Resistivity of potentiometer wire is `10^(-7)` ohm metre and its area of cross-section is `10^(-6)m^2`. When a current `l=0.1A` flows through the wire, its potential gradient is:

Options

A. `10^(-2)V//m`
B. `10^(-4)V//m`
C. 0.1V/m
D. 10V/m

Answer

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

### Step 1: Understand the given parameters
We are given:
- Resistivity of the wire, \( \rho = 10^{-7} \, \Omega \, \text{m} \)
- Area of cross-section, \( A = 10^{-6} \, \text{m}^2 \)
- Current flowing through the wire, \( I = 0.1 \, \text{A} \)

### Step 2: Use the formula for resistance
The resistance \( R \) of a wire can be calculated using the formula:
\[ R = \frac{\rho L}{A} \]
However, we are interested in the resistance per unit length, \( \frac{R}{L} \):
\[ \frac{R}{L} = \frac{\rho}{A} \]

### Step 3: Substitute the values into the formula
Substituting the values of \( \rho \) and \( A \):
\[ \frac{R}{L} = \frac{10^{-7}}{10^{-6}} = 10^{-1} \, \Omega/\text{m} \]

### Step 4: Relate potential gradient to current and resistance
The potential gradient \( E \) is defined as the potential difference per unit length:
\[ E = \frac{V}{L} \]
Using Ohm's law, the potential difference \( V \) can be expressed as:
\[ V = I \cdot R \]
Thus,
\[ E = \frac{I \cdot R}{L} \]

### Step 5: Substitute \( R \) in terms of \( \frac{R}{L} \)
Now substituting \( R \) with \( \frac{R}{L} \):
\[ E = I \cdot \frac{R}{L} \]
Substituting the values:
\[ E = 0.1 \cdot 10^{-1} = 0.01 \, \text{V/m} \]

### Step 6: Convert to standard form
Expressing \( 0.01 \) in standard form:
\[ E = 10^{-2} \, \text{V/m} \]

### Final Answer
The potential gradient is:
\[ E = 10^{-2} \, \text{V/m} \]

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