Answer:
21.6 A
Step-by-step explanation:
n = number density of free electrons in copper = 8.5 x 10²² cm⁻³ = 8.5 x 10²⁸ m⁻³
e = magnitude of charge on electron = 1.6 x 10⁻¹⁹ C
d = diameter of copper wire = 0.790 mm = 0.790 x 10⁻³ m
Area of cross-section of copper wire is given as
A = (0.25) πd²
A = (0.25) (3.14) (0.790 x 10⁻³)²
A = 4.89 x 10⁻⁷ m²
v = drift speed = 3.25 mm/s = 3.25 x 10⁻³ m /s
the electric current is given as
i = n e A v
i = (8.5 x 10²⁸) (1.6 x 10⁻¹⁹) (4.89 x 10⁻⁷ ) (3.25 x 10⁻³)
i = 21.6 A