When current is 4 A in a conductor the drift velocity is `v_(d)`. When the current is 3 A the drift velocity will be

Options

A. `(3)/(4)v_(d)`
B. `v_(d)`
C. `(4)/(3)v_(d)`
D. None of these

Answer

To solve the problem, we need to understand the relationship between current (I) and drift velocity (v_d) in a conductor. The relationship is given by the formula:

\[ I = n \cdot e \cdot A \cdot v_d \]

where:
- \( I \) is the current,
- \( n \) is the number density of charge carriers (electrons),
- \( e \) is the charge of an electron,
- \( A \) is the cross-sectional area of the conductor,
- \( v_d \) is the drift velocity.

### Step-by-Step Solution:

1. **Identify the Given Values:**
- When the current \( I_1 = 4 \, \text{A} \), the drift velocity is \( v_d \).
- We need to find the drift velocity \( v_{d2} \) when the current \( I_2 = 3 \, \text{A} \).

2. **Assume Constants:**
- Since nothing is mentioned about changes in the conductor, we can assume that the number density of electrons \( n \), the charge of an electron \( e \), and the area \( A \) remain constant.

3. **Establish the Relationship:**
- Since \( I \) is directly proportional to \( v_d \) (as \( n \), \( e \), and \( A \) are constants), we can write:
\[ \frac{I_2}{I_1} = \frac{v_{d2}}{v_{d1}} \]

4. **Substitute the Known Values:**
- Substitute \( I_1 = 4 \, \text{A} \) and \( I_2 = 3 \, \text{A} \):
\[ \frac{3}{4} = \frac{v_{d2}}{v_d} \]

5. **Solve for \( v_{d2} \):**
- Rearranging the equation gives:
\[ v_{d2} = v_d \cdot \frac{3}{4} \]

6. **Conclusion:**
- Therefore, when the current is 3 A, the drift velocity \( v_{d2} \) is:
\[ v_{d2} = \frac{3}{4} v_d \]

### Final Answer:
The drift velocity when the current is 3 A will be \( \frac{3}{4} v_d \).

---