In a wire of circular cross-section with radius 'r' free electrons travel with a drift velocity v, when a current I flows through the wire. The current in another wire of half the radius and of the same material when the drift velocity is 2v

Options

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

Answer

To solve the problem, we need to find the current \( I_2 \) in the second wire, which has half the radius of the first wire and a drift velocity of \( 2v \).

### Step-by-Step Solution:

1. **Identify the Current in the First Wire:**
The current \( I_1 \) in the first wire can be expressed using the formula:
\[ I_1 = n \cdot e \cdot A_1 \cdot v \]
where:
- \( n \) = number of free electrons per unit volume,
- \( e \) = charge of an electron,
- \( A_1 \) = cross-sectional area of the first wire,
- \( v \) = drift velocity.

2. **Calculate the Cross-Sectional Area of the First Wire:**
The area \( A_1 \) for a wire with radius \( r \) is given by:
\[ A_1 = \pi r^2 \]

3. **Identify the Current in the Second Wire:**
The current \( I_2 \) in the second wire can be expressed as:
\[ I_2 = n \cdot e \cdot A_2 \cdot (2v) \]
where \( A_2 \) is the cross-sectional area of the second wire.

4. **Calculate the Cross-Sectional Area of the Second Wire:**
The radius of the second wire is \( \frac{r}{2} \), so its area \( A_2 \) is:
\[ A_2 = \pi \left(\frac{r}{2}\right)^2 = \pi \cdot \frac{r^2}{4} \]

5. **Substituting Areas into the Current Equation:**
Now substituting \( A_1 \) and \( A_2 \) into the current equations:
\[ I_1 = n \cdot e \cdot \pi r^2 \cdot v \]
\[ I_2 = n \cdot e \cdot \left(\pi \cdot \frac{r^2}{4}\right) \cdot (2v) \]

6. **Simplifying the Expression for \( I_2 \):**
\[ I_2 = n \cdot e \cdot \pi \cdot \frac{r^2}{4} \cdot 2v = n \cdot e \cdot \pi r^2 \cdot \frac{2v}{4} = n \cdot e \cdot \pi r^2 \cdot \frac{v}{2} \]

7. **Finding the Ratio of Currents:**
Now, we can find the ratio of \( I_1 \) to \( I_2 \):
\[ \frac{I_1}{I_2} = \frac{n \cdot e \cdot \pi r^2 \cdot v}{n \cdot e \cdot \pi r^2 \cdot \frac{v}{2}} = \frac{1}{\frac{1}{2}} = 2 \]

8. **Expressing \( I_2 \) in Terms of \( I_1 \):**
Since \( I_1 = I \):
\[ \frac{I}{I_2} = 2 \implies I_2 = \frac{I}{2} \]

### Final Answer:
The current in the second wire is:
\[ I_2 = \frac{I}{2} \]