Answer:
-3.71 * 10^-4 m/s
Step-by-step explanation:
A 10 gauge copper wire has a standard cross sectional area of 5.26 [tex]mm^{2}[/tex]
= [tex]m^{2}[/tex]
density of copper, is 8.80 × 103 kg/m3
The atomic mass of copper is 63.54 * 10^-3 Kg/mol.
Avogadro’s number, 6.02 × 1023 atoms/mol
Density of copper's free electrons =[tex]\frac{1 e^{-} }{atom} * \frac{6.02*10^{23} atoms}{mol} *\frac{1 mol}{63.54 *10^{-3} Kg} *\frac{8.80*10^{3} kg}{m^{3} } \\\\[/tex]
= 8.3374 * 10^28 [tex]e^{-} /m^{3}[/tex]
drift velocity of the electrons =
[tex]=\frac{I}{nqa} =\frac{26}{8.3374 * 10 ^{28}*(-1.60*10^{-19})* 5.26 * 10^{-6}} \\\\= -3.71 * 10^{-4}[/tex]
The drift velocity of the electrons is , [tex]v_{d}=3.7*10^{-4} m/s[/tex]
The drift velocity is given as,
[tex]v_{d}=\frac{I}{neA}[/tex]
Where,
The cross sectional area of standard gauge wire is [tex]5.26mm^{2}[/tex]
Given that, [tex]A=5.26mm^{2}=5.26*10^{-6} m^{2} ,I=26A,n=8.33*10^{28} ,e=1.6*10^{-19}[/tex]
Substitute values in above relation,
[tex]v_{d}=\frac{26}{8.33*10^{28}*1.6*10^{-19}*5.26*10^{-6} } \\\\v_{d}=3.7*10^{-4} m/s[/tex]
Learn more about the drift velocity here:
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