A current 'i' flow through a metallic wire of radius 'r' and area 'A', the free electrons in it drift with a velocity `'v_d'`:
If the drift velocity is doubled, then the current will become:

Options

A. four times
B. doubled
C. halved
D. none of these

Explanation

To solve the problem, we need to understand the relationship between current, drift velocity, and other parameters in a metallic wire.

### Step-by-Step Solution:

1. **Identify the formula for current (I)**:
The current \( I \) flowing through a metallic wire is given by the formula:
\[ I = n \cdot e \cdot A \cdot v_d \]
where:
- \( n \) = number of free charge carriers (electrons) per unit volume,
- \( e \) = charge of an electron,
- \( A \) = cross-sectional area of the wire,
- \( v_d \) = drift velocity of the electrons.

2. **Understand the effect of doubling the drift velocity**:
According to the problem, the drift velocity \( v_d \) is doubled. Therefore, we can express the new drift velocity as:
\[ v_d' = 2v_d \]

3. **Substitute the new drift velocity into the current formula**:
We can now substitute \( v_d' \) into the current formula to find the new current \( I' \):
\[ I' = n \cdot e \cdot A \cdot v_d' = n \cdot e \cdot A \cdot (2v_d) \]

4. **Simplify the expression for the new current**:
By simplifying the expression, we get:
\[ I' = 2(n \cdot e \cdot A \cdot v_d) = 2I \]
This shows that the new current \( I' \) is twice the original current \( I \).

5. **Conclusion**:
Therefore, if the drift velocity is doubled, the current will also become:
\[ I' = 2I \]

### Final Answer:
The current will become **doubled**.

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