A long wire carries a steady curent . It is bent into a circle of one turn and the magnetic field at the centre of the coil is `B`. It is then bent into a circular loop of `n` turns. The magnetic field at the centre of the coil will be

Options

A. ` 2n B`
B. ` n^(2) B`
C. `nB`
D. ` 2 n ^(2) B`

Explanation

To solve the problem, we need to find the magnetic field at the center of a circular loop made from a wire that originally carried a steady current and was bent into a single turn, then into `n` turns.

### Step-by-Step Solution:

1. **Understand the Magnetic Field of a Single Turn:**
The magnetic field \( B \) at the center of a 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. **Relate the Length of the Wire:**
When the wire is bent into a single turn, the length of the wire is equal to the circumference of the circle:
\[ L = 2\pi R \]
When the wire is bent into \( n \) turns, the total length of the wire remains the same, and we can express it as:
\[ L = n \cdot 2\pi R' \]
where \( R' \) is the radius of the new circular loop with \( n \) turns.

3. **Set the Lengths Equal:**
Since the length of the wire does not change, we can set the two expressions for length equal to each other:
\[ 2\pi R = n \cdot 2\pi R' \]
Simplifying this gives:
\[ R' = \frac{R}{n} \]

4. **Calculate the New Magnetic Field:**
Now, we can substitute \( R' \) back into the formula for the magnetic field for \( n \) turns:
\[ B' = \frac{\mu_0 I}{2R'} \]
Substituting \( R' = \frac{R}{n} \) into the equation gives:
\[ B' = \frac{\mu_0 I}{2 \left(\frac{R}{n}\right)} = \frac{\mu_0 I n}{2R} \]

5. **Relate the New Magnetic Field to the Original:**
From the original magnetic field \( B = \frac{\mu_0 I}{2R} \), we can express \( B' \) in terms of \( B \):
\[ B' = n \cdot B \]

6. **Final Expression:**
Therefore, the magnetic field at the center of the coil with \( n \) turns is:
\[ B' = n \cdot B \]

### Conclusion:
The magnetic field at the center of the coil when the wire is bent into \( n \) turns is \( nB \).