An electron takes 40 × 10 3 s to drift from one end of a metal wire of length 2 m to its other end. The area of cross-section of the wire is 4 mm 2 and it is carrying a current of 1 . 6 A . The number density of free electrons in the metal wire is

Step-by-step solution

From the given data, drift velocity can be written as, v = 2   m 40 × 10 3   s = 1 2 × 10 4   m   s - 1 Now for current we can write, I = n e A v ⇒ n = I e A v ⇒ n = 1 . 6 1 . 6 × 10 - 19 4 × 10 - 6 1 2 × 10 4 = 5 × 10 28   m - 3