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

<p>Using the formula:</p> <p>$ I = n e A v $</p> <p>where $I$ is the current, $n$ is the number density of free electrons, $e$ is the charge of an electron, $A$ is the cross-sectional area of the wire, and $v$ is the drift speed of the electrons.</p> <p>We can isolate $v$ to find:</p> <p>$ v = \frac{I}{n e A}$</p> <p>Substituting the given values:</p> <p>$ v = \frac{3.2 \, \text{A}}{8 \times 10^{28} \, \text{m}^{-3} \times 1.6 \times 10^{-19} \, \text{C} \times 2 \times 10^{-6} \, \text{m}^2}$</p> <p>This simplifies to:</p> <p>$ v = \frac{3.2}{16 \times 1.6 \times 10^3} \, \text{ms}^{-1} = 125 \times 10^{-6} \, \text{ms}^{-1} $</p> <p>So, the drift speed of the electrons is indeed $125 \times 10^{-6} \, \text{ms}^{-1}$.</p>