Question

A current of 1 A is drawn by a filament of an electric bulb. Number of electrons passing through a cross section of the filament in 16 seconds would be roughly:
A. ${10^{20}}$
B. ${10^{16}}$
C. ${10^{18}}$
D. ${10^{23}}$

Answer 1

Rate of flow of charge is considered current. Here electrons are passing through a cross section in a particular direction. We consider the flow of current in direction opposite to electron flow as electrons have negative charge. Amount of current is given and time of flow of electrons in filament is also given. From this we can find out the time rate of flow of charge.

Formula used:
\eqalign{ & i = \dfrac{{dq}}{{dt}} \cr & q = ne \cr}

Complete step by step answer:
When charge is under motion current is produced. Here in the case of a bulb connected to a potential difference this will cause electrons to flow. Now rate of flow of these electrons is considered as magnitude of current which is given as 1 ampere
Let us assume total number of electrons flowing be ‘n’
Charge of an electron be ‘e’ neglecting negative sign as we need only magnitude
Magnitude of total electron charge would be ‘q’
$q = ne$
$e = 1.6 \times {10^{ - 19}}coulombs$
$i = \dfrac{{dq}}{{dt}}$
Time is given as 16 seconds and $q = ne$
\eqalign{ & i = \dfrac{q}{{16}} \cr & \Rightarrow 1 = \dfrac{{ne}}{{16}} \cr & \Rightarrow 1 = \dfrac{{n \times 1.6 \times {{10}^{ - 19}}}}{{16}} \cr & \Rightarrow 1 = n \times {10^{ - 20}} \cr & \Rightarrow n = {10^{20}} \cr}
Total number of electrons passing through cross section would be ${10^{20}}$

So, the correct answer is “Option A”.

Additional Information: In a particular conductor current is same throughout the conductor and doesn’t vary with cross section but there is current density which is ratio of current flowing and cross section. This current density varies with cross section.

Note: If charge is given as a function of time then we should differentiate that charge function with respect to time to get current and if current is given as a function of time then we can integrate with respect to time to find current. We can also find the average value of current and r.m.s value of current with that current function.