When current is passed through a circular wire prepared from a conducting material, the magnetic field produced at its centre is B. Now a loop having two turns is prepared from the same wire and the same current is passed through it. The magnetic field at its centre will be :

Options

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

Explanation

To solve the problem, we need to understand how the magnetic field produced by a circular loop of wire changes when the number of turns in the loop is altered. Let's break down the solution step by step.

### Step-by-Step Solution:

1. **Magnetic Field of a Single Loop**:
The magnetic field \( B \) at the center of a single circular loop of radius \( r \) carrying a current \( I \) is given by the formula:
\[ B = \frac{\mu_0 I}{2r} \]
where \( \mu_0 \) is the permeability of free space.

2. **Magnetic Field of a Loop with Two Turns**:
When we create a loop with two turns (i.e., a coil with two loops) from the same wire and pass the same current \( I \) through it, the magnetic field at the center of this new loop can be calculated. The magnetic field at the center of a coil with \( n \) turns is given by:
\[ B' = n \cdot \frac{\mu_0 I}{2r} \]
Here, \( n = 2 \) for our case.

3. **Substituting the Values**:
Since we have two turns, we substitute \( n = 2 \) into the equation:
\[ B' = 2 \cdot \frac{\mu_0 I}{2r} = \frac{\mu_0 I}{r} \]

4. **Relating to the Original Magnetic Field**:
From the first step, we know that:
\[ B = \frac{\mu_0 I}{2r} \]
Therefore, we can express \( B' \) in terms of \( B \):
\[ B' = 2B \]

5. **Conclusion**:
The magnetic field at the center of the loop with two turns is twice the magnetic field produced by a single loop. Thus, the magnetic field at the center of the two-turn loop is:
\[ B' = 2B \]

### Final Answer:
The magnetic field at the center of the loop with two turns is \( 2B \).