On meter length of wires carriers a constant current. The wire is bent to from a circular loop. The magnetic field at the centre of this loop is `B`. The same is now bent to form a circular loop of smaller radius to have four turns in the loop. The magnetic field at the centre of this loop `B`. The same is now bent to form a circular loop of smaller radius of have four turns in the loop. The magnetic field at the centre of this new loop is

Options

A. `B/2`
B. `4B`
C. `B/4`
D. `16B`

Answer

To solve the problem, we will analyze the magnetic field produced by a circular loop of wire carrying a current.

### Step-by-Step Solution:

1. **Understanding the Initial Setup**:
- We have a wire of length 1 meter carrying a constant current, which is bent to form a circular loop.
- The magnetic field at the center of this loop is given as \( B \).

2. **Magnetic Field Formula**:
- The magnetic field \( B \) at the center of a circular loop of radius \( R \) with \( N \) turns and carrying current \( I \) is given by the formula:
\[ B = \frac{\mu_0 N I}{2R} \]
- Here, \( \mu_0 \) is the permeability of free space.

3. **First Loop**:
- For the first loop, we have \( N = 1 \) (one turn) and the length of the wire is \( 2\pi R = 1 \) meter.
- Therefore, we can express the radius \( R \) as:
\[ R = \frac{1}{2\pi} \]
- Substituting \( N = 1 \) into the magnetic field formula gives:
\[ B = \frac{\mu_0 I}{2R} \]

4. **Second Loop**:
- The wire is now bent to form a smaller circular loop with 4 turns.
- Let the new radius be \( R' \). The total length of the wire remains the same, so:
\[ 2\pi R' = 4 \times 2\pi R \]
- This simplifies to:
\[ R' = \frac{R}{4} \]

5. **Magnetic Field for the Second Loop**:
- Now, substituting \( N = 4 \) and \( R' = \frac{R}{4} \) into the magnetic field formula:
\[ B' = \frac{\mu_0 (4) I}{2R'} = \frac{\mu_0 (4) I}{2 \times \frac{R}{4}} = \frac{\mu_0 (4) I \times 4}{2R} = \frac{16 \mu_0 I}{2R} \]
- Since \( B = \frac{\mu_0 I}{2R} \), we can express \( B' \) in terms of \( B \):
\[ B' = 16B \]

### Final Answer:
The magnetic field at the center of the new loop with 4 turns is \( 16B \).