A current of 10 A exists in a wire of cross-sectional area of 5 mm 2 with a drift velocity of 2 × 10 - 3 m s - 1 . The number of free electrons in each cubic meter of the wire is

Step-by-step solution

i = 10   A ,   A = 5   mm 2 = 5 × 10 - 6   m 2 and v d = 2 × 10 - 3   m   s - 1 We know, i = n e A v d ∴ 10 = n × 1 . 6 × 10 - 19 × 5 × 10 - 6 × 2 × 10 - 3 ⇒ n = 0 . 625 × 10 28 = 625 × 10 25