10^{ - 13}}C$
Time period required in crossing is,
$t = 1 \times {10^{ - 3}}\sec $
So the electrical current is defined as,
$i = \dfrac{q}{t}$
Putting values we get,
$i = \dfrac{{ - 1.6 \times {{10}^{ - 13}}C}}{{{{10}^{ - 3}}\sec }}$
$\therefore i = - 1.6 \times {10^{ - 10}}A$
Here, negative charge signifies that current is flowing in the opposite direction of the electron.

So the correct answer is D.

Note: It should be remembered that, although the electric current has both magnitude and direction, it is still not considered as a vector because electrical currents addition do not follow triangle law of vector addition, so it cannot be regarded as a vector. As long as potential difference is maintained, the electrical current continues to flow in the circuit. Also electrical current always flows from high potential to low potential or it always flows in the direction of the applied electrical field.