A current of $2 \mathrm{~A}$ flows through a wire of cross-sectional area $25.0 \mathrm{~mm}^{2}$. The number of free electrons in a cubic meter are $2.0 \times 10^{28}$. The drift velocity of the electrons is __________ $\times 10^{-6} \mathrm{~ms}^{-1}$ (given, charge on electron $=1.6 \times 10^{-19} \mathrm{C}$ ).

<p>The drift velocity $v_d$ can be found using the formula for current $I$ in a conductor:</p> <p>$I = nqAv_d$,</p> <p>where:</p> <ul> <li>$n$ is the number density of free electrons (number of free electrons per unit volume),</li> <li>$q$ is the charge of an electron,</li> <li>$A$ is the cross-sectional area of the conductor, and</li> <li>$v_d$ is the drift velocity of the electrons.</li> </ul> <p>We can rearrange the above formula to solve for $v_d$:</p> <p>$v_d = \frac{I}{nqA}$.</p> <p>Given that $I = 2 \, \text{A}$, $n = 2.0 \times 10^{28} \, \text{m}^{-3}$, $q = 1.6 \times 10^{-19} \, \text{C}$, and $A = 25.0 \, \text{mm}^{2} = 25.0 \times 10^{-6} \, \text{m}^{2}$, we can substitute these values into the formula to find $v_d$:</p> <p>$v_d = \frac{2}{(2.0 \times 10^{28})(1.6 \times 10^{-19})(25.0 \times 10^{-6})}$</p> <p>$v_d = 25 \times 10^{-6} \, \text{ms}^{-1}$.</p> <p>Therefore, the drift velocity of the electrons is $25 \times 10^{-6} \, \text{ms}^{-1}$.</p>