Question

A 10 gauge copper wire carries a current of 26 A. Assuming one free electron per copper atom, calculate the magnitude of the drift velocity of the electrons.

Answer 1

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 $mm^{2}$

= $m^{2}$

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 =$\frac{1 e^{-} }{atom} * \frac{6.02*10^{23} atoms}{mol} *\frac{1 mol}{63.54 *10^{-3} Kg} *\frac{8.80*10^{3} kg}{m^{3} }$

= 8.3374 * 10^28 $e^{-} /m^{3}$

drift velocity of the electrons =

$=\frac{I}{nqa} =\frac{26}{8.3374 * 10 ^{28}*(-1.60*10^{-19})* 5.26 * 10^{-6}} \\\\= -3.71 * 10^{-4}$