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Question: Copper has one conduction electron per atom. Its density is \[8.89g/{m^3}\] and its atomic mass is \...

Copper has one conduction electron per atom. Its density is 8.89g/m38.89g/{m^3} and its atomic mass is 6.3g/mol6.3g/mol. if a copper wire of diameter 1.0mm1.0mm carries a current of 2.0A2.0A. What is the drift speed of the electrons in the wire?
A. 1.9×1041.9 \times {10^{ - 4}}
B. 2.9×1042.9 \times {10^{ - 4}}
C. 1.9×1041.9 \times {10^4}
D. 2.9×1042.9 \times {10^4}

Explanation

Solution

The speed of electricity is determined as the current of an electric charge traveling through a metal wire. The speed of the electrons in the wire is based on the number of electrons passing by one section of the wire per second, the density of the electron, and the cross-section area of the wire.
The current of the wire related to I=ΔQΔtI = \dfrac{{\Delta Q}}{{\Delta t}}

Formula used:
From the formula of the current of the wire, we can calculate the speed of the electron in the wire Vd=IAqne{V_d} = \dfrac{I}{{Aq{n_e}}}

Complete step by step solution:
Let the density of the copper \rho = 8.89g/c{m^3}$$$$ = 8.89 \times {10^6}g/{m^3}
The atomic mass of the copper is M=63.54g/molM = 63.54g/mol the diameter of the copper wire d = 1.0mm$$$$ = {10^{ - 3}}m
And, carried current of I=2.0AI = 2.0A
We know, Avogadro constant NA=6.023×1023{N_A} = 6.023 \times {10^{23}}
\therefore Copper atoms per unit volume ncu=ρNAM{n_{cu}} = \dfrac{{\rho {N_A}}}{M}
{n_{cu}} = \dfrac{{8.89 \times {{10}^6} \times 6.023 \times {{10}^{23}}}}{{63.54}}$$$$ = 8.43 \times {10^{28}}$$$$Cu atoms per m3{m^3}
One copper atom gives one conduction electron.
\therefore Number of conduction electrons ne=8.43×1028{n_e} = 8.43 \times {10^{28}} electron m3{m^3}
The cross-section area of the wire A=πd24A = \dfrac{{\pi {d^2}}}{4}
\Rightarrow A = \dfrac{{\pi {{\left( {{{10}^{ - 3}}} \right)}^2}}}{4}$$$$ = 7.85 \times {10^{ - 7}}{m^2}
\therefore The drift speed of the electrons in the wire Vd=IAqne{V_d} = \dfrac{I}{{Aq{n_e}}}
\Rightarrow {V_d} = \dfrac{2}{{\left( {7.85 \times {{10}^{ - 7}}} \right)\left( {1.6 \times {{10}^{ - 19}}} \right)\left( {8.43 \times {{10}^{28}}} \right)}}$$$$ = 1.9 \times {10^{ - 4}}m/s
\therefore The correct answer is option A.

Note:
There are three different velocities present when the electrical current travels through a metal wire.

  1. The individual electron velocity. 2) The electron drift velocity. 3) The signal velocity.
    The speed of an electron in a wire is slow compared to the speed of light in air. The resistance of the wire has only a small effect almost negligible on the speed of a signal in a transmission line.