The length of potentiometer wire is 1m and its resistance is `4Omega`. Acurrent of 5mA is flowing in it. An unknown emf is balanced on 40 cm length of this wire. The unknown emf is x mv. What is the value of .x. ?

Answer

To solve the problem step by step, we will follow the outlined approach in the video transcript.

### Step 1: Identify the given values
- Length of the potentiometer wire (L) = 1 m
- Resistance of the wire (R) = 4 Ω
- Current flowing through the wire (I) = 5 mA = 0.005 A
- Length at which the unknown EMF is balanced (l1) = 40 cm = 0.4 m

### Step 2: Calculate the total potential (voltage) across the potentiometer wire using Ohm's law
Ohm's law states that \( V = I \times R \).

\[ V = 0.005 \, \text{A} \times 4 \, \Omega = 0.02 \, \text{V} = 20 \, \text{mV} \]

### Step 3: Calculate the potential gradient (k) of the wire
The potential gradient (k) is defined as the potential per unit length of the wire.

\[ k = \frac{V}{L} = \frac{20 \, \text{mV}}{1 \, \text{m}} = 20 \, \text{mV/m} \]

### Step 4: Calculate the unknown EMF (E) using the potential gradient
The unknown EMF can be calculated using the formula:

\[ E = k \times l1 \]

Substituting the values we have:

\[ E = 20 \, \text{mV/m} \times 0.4 \, \text{m} = 8 \, \text{mV} \]

### Step 5: Conclusion
The unknown EMF is 8 mV, thus the value of \( x \) is:

\[ \boxed{8} \]

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