The current in the primary circuit of a potentiometer is `0.2 A`. The specific resistance and cross-section of the potentiometer wire are `4xx10^(-7)` ohm meter and `8xx10^(-7)m^(2)` respectively. The potential gradient will be equal to -

Options

A. `0.2 v//m`
B. `1 v//m`
C. `0.5 v//m`
D. `0.1 v//m`

Answer

To find the potential gradient in the potentiometer wire, we can follow these steps:

### Step 1: Identify the given values
- Current (I) = 0.2 A
- Specific resistance (ρ) = \(4 \times 10^{-7} \, \Omega \cdot m\)
- Cross-sectional area (A) = \(8 \times 10^{-7} \, m^2\)

### Step 2: Use the formula for resistance
The resistance (R) of the wire can be calculated using the formula:
\[ R = \frac{\rho \cdot L}{A} \]
where:
- \(R\) is the resistance,
- \(ρ\) is the specific resistance,
- \(L\) is the length of the wire,
- \(A\) is the cross-sectional area.

### Step 3: Substitute the known values into the resistance formula
Substituting the values into the resistance formula gives:
\[ R = \frac{(4 \times 10^{-7}) \cdot L}{(8 \times 10^{-7})} \]
This simplifies to:
\[ R = \frac{4L}{8} = \frac{L}{2} \quad \text{(Equation 1)} \]

### Step 4: Calculate the potential difference (V)
Using Ohm's law, the potential difference (V) across the wire can be calculated as:
\[ V = I \cdot R \]
Substituting Equation 1 into this gives:
\[ V = I \cdot \frac{L}{2} \]
Substituting \(I = 0.2 A\):
\[ V = 0.2 \cdot \frac{L}{2} = 0.1L \]

### Step 5: Calculate the potential gradient (V/L)
The potential gradient (k) is defined as the potential difference per unit length:
\[ k = \frac{V}{L} \]
Substituting the expression for V:
\[ k = \frac{0.1L}{L} = 0.1 \, V/m \]

### Conclusion
Thus, the potential gradient is:
\[ \text{Potential Gradient} = 0.1 \, V/m \]