Area of cross-section of a copper wire is equal to area of a square of $2 \mathrm{~mm}\( length It carries a current of \)8 \mathrm{~A}$ Find drift velocity of electrons (Density of free electrons in copper $\left.=8 \times 10^{28} \mathrm{~m}^{-3}\right)$ (A) \(1.56 \times 10^{-2} \mathrm{~ms}^{-1}\) (B) \(1.56 \times 10^{-4} \mathrm{~ms}^{-1}\) (C) \(3.12 \times 10^{-2} \mathrm{~ms}^{-1}\) (D) \(3.12 \times 10^{-3} \mathrm{~ms}^{-1}\)

Given the area of cross-section of the copper wire is that of a square with 2 mm side length, we can convert this into square meters: Area = (2 mm)² = (2 x 10^{-3} m)² = 4 x 10^{-6} m²

Now, we can calculate the drift velocity using the formula mentioned above: \(v_d = \frac{I}{nAe}\) Plugging in the given values, we have: \(v_d = \frac{8 \,\text{A}}{(8 \times 10^{28} \,\text{m}^{-3})(4 \times 10^{-6} \,\text{m}^2)(1.6 \times 10^{-19} \,\text{C})}\) \(v_d = \frac{8}{(8 \times 10^{28})(4 \times 10^{-6})(1.6 \times 10^{-19})}\) \(v_d = \frac{1}{(4)(1.6 \times 10^{-13})} \,\text{m/s}\) \(v_d = \frac{1}{(6.4 \times 10^{-13})} \,\text{m/s}\) \(v_d = 1.56 \times 10^{-12} \,\text{m/s}\) Now, we need to convert the drift velocity into ms^{-1}. \(v_d = 1.56 \times 10^{-12} \times 10^9 \,\text{ms}^{-1}\) \(v_d = 1.56 \times 10^{-3} \,\text{ms}^{-1}\) The correct option is: (D) \(1.56 \times 10^{-3} \,\text{ms}^{-1}\)