Question

Ten million electrons pass from point P to point Q in one microsecond. The current and its direction is?
A) $1.6\times {10}^{-14}A$ $\text{From point P to point Q}$
B) $3.2\times 10^{-14}A$ $\text{From point P to point Q}$
C) $1.6\times {10}^{-14}A$ $\text{From point Q to point P}$
D) $3.2\times 10^{-14}A$ $\text{From point Q to point P}$

Answer 1

Hint: Current is the flow of electrons, but current and electrons flow in the opposite direction. Current flows from positive to negative and electrons flow from negative to positive. The current is determined by the number of electrons passing through the cross-section of a conductor in one second.

Complete step by step solution:
A large current, such as that used to start a truck engine, moves a large amount of charge in a small time, whereas a small current, such as that used to operate a hand-held calculator, moves a small amount of charge over a long period of time. In equation form, electric current I is defined to be:
$$I={\displaystyle \frac{\mathrm{\Delta }Q}{\mathrm{\Delta }t}}$$
where $$\mathrm{\Delta }Q$$ is the amount of charge passing through a given area in time $$\mathrm{\Delta }t=t$$. (As in previous chapters, initial time is often taken to be zero, in which case $$\mathrm{\Delta }t=t$$. The SI unit for current is the ampere (A), named for the French physicist André-Marie Ampère.
Since $$I={\displaystyle \frac{\mathrm{\Delta }Q}{\mathrm{\Delta }t}}$$ we see that an ampere is one coulomb per second: One ampere is equivalent to one coulomb of charge flowing per second,
Here, number of electrons,
$$n=10000000={10}^7$$
Total charge on ten million electrons is Q = ne
Where
$\Rightarrow e=1.6\times {10}^{-19}$
$\Rightarrow Q=10^7\times 1.6\times {10}^{-19}$
$\therefore Q=1.6\times 10^{-12}C$
Time taken by ten million electrons to pass from point P to point Q is
$\Rightarrow t=1\mu s$
$\Rightarrow t={10}^{-6}$
$\Rightarrow \text{The current}$ $I={\displaystyle \frac{\mathrm{\Delta }Q}{\mathrm{\Delta }t}}$
$\Rightarrow I={\displaystyle \frac{1.6\times {10}^{-12}}{{10}^{-6}}}$
$\therefore I=1.6\times 10^{-6}A$
Since the direction of the current is always opposite to the direction of flow of electrons. Therefore, due to flow of electrons from point P to point Q the current will flow from point Q to point P.

Note: One electron has a charge of $$1.602\times {10}^{-19}coulombs$$. Consequently, you can find the number of electrons in 1 coulomb of charge by taking the reciprocal of this number.
$$1coulomb={\displaystyle \frac{1}{1.602\times {10}^{-19}}}=6.242\times {10}^{18}Electrons.$$An electric current of 1 ampere is equal to 1 coulomb of charge passing a point in a circuit every second:
Therefore, a current of 1 ampere = $$6.242\times {10}^{18}$$electrons moving past any point in a circuit every second.