In conducting wire of radius `5mm`, resistivity `p-1.1xx10^(-8)Omega//m` and current of `5A` is flowing. Drift velocity of free electron is `1.1xx10^(-3)m//s` find out mobility of free elctron.

Options

A. `1.57m^(2)` volt/sec.
B. `1.25m^(2)` volt/sec.
C. `1.2m^(2)` volt/sec.
D. `2 m^(2)` volt/sec.

Answer

To find the mobility of free electrons in the conducting wire, we can follow these steps:

### Step 1: Identify the given values
- Radius of the wire, \( r = 5 \, \text{mm} = 5 \times 10^{-3} \, \text{m} \)
- Resistivity, \( \rho = 1.1 \times 10^{-8} \, \Omega \cdot \text{m} \)
- Current, \( I = 5 \, \text{A} \)
- Drift velocity, \( v_d = 1.1 \times 10^{-3} \, \text{m/s} \)

### Step 2: Calculate the cross-sectional area of the wire
The area \( A \) of the wire can be calculated using the formula for the area of a circle:
\[ A = \pi r^2 \]
Substituting the radius:
\[ A = \pi (5 \times 10^{-3})^2 = \pi \times 25 \times 10^{-6} \, \text{m}^2 = 25\pi \times 10^{-6} \, \text{m}^2 \]

### Step 3: Calculate the electric field \( E \)
The electric field \( E \) can be expressed in terms of resistivity, current, and area:
\[ E = \rho \frac{I}{A} \]
Substituting the values:
\[ E = 1.1 \times 10^{-8} \, \Omega \cdot \text{m} \cdot \frac{5 \, \text{A}}{25\pi \times 10^{-6} \, \text{m}^2} \]
Calculating the denominator:
\[ E = 1.1 \times 10^{-8} \cdot \frac{5}{25\pi \times 10^{-6}} = 1.1 \times 10^{-8} \cdot \frac{5}{25\pi} \cdot 10^6 \]
\[ E = \frac{1.1 \times 5 \times 10^{-2}}{25\pi} \approx \frac{5.5 \times 10^{-2}}{25\pi} \]

### Step 4: Calculate the mobility \( \mu \)
The mobility \( \mu \) of free electrons is given by:
\[ \mu = \frac{v_d}{E} \]
Substituting \( E \) from the previous step:
\[ \mu = \frac{1.1 \times 10^{-3}}{E} \]
Now substituting the expression for \( E \):
\[ \mu = \frac{1.1 \times 10^{-3}}{\frac{1.1 \times 5 \times 10^{-2}}{25\pi}} = \frac{1.1 \times 10^{-3} \cdot 25\pi}{1.1 \times 5 \times 10^{-2}} \]
The \( 1.1 \) cancels out:
\[ \mu = \frac{25\pi \times 10^{-3}}{5 \times 10^{-2}} = \frac{25\pi}{5} \times 10^{-1} = 5\pi \times 10^{-1} \]
Calculating \( 5\pi \):
\[ \mu \approx 5 \times 3.14 \times 10^{-1} \approx 15.7 \times 10^{-1} \approx 1.57 \, \text{m}^2/\text{V} \cdot \text{s} \]

### Final Answer
The mobility of the free electron is approximately:
\[ \mu \approx 1.57 \, \text{m}^2/\text{V} \cdot \text{s} \]