Question

Determine the number of electrons per second that pass through the cross section of a wire that carry a current of 3.45 A magnitude.6.77E+19 electrons/s1.08E+01 electrons/s4.64E-20 electrons/s5.53E-19 electrons/s2.15E+19 electrons/s

Answer 1

Given:

The current passing through the wire, I=3.45 A

To find:

The number of electrons passing per second.

Step-by-step explanation:

The electric current is defined as the time rate of flow of charge through a point.

That is electric current is the total charge flowing through a point per second.

Thus the current through the given cross-section is

`\begin{gathered} I=(Q)/(t) \\ \implies3.45\text{ A}=\frac{Q}{1\text{ s}} \\ \implies Q=3.45\text{ C} \end{gathered}`

Where Q is the total charge passing through the cross-secion per esecond.

The charge of one electron is e=1.602×10⁻¹⁹ C

Thus he ctotal number of electrons passing through the cross-section per second is given by,

`n=(Q)/(e)`

On substituting the known values,

`\begin{gathered} n=(3.45)/(1.602*10^(-19)) \\ =2.16*10^(19)\text{ } \end{gathered}`

Final answer:

Thus the number of electrons passing through the cross-section per second is 2.15E+19 electrons/s