Question

An aluminium wire with a cross-sectional area of 2.6 × 10−6 m2 carries a current of 1.9 A . Find the drift speed of the electrons in the wire. Assume that each atom supplies one electron. Aluminium has a molecular weight of 26.98 g/mol and a density of 2.7 g/cm3 . Avogadro’s number is 6.022 × 1023 and the fundamental charge is 1.602 × 10−19 C . Answer in units of m/s.

Answer 1

Answer:

$v_d = 7.6 \times 10^{-5} m/s$

Explanation:

As we know that electric current is given as

$i = neA v_d$

here we know that

$A = 2.6 \times 10^{-6} m^2$

i = 1.9 A

$e = 1.6 \times 10^{-19} C$

Molar mass of the wire is

M = 26.98 g/mol

density of the wire is

$\rho = 2.7 g/cm^3$

now let the mass of the wire is M gram

so number of atoms of Al in the wire is given as

$N = \frac{M}{26.98}(6.02 \times 10^{23})$

now number density is given as

$n = \frac{N}{V}$

$n = \frac{\rho}{26.98}(6.02 \times 10^{23})$

$n = 6.02 \times 10^{22} per cm^3$

$n = 6.02 \times 10^{28} per m^3$

now we will have

$i = neAv_d$

$1.9 = (6.02 \times 10^{28})(1.6 \times 10^{-19})(2.6 \times 10^{-6})v_d$

$v_d = 7.6 \times 10^{-5} m/s$