A current of 5 A passss through a copper conductor (resistivity = `1.7 xx 10^(-8) Omega m`) of radius of cross -section 5mm. Find the mobility of the charges if their drift velocity is `1.1 xx 10^(-3) m//s`.

Options

A. `1.0 m^(2)//Vs`
B. `1.3 m^(2)//Vs`
C. `1.5 m^(2)//Vs`
D. `1.8 m^(2)//Vs`

Explanation

To solve the problem, we need to find the mobility of the charges in the copper conductor. We will use the relationship between current, drift velocity, electric field, and mobility.

### Step-by-Step Solution:

1. **Given Data:**
- Current, \( I = 5 \, \text{A} \)
- Resistivity, \( \rho = 1.7 \times 10^{-8} \, \Omega \, \text{m} \)
- Radius of cross-section, \( r = 5 \, \text{mm} = 5 \times 10^{-3} \, \text{m} \)
- Drift velocity, \( v_d = 1.1 \times 10^{-3} \, \text{m/s} \)

2. **Calculate the Area of Cross-section (A):**
\[ A = \pi r^2 = \pi (5 \times 10^{-3})^2 = \pi (25 \times 10^{-6}) \, \text{m}^2 = 25\pi \times 10^{-6} \, \text{m}^2 \]

3. **Calculate the Electric Field (E):**
The electric field can be calculated using the formula:
\[ E = \frac{I \cdot \rho}{A} \]
Substituting the values:
\[ E = \frac{5 \times (1.7 \times 10^{-8})}{25\pi \times 10^{-6}} \]
Simplifying:
\[ E = \frac{8.5 \times 10^{-8}}{25\pi \times 10^{-6}} = \frac{8.5}{25\pi} \times 10^{-2} \, \text{V/m} \]

4. **Calculate Mobility (μ):**
The mobility can be calculated using the relationship:
\[ \mu = \frac{v_d}{E} \]
Substituting the values:
\[ \mu = \frac{1.1 \times 10^{-3}}{\frac{8.5}{25\pi} \times 10^{-2}} \]
Simplifying:
\[ \mu = \frac{1.1 \times 10^{-3} \times 25\pi}{8.5 \times 10^{-2}} = \frac{27.5\pi \times 10^{-3}}{8.5 \times 10^{-2}} \]
\[ \mu = \frac{27.5\pi}{8.5} \times 10^{-1} \]
Approximating \( \pi \approx 3.14 \):
\[ \mu \approx \frac{27.5 \times 3.14}{8.5} \times 10^{-1} \approx 1.015 \, \text{m}^2/\text{V.s} \]

### Final Answer:
The mobility of the charges is approximately \( \mu \approx 1.015 \, \text{m}^2/\text{V.s} \).