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 find the mobility of the charges in the copper conductor, we can follow these steps:

### Step 1: Identify the given values
- Current (I) = 5 A
- Resistivity (ρ) = \(1.7 \times 10^{-8} \, \Omega \cdot m\)
- Radius of the conductor (r) = 5 mm = \(5 \times 10^{-3} \, m\)
- Drift velocity (v_d) = \(1.1 \times 10^{-3} \, m/s\)

### Step 2: Calculate the cross-sectional area (A) of the conductor
The cross-sectional area of a circular conductor is given by the formula:
\[ A = \pi r^2 \]
Substituting the radius:
\[ A = \pi (5 \times 10^{-3})^2 \]
\[ A = \pi (25 \times 10^{-6}) \]
\[ A \approx 7.85 \times 10^{-5} \, m^2 \]

### Step 3: Calculate the electric field (E) using Ohm's Law
From Ohm's Law, we know:
\[ V = IR \]
Where \( R \) can be expressed as:
\[ R = \frac{\rho L}{A} \]
Thus,
\[ V = I \cdot \frac{\rho L}{A} \]
We can relate the voltage to the electric field:
\[ V = E \cdot L \]
So,
\[ E = \frac{I \cdot \rho}{A} \]
Now substituting the values:
\[ E = \frac{5 \cdot (1.7 \times 10^{-8})}{7.85 \times 10^{-5}} \]
Calculating this gives:
\[ E \approx 1.08 \times 10^{-3} \, V/m \]

### Step 4: Calculate the mobility (μ)
The mobility of the charge carriers is given by the relationship:
\[ \mu = \frac{v_d}{E} \]
Substituting the values:
\[ \mu = \frac{1.1 \times 10^{-3}}{1.08 \times 10^{-3}} \]
Calculating this gives:
\[ \mu \approx 1.0185 \, m^2/Vs \]
Rounding off, we can say:
\[ \mu \approx 1 \, m^2/Vs \]

### Final Answer
The mobility of the charges is approximately \( 1 \, m^2/Vs \).

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