The drift velocity of a current carrying conductor is `V_d` if current flowing through the wire is halved then the value of drift velocity is:

Options

A. remains same
B. four times
C. doubled
D. halved

Explanation

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

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

Where:
- \( I \) is the current,
- \( n \) is the number of charge carriers per unit volume,
- \( A \) is the cross-sectional area of the conductor,
- \( e \) is the charge of an electron,
- \( V_d \) is the drift velocity.

### Step-by-Step Solution:

1. **Understand the relationship**: From the formula \( I = n \cdot A \cdot e \cdot V_d \), we can see that the current \( I \) is directly proportional to the drift velocity \( V_d \) when \( n \), \( A \), and \( e \) are constant.

2. **Halve the current**: According to the problem, the current flowing through the wire is halved. This means:
\[ I' = \frac{I}{2} \]
where \( I' \) is the new current.

3. **Relate the new current to drift velocity**: Since \( I \) is directly proportional to \( V_d \), we can express the new drift velocity \( V_d' \) in terms of the original drift velocity \( V_d \):
\[ I' = n \cdot A \cdot e \cdot V_d' \]

4. **Substituting the new current**: Now substituting \( I' \) into the equation:
\[ \frac{I}{2} = n \cdot A \cdot e \cdot V_d' \]

5. **Expressing \( V_d' \)**: Since \( I = n \cdot A \cdot e \cdot V_d \), we can replace \( I \) in the equation:
\[ \frac{n \cdot A \cdot e \cdot V_d}{2} = n \cdot A \cdot e \cdot V_d' \]

6. **Canceling common terms**: We can cancel \( n \), \( A \), and \( e \) from both sides (assuming they are not zero):
\[ \frac{V_d}{2} = V_d' \]

7. **Conclusion**: Thus, the new drift velocity \( V_d' \) is:
\[ V_d' = \frac{V_d}{2} \]

### Final Answer:
The drift velocity is halved. Therefore, the correct option is that the drift velocity becomes \( \frac{V_d}{2} \).