The magnetic induction at the centre of a solenoid is `B`. If the length of the solenoid is reduced to half and the same wire is would in two layers the new magnetic induction is

Options

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

Answer

To solve the problem, we need to analyze the changes in the solenoid's dimensions and how they affect the magnetic induction (B) at its center.

### Step-by-Step Solution:

1. **Understand the initial conditions**:
- The magnetic induction at the center of the solenoid is given as \( B \).
- The formula for the magnetic induction at the center of a solenoid is:
\[ B = \frac{\mu_0 n I}{2} \]
where \( \mu_0 \) is the permeability of free space, \( n \) is the number of turns per unit length, and \( I \) is the current through the wire.

2. **Initial length and turns**:
- Let the initial length of the solenoid be \( L \) and the number of turns be \( N \).
- Therefore, the number of turns per unit length \( n \) is:
\[ n = \frac{N}{L} \]

3. **Reducing the length**:
- The length of the solenoid is reduced to half, so the new length \( L' \) is:
\[ L' = \frac{L}{2} \]

4. **Winding the same wire in two layers**:
- Since the same wire is used, the total number of turns \( N \) remains the same.
- However, when the solenoid is wound in two layers, the effective number of turns per unit length \( n' \) doubles because the same number of turns is now spread over half the length:
\[ n' = \frac{N}{L'} = \frac{N}{L/2} = \frac{2N}{L} = 2n \]

5. **New magnetic induction**:
- Substitute \( n' \) into the formula for magnetic induction:
\[ B' = \frac{\mu_0 n' I}{2} = \frac{\mu_0 (2n) I}{2} = \mu_0 n I = 2B \]

6. **Final calculation**:
- Since we have reduced the length and doubled the number of turns per unit length, the new magnetic induction \( B' \) becomes:
\[ B' = 4B \]

### Conclusion:
The new magnetic induction at the center of the solenoid after reducing its length to half and winding the same wire in two layers is:
\[ B' = 4B \]