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Question: How many electrons per second pass through a section of wire carrying a current of 0.7 A?...

How many electrons per second pass through a section of wire carrying a current of 0.7 A?

Explanation

Solution

To solve this problem, the relationship between the quantities current and charge should be established. After we obtain the relationship, the charge obtained must be divided by the fundamental charge of an electron to obtain the number of electrons.

Step-by-step solution:
The current is defined as the rate of flow of charges through an unit cross-sectional area in a given time.
Current, I=qtI = \dfrac{q}{t}
where q = charge and t = time. The SI unit of current is ampere.
When we say the flow of charges in a conductor, they are the free electrons in the conductor. The electron has a charge known as fundamental charge, called ee.
The value of fundamental charge, e=1⋅6×10−19Ce = 1 \cdot 6 \times {10^{ - 19}}C
Thus, the net charge, q is the product of the number of electrons and the fundamental charge.
q=neq = ne
Here, the number n should always be a positive natural number since the charge is quantised i.e. charges exist in integral multiples of the fundamental charge, e.
Substituting in the current, we get –
I=netI = \dfrac{{ne}}{t}
Given,
Time, t = 1 sec Current, I=0⋅7AI = 0 \cdot 7A Fundamental charge, e=1⋅6×10−19Ce = 1 \cdot 6 \times {10^{ - 19}}C
Substituting, we get –
I=netI = \dfrac{{ne}}{t}
0⋅7=n×1⋅6×10−1910 \cdot 7 = \dfrac{{n \times 1 \cdot 6 \times {{10}^{ - 19}}}}{1}
Solving for n,
n=0⋅71⋅6×10−19=0⋅437×1019=4⋅4×1018n = \dfrac{{0 \cdot 7}}{{1 \cdot 6 \times {{10}^{ - 19}}}} = 0 \cdot 437 \times {10^{19}} = 4 \cdot 4 \times {10^{18}}
Hence, the number of electrons per second passing through the wire are – 4⋅4×10184 \cdot 4 \times {10^{18}}

Note: There is an alternative solution to this problem. For that, we have to define the units, coulomb and ampere.
One ampere is equal to one coulomb per second.
1A=1C1sec⁡1A = \dfrac{{1C}}{{1\sec }}
Also,
One coulomb of charge is equal to the charge contained by 1e\dfrac{1}{e} number of electrons, which is equal to the number 6⋅25×10186 \cdot 25 \times {10^{18}}.
Therefore, in one second, 6⋅25×10186 \cdot 25 \times {10^{18}} electrons constitute the current one ampere, for 0.7 A, the number of electrons will be –
n=0⋅7×6⋅25×1018=4⋅4×1018n = 0 \cdot 7 \times 6 \cdot 25 \times {10^{18}} = 4 \cdot 4 \times {10^{18}}
Hence, we get the same answer.