2 mA current is flowing through a potentiometer wire of 5m and `5 Omega` resistance. The potential gradient is :

Options

A. `1.6 xx 10^(-3)(V)/(m)`
B. `2 xx 10^(-3)(V)/(m)`
C. `2.5 xx 10^(-2)(V)/(m)`
D. `2.3 xx 10^(3)(V)/(m)`

Explanation

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

### Step 1: Understand the Concept of Potential Gradient
The potential gradient (V_g) is defined as the voltage drop (V) per unit length (L) of the wire. Mathematically, it is given by:
\[ V_g = \frac{V}{L} \]

### Step 2: Calculate the Total Resistance of the Wire
The resistance of the potentiometer wire is given as \( R = 5 \, \Omega \).

### Step 3: Calculate the Current Flowing Through the Wire
The current flowing through the potentiometer wire is given as \( I = 2 \, \text{mA} = 2 \times 10^{-3} \, \text{A} \).

### Step 4: Calculate the Voltage Drop Across the Wire
Using Ohm's Law, the voltage drop (V) across the wire can be calculated as:
\[ V = I \times R \]
Substituting the values:
\[ V = (2 \times 10^{-3} \, \text{A}) \times (5 \, \Omega) \]
\[ V = 10 \times 10^{-3} \, \text{V} = 0.01 \, \text{V} \]

### Step 5: Calculate the Length of the Wire
The length of the potentiometer wire is given as \( L = 5 \, \text{m} \).

### Step 6: Calculate the Potential Gradient
Now, we can find the potential gradient using the formula from Step 1:
\[ V_g = \frac{V}{L} \]
Substituting the values we calculated:
\[ V_g = \frac{0.01 \, \text{V}}{5 \, \text{m}} \]
\[ V_g = 0.002 \, \text{V/m} = 2 \times 10^{-3} \, \text{V/m} \]

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