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

The relation between current, number density of free electrons, cross-sectional area, and drift velocity is $$I = nAv_d e$$, where $$n$$ is the number of free electrons per cubic metre, $$A$$ is the cross-sectional area, $$v_d$$ is the drift velocity, and $$e = 1.6 \times 10^{-19}$$ C is the electron charge.

Given: $$I = 10$$ A, $$A = 5 \text{ mm}^2 = 5 \times 10^{-6} \text{ m}^2$$, $$v_d = 2 \times 10^{-3}$$ m/s. Solving for $$n$$: $$n = \frac{I}{Av_d e} = \frac{10}{5 \times 10^{-6} \times 2 \times 10^{-3} \times 1.6 \times 10^{-19}}$$.

Computing the denominator: $$5 \times 10^{-6} \times 2 \times 10^{-3} = 10^{-8}$$, and $$10^{-8} \times 1.6 \times 10^{-19} = 1.6 \times 10^{-27}$$. Therefore $$n = \frac{10}{1.6 \times 10^{-27}} = 6.25 \times 10^{27} = 625 \times 10^{25}$$.

The correct answer is option 2: $$625 \times 10^{25}$$.