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Step-by-Step Solution
Step 1: Identify the Known Quantities
• Current, $I = 5\,\text{A}$
• Resistivity of copper, $\rho = 1.7 \times 10^{-8}\,\Omega\,\text{m}$
• Radius of the conductor, $r = 5\,\text{mm} = 5 \times 10^{-3}\,\text{m}$
• Drift velocity of charges, $v_d = 1.1 \times 10^{-3}\,\text{m/s}$
• We need to find the mobility of charges, $\mu$.
Step 2: Recall the Relationship for Mobility
Mobility $\mu$ is defined as
$$
\mu = \frac{v_d}{E},
$$
where $v_d$ is the drift velocity and $E$ is the electric field in the conductor.
Step 3: Express the Electric Field in Terms of Resistivity and Current Density
The electric field $E$ inside a conductor can be written as
$$
E = \rho \, J,
$$
where $J$ is the current density. The current density $J$ is given by
$$
J = \frac{I}{A},
$$
where $A$ is the cross-sectional area of the conductor.
Step 4: Calculate the Cross-Sectional Area
Since the conductor is circular in cross-section with radius $r$,
$$
A = \pi r^2 = \pi \,(5 \times 10^{-3}\,\text{m})^2 = \pi \times 25 \times 10^{-6}\,\text{m}^2 = 25\pi \times 10^{-6}\,\text{m}^2.
$$
Step 5: Determine the Current Density $J$
$$
J = \frac{I}{A}
= \frac{5\,\text{A}}{25\pi \times 10^{-6}\,\text{m}^2}
= \frac{5}{25\pi} \times 10^{6}\,\text{A/m}^2
= \frac{1}{5\pi} \times 10^{6}\,\text{A/m}^2.
$$
Step 6: Compute the Electric Field $E$
$$
E = \rho \, J
= \bigl(1.7 \times 10^{-8}\,\Omega\,\text{m}\bigr)
\times \left(\frac{1}{5\pi} \times 10^{6}\,\text{A/m}^2\right).
$$
Keep the expression symbolic for clarity in the next step.
Step 7: Calculate the Mobility $\mu$
Using
$$
\mu = \frac{v_d}{E},
$$
we substitute $v_d = 1.1 \times 10^{-3}\,\text{m/s}$ and $E$ from above. Numerically, this gives approximately
$$
\mu \approx 1.0\,\text{m}^2/\text{V}\cdot\text{s}.
$$
Step 8: Final Answer
Thus, the mobility of the charges in the given copper conductor is
$$
\mu \approx 1.0\,\text{m}^2/\text{V}\cdot\text{s}.
$$
