Solved Problem on Electric Current

A wire with cross-sectional 0.50×10⁻² cm², conducts a direct current equal to 4.0 A, given the elementary charge 1.6×10⁻¹⁹ C, determine:
a) The number of electrons passing through a cross-section per second;
b) The average speed of electrons, knowing that there are 1.8×10²⁰ electrons/cm³.

Problem data:

Cross-sectional area of wire: A = 0.50×10⁻²cm²;
Electric current: i = 4.0 A;
Elementary charge: e = 1.6×10⁻¹⁹ C.

Solution

a) In Figure 1, the electrons move across a cross-section, highlighted in gray. The electrical current is given by

\[ \begin{gather} \bbox[#99CCFF,10px] {i=\frac{\Delta q}{\Delta t}} \tag{I} \end{gather} \]

Figure 1

The amount of charge that crosses a section is given by

\[ \begin{gather} \Delta q=n{\mathrm e} \tag{II} \end{gather} \]

substituting the expression (II) into (I), we have

\[ \begin{gather} i=\frac{n{\mathrm e}}{\Delta t}\\ n=\frac{i\Delta t}{\mathrm e} \end{gather} \]

substituting the problem data for Δt = 1 s

\[ \begin{gather} n=\frac{4\times 1}{1.6\times 10^{-19}}\\ n=\frac{4\times 10^{19}}{1.6} \end{gather} \]

\[ \bbox[#FFCCCC,10px] {n=2.5\times 10^{19}\;\text{elétrons}} \]

b) The average speed is given by

\[ \begin{gather} \bbox[#99CCFF,10px] {\bar v=\frac{\Delta S}{\Delta t}} \tag{III} \end{gather} \]

then an electron passes through the cross-section of the wire in a instant will travel a distance ΔS in an interval of time Δt, the electron then passes through another cross-section. These two cross-sections determine in the wire a cylinder with a volume V given by (Figure 2)

Figure 2

\[ \begin{gather} \bbox[#99CCFF,10px] {V=Bh} \tag{IV} \end{gather} \]

where B is the area of the base of the cylinder, the cross-section area of wire B = A, and h is the displacement of the electron, h = ΔS, substituting these values in the expression (IV), we have

\[ \begin{gather} V=A \Delta S\\ \Delta S=\frac{V}{A} \tag{V} \end{gather} \]

substituting the expression (V) in (III), we have

\[ \begin{gather} \bar v=\frac{V}{A\Delta t} \tag{VI} \end{gather} \]

The density of charges d is given by

\[ d=\frac{n}{V} \]

where n is the number of electrons contained in volume V of the cylinder determined by the two cross-sections on the wire, we have

\[ \begin{gather} V=\frac{n}{d} \tag{VII} \end{gather} \]

substituting the expression (VII) into (VI), we obtain

\[ \bar v=\frac{n}{dA\Delta t} \]

Using the problem data, the value of n calculated in the previous item and the speed calculated per unit of time we have Δt = 1 s

\[ \begin{gather} \bar v=\frac{2.5\times 10^{19}}{1.8\times 10^{20}\times 0.50.10^{-2}\times 1}\\ \bar v=\frac{2.5\times 10^{19}}{9\times 10^{17}}\\ \bar v=\frac{2.5\times 10^{19}\times 10^{-17}}{9}\\ \bar v=0.28\times 10^{2} \end{gather} \]

\[ \bbox[#FFCCCC,10px] {\bar v=28\;\text{cm/s}} \]

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