An electron in potentiometer experiences a force `2.4xx10^(-19)N`. The length of potentiometer wire is 6m. The emf of the battery connected across the wire is (electronic charge `=1.6xx10^(-19)C`)

Options

A. 6 V
B. 9 V
C. 12 V
D. 15 V

Answer

To find the EMF of the battery connected across the potentiometer wire, we can follow these steps:

### Step 1: Understand the relationship between force, electric field, and charge
The force \( F \) experienced by a charge \( Q \) in an electric field \( E \) is given by the equation:
\[ F = Q \cdot E \]

### Step 2: Express the electric field in terms of EMF and length
The electric field \( E \) across the potentiometer wire can also be expressed in terms of the EMF \( E \) and the length \( L \) of the wire:
\[ E = \frac{E_{MF}}{L} \]
where \( E_{MF} \) is the electromotive force (EMF) of the battery.

### Step 3: Substitute the expression for electric field into the force equation
Substituting the expression for \( E \) into the force equation, we have:
\[ F = Q \cdot \left(\frac{E_{MF}}{L}\right) \]

### Step 4: Rearrange the equation to solve for EMF
Rearranging the equation to solve for EMF \( E_{MF} \):
\[ E_{MF} = \frac{F \cdot L}{Q} \]

### Step 5: Substitute the known values into the equation
Given:
- \( F = 2.4 \times 10^{-19} \, \text{N} \)
- \( L = 6 \, \text{m} \)
- \( Q = 1.6 \times 10^{-19} \, \text{C} \)

Substituting these values into the equation:
\[ E_{MF} = \frac{(2.4 \times 10^{-19} \, \text{N}) \cdot (6 \, \text{m})}{1.6 \times 10^{-19} \, \text{C}} \]

### Step 6: Calculate the EMF
Calculating the numerator:
\[ 2.4 \times 10^{-19} \times 6 = 14.4 \times 10^{-19} \]
Now, divide by \( Q \):
\[ E_{MF} = \frac{14.4 \times 10^{-19}}{1.6 \times 10^{-19}} = 9 \, \text{V} \]

### Final Answer
The EMF of the battery is:
\[ E_{MF} = 9 \, \text{V} \]