A current `I` flows through a uniform wire of diameter `d` when the mean electron drift velocity is `v`. The same current will flow though a wire of diameter `d//2` made of the same material if the mean drift velocity of the electron is

Options

A. `v//4`
B. `v//2`
C. `2v`
D. `4v`

Explanation

To solve the problem, we need to find the mean drift velocity of electrons in a wire of diameter \( \frac{d}{2} \) when the same current \( I \) flows through it. We will use the formula for current in terms of drift velocity, which is given by:

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

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

### Step 1: Calculate the cross-sectional area of the first wire
The diameter of the first wire is \( d \). The cross-sectional area \( A_1 \) is given by:

\[ A_1 = \frac{\pi d^2}{4} \]

### Step 2: Write the equation for the current in the first wire
Using the formula for current, we have:

\[ I = n \cdot A_1 \cdot e \cdot v \]

Substituting \( A_1 \):

\[ I = n \cdot \left(\frac{\pi d^2}{4}\right) \cdot e \cdot v \]

### Step 3: Calculate the cross-sectional area of the second wire
The diameter of the second wire is \( \frac{d}{2} \). The cross-sectional area \( A_2 \) is given by:

\[ A_2 = \frac{\pi \left(\frac{d}{2}\right)^2}{4} = \frac{\pi d^2}{16} \]

### Step 4: Write the equation for the current in the second wire
For the second wire, the current is given by:

\[ I = n \cdot A_2 \cdot e \cdot v_d' \]

Substituting \( A_2 \):

\[ I = n \cdot \left(\frac{\pi d^2}{16}\right) \cdot e \cdot v_d' \]

### Step 5: Set the two equations for current equal to each other
Since the current \( I \) is the same in both wires, we can set the two equations equal:

\[ n \cdot \left(\frac{\pi d^2}{4}\right) \cdot e \cdot v = n \cdot \left(\frac{\pi d^2}{16}\right) \cdot e \cdot v_d' \]

### Step 6: Cancel common terms
We can cancel \( n \), \( e \), and \( \pi \) from both sides:

\[ \frac{d^2}{4} \cdot v = \frac{d^2}{16} \cdot v_d' \]

### Step 7: Solve for \( v_d' \)
Now, we can solve for \( v_d' \):

\[ \frac{d^2}{4} \cdot v = \frac{d^2}{16} \cdot v_d' \]

Multiplying both sides by 16:

\[ 4d^2 \cdot v = d^2 \cdot v_d' \]

Dividing both sides by \( d^2 \):

\[ 4v = v_d' \]

### Conclusion
Thus, the mean drift velocity \( v_d' \) in the second wire (with diameter \( \frac{d}{2} \)) is:

\[ v_d' = 4v \]