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

Answer

To solve the problem, we need to find the magnetic field at the center of a circular loop of wire when it is bent into `n` turns from its original single turn configuration.

### Step-by-Step Solution:

1. **Understanding the Magnetic Field for One Turn**:
The magnetic field at the center of a circular loop of wire carrying a current `I` is given by the formula:
\[ B = \frac{\mu_0 I}{2R} \]
where \( R \) is the radius of the loop and \( \mu_0 \) is the permeability of free space.

2. **Given Magnetic Field for One Turn**:
According to the problem, when the wire is bent into a single turn (one loop), the magnetic field at the center is given as \( B \). Therefore, we can express this as:
\[ B = \frac{\mu_0 I}{2R} \]

3. **Length of the Wire**:
The length of the wire used to form the single turn is the circumference of the circle:
\[ L = 2\pi R \]

4. **Bending the Wire into `n` Turns**:
When the wire is bent into `n` turns, the total length of the wire remains the same. The new radius \( R' \) for the `n` turns can be calculated using the total length of the wire:
\[ L = n \cdot 2\pi R' = 2\pi R \]
From this, we can solve for \( R' \):
\[ R' = \frac{R}{n} \]

5. **Magnetic Field for `n` Turns**:
The magnetic field at the center of the new circular loop with `n` turns can be expressed as:
\[ B' = \frac{\mu_0 I}{2R'} \]
Substituting \( R' = \frac{R}{n} \) into this equation gives:
\[ B' = \frac{\mu_0 I}{2 \left(\frac{R}{n}\right)} = \frac{\mu_0 I n}{2R} \]

6. **Relating \( B' \) to \( B \)**:
We can relate \( B' \) to the original magnetic field \( B \):
\[ B' = n \cdot \frac{\mu_0 I}{2R} = nB \]

7. **Final Result**:
Thus, the magnetic field at the center of the coil when the wire is bent into `n` turns is:
\[ B' = nB \]

### Summary:
The magnetic field at the center of the coil when the wire is bent into `n` turns is \( B' = nB \).