A current flows in a wire of circular cross section with the free electrons travelling with drift velocity `vecV`. If an equal current flows in a wire of twice the radius, new drift velocity is

Options

A. `vecV`
B. `(vecV)/2`
C. `(vecV)/4`
D. `2vecV`

Explanation

To solve the problem of finding the new drift velocity when the radius of the wire is doubled while maintaining the same current, we can follow these steps:

### Step-by-Step Solution

1. **Understand the Drift Velocity Formula**:
The drift velocity \( V_d \) of electrons in a conductor is given by the formula:
\[ V_d = \frac{I}{n \cdot A \cdot e} \]
where:
- \( I \) is the current,
- \( n \) is the number of charge carriers per unit volume,
- \( A \) is the cross-sectional area of the wire,
- \( e \) is the charge of an electron (approximately \( 1.6 \times 10^{-19} \) coulombs).

2. **Calculate the Cross-Sectional Area**:
The cross-sectional area \( A \) of a wire with a circular cross-section is given by:
\[ A = \pi r^2 \]
If the original radius is \( r \), then for a wire with twice the radius \( (2r) \):
\[ A_2 = \pi (2r)^2 = \pi (4r^2) = 4A_1 \]
Thus, the new area \( A_2 \) is four times the original area \( A_1 \).

3. **Set Up the Equation for Drift Velocities**:
Since the current \( I \) remains the same in both wires, we can write the drift velocities for the two scenarios:
\[ V_{d1} = \frac{I}{n \cdot A_1 \cdot e} \]
\[ V_{d2} = \frac{I}{n \cdot A_2 \cdot e} \]

4. **Substituting the Area**:
Substitute \( A_2 = 4A_1 \) into the equation for \( V_{d2} \):
\[ V_{d2} = \frac{I}{n \cdot (4A_1) \cdot e} = \frac{I}{4(n \cdot A_1 \cdot e)} \]
This shows that:
\[ V_{d2} = \frac{1}{4} V_{d1} \]

5. **Conclusion**:
Therefore, the new drift velocity \( V_{d2} \) when the radius of the wire is doubled is:
\[ V_{d2} = \frac{V_{d1}}{4} \]

### Final Answer
If the original drift velocity is \( V_d \), then the new drift velocity when the radius is doubled is:
\[ V_{d2} = \frac{V_d}{4} \]