An aluminum wire having a cross-sectional area equal to 3.90 10-6 m2 carries a current of 6.00 A. The density of aluminum is 2.70 g/cm3. Assume each aluminum atom supplies one conduction electron per atom. Find the drift speed of the electrons in the wire.

Vd = 1.597 ×10⁻⁴ m/s

Explanation:

Given: A = 3.90×10⁻⁶ m², I = 6.00 A, ρ = 2.70 g/cm³

To find:

Drift Velocity Vd=?

Solution:

the formula is Vd = I/nqA         (n is the number of charge per unit volume)

n = No. of electron in a mole ( Avogadro's No.) / Volume

Volume = Molar mass / density   ( molar mass of Al =27 g)

V = 27 g / 2.70 g/cm³ = 10 cm³ = 1 × 10 ⁻⁵ m³

n= (6.02 × 10 ²³) / (1 × 10 ⁻⁵ m³)

n= 6.02 × 10 ²⁸

Now

Vd = (6A) / (  6.02 × 10 ²⁸ ×  1.6 × 10⁻¹⁹ C ×  3.9×10⁻⁶ m²)

Vd = 1.597 ×10⁻⁴ m/s

answer: i pick a

explanation:

this means that the temperature of both substances will be equal