The number density of free electrons in copper is nearly $$8 \times 10^{28}$$ m$$^{-3}$$. A copper wire has its area of cross-section $$= 2 \times 10^{-6}$$ m$$^2$$ and is carrying a current of 3.2 A. The drift speed of the electrons is _______ $$\times 10^{-6}$$ m s$$^{-1}$$.

We have number density $$n = 8 \times 10^{28}$$ m$$^{-3}$$, area $$A = 2 \times 10^{-6}$$ m$$^2$$, and current $$I = 3.2$$ A.

The relation between current and drift speed is:

$$I = neAv_d$$

Solving for drift speed:

$$v_d = \frac{I}{neA}$$

$$v_d = \frac{3.2}{8 \times 10^{28} \times 1.6 \times 10^{-19} \times 2 \times 10^{-6}}$$

$$v_d = \frac{3.2}{8 \times 1.6 \times 2 \times 10^{28-19-6}}$$

$$v_d = \frac{3.2}{25.6 \times 10^{3}}$$

$$v_d = \frac{3.2}{25600} = 1.25 \times 10^{-4} \text{ m/s} = 125 \times 10^{-6} \text{ m/s}$$

So, the answer is $$125$$.