Answer –

It is given that

Number density of free electrons in a copper conductor is n =  8.5 x 10 ²⁸ m ^{– 3}

Assume that the Length of the copper wire is denoted by l and we have l = 3.0 m

Let A denote the area of cross – section of the wire = 2.0 x 10 ^{– 6} m ²

Value of the current carried by the wire is  I = 3.0 A

The current in the wire is given  by the equation –

I = n A e V _d

Where,                    

e = electric charge = 1.6 x 10 ^{– 19} C

V _d is the drift velocity of electrons and is given by

V _d  = l/t

So, the expression of current becomes –

\[I=nAe\frac{l}{t}\]

\[\therefore t=\frac{nAel}{I}=\frac{3\times 8.5\times {{10}^{28}}\times 2\times {{10}^{-6}}\times 1.6\times {{10}^{-19}}}{3.0}\]

\[t=2.7\times {{10}^{4}}\sec \]