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Step-by-Step Solution
Step 1: List the Given Data
• Current, $I = 1.5\,\text{A}$
• Electron density, $n = 9 \times 10^{28}\,\text{m}^{-3}$
• Charge of an electron, $e = 1.6 \times 10^{-19}\,\text{C}$
• Cross-sectional area, $A = 5\,\text{mm}^2 = 5 \times 10^{-6}\,\text{m}^2$
Step 2: Use the Formula Relating Drift Velocity and Current
The relation between current $I$, number density $n$, charge $e$, cross-sectional area $A$, and drift velocity $v_d$ is
$I = n\,e\,A\,v_d.$
Rearranging to find the drift velocity $v_d$:
$v_d = \frac{I}{n\,e\,A}.$
Step 3: Substitute the Values
Substitute the known values into the formula:
$
v_d = \frac{1.5}{\bigl(9 \times 10^{28}\bigr)\,\bigl(1.6 \times 10^{-19}\bigr)\,\bigl(5 \times 10^{-6}\bigr)}.
$
Step 4: Simplify the Denominator
1. First multiply $n$ and $e$:
$
\bigl(9 \times 10^{28}\bigr)\,\bigl(1.6 \times 10^{-19}\bigr) = 14.4 \times 10^{9} = 1.44 \times 10^{10}.
$
2. Now multiply by $A$:
$
\bigl(1.44 \times 10^{10}\bigr)\,\bigl(5 \times 10^{-6}\bigr) = 7.2 \times 10^{4}.
$
Step 5: Compute the Drift Velocity
Hence,
$
v_d = \frac{1.5}{7.2 \times 10^{4}} \approx 2.08 \times 10^{-5}\,\text{m/s}.
$
Step 6: Convert to mm/s (If Required)
To convert from $\,\text{m/s}$ to $\,\text{mm/s}$, multiply by $1000$:
$
v_d = 2.08 \times 10^{-5}\,\text{m/s} \times 1000 = 0.0208\,\text{mm/s}.
$
Rounded suitably, $v_d \approx 0.02\,\text{mm/s}.$
Final Answer
The drift velocity in mm/s is approximately $0.02.$
