The magnetic field of a given length of wire carrying a current of a single turn circular coil at centre is B, then its value for two turns for the same wire when same current passing through it is

Options

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

Answer

To solve the problem, we need to analyze the relationship between the magnetic field produced by a circular coil and the parameters involved, specifically the number of turns and the radius of the coil.

### Step-by-Step Solution:

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

2. **Single Turn Coil**:
For a single turn coil (where \( N = 1 \)), the magnetic field at the center is:
\[ B = \frac{\mu_0 I}{2R} \]
According to the problem, this magnetic field is given as \( B \).

3. **Two Turns Coil**:
Now, we take the same length of wire and make two turns. The length of the wire remains the same, but the radius will change.

- The length of wire \( L \) can be expressed as:
\[ L = 2 \pi R \]
for the single turn.

- For two turns, the total length of wire used is still \( L \), but now we have:
\[ L = 2 \times 2 \pi r' = 4 \pi r' \]
where \( r' \) is the new radius for the two turns.

4. **Finding the New Radius**:
Since the length of the wire remains constant, we can set the two equations equal:
\[ 2 \pi R = 4 \pi r' \]
Simplifying this gives:
\[ R = 2r' \quad \Rightarrow \quad r' = \frac{R}{2} \]

5. **Calculating the New Magnetic Field**:
Now substituting \( N = 2 \) and \( R = 2r' \) into the magnetic field formula:
\[ B' = \frac{\mu_0 (2) I}{2r'} = \frac{\mu_0 (2) I}{2 \left(\frac{R}{2}\right)} = \frac{\mu_0 (2) I}{R} \]
This can be rewritten as:
\[ B' = 4 \left(\frac{\mu_0 I}{2R}\right) = 4B \]

6. **Conclusion**:
Therefore, the magnetic field for two turns of the wire carrying the same current is:
\[ B' = 4B \]

### Final Answer:
The value of the magnetic field for two turns of the same wire carrying the same current is \( 4B \).