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Step-by-Step Solution
Step 1: Identify the Known Quantities
• Current, $I = 10\,\text{A}$
• Cross-sectional area, $A = 5\,\text{mm}^2 = 5 \times 10^{-6}\,\text{m}^2$
• Drift velocity, $v_d = 2 \times 10^{-3}\,\text{m s}^{-1}$
• Charge of an electron, $e = 1.6 \times 10^{-19}\,\text{C}$
Step 2: Write the Relevant Formula
The current $I$ in a conductor due to moving electrons is given by:
$I = n \, e \, A \, v_d$
where $n$ is the number of free electrons per cubic meter.
Step 3: Rearrange the Formula to Solve for n
Rearranging, we get:
$n = \frac{I}{e \, A \, v_d}$
Step 4: Substitute the Numerical Values
Substitute $I = 10\,\text{A}$, $e = 1.6 \times 10^{-19}\,\text{C}$, $A = 5 \times 10^{-6}\,\text{m}^2$, and $v_d = 2 \times 10^{-3}\,\text{m s}^{-1}$:
$
n = \frac{10}{(1.6 \times 10^{-19}) \times (5 \times 10^{-6}) \times (2 \times 10^{-3})}
$
Step 5: Calculate and Simplify
First, multiply the denominator:
$
(1.6 \times 10^{-19}) \times (5 \times 10^{-6}) \times (2 \times 10^{-3}) = 1.6 \times 5 \times 2 \times 10^{-19 -6 -3} = 16 \times 10^{-28} = 1.6 \times 10^{-27}
$
Now, divide:
$
n = \frac{10}{1.6 \times 10^{-27}} = \frac{10}{1.6} \times 10^{27} \approx 6.25 \times 10^{27}
$
Finally, we can also write this as:
$
n \approx 625 \times 10^{25}
$
Final Answer
The number of free electrons per cubic meter in the wire is $625 \times 10^{25}$.
