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

Answer

To solve the problem, we need to analyze the magnetic field produced by a wire bent into a circular loop and how it changes when the wire is bent into multiple turns.

### Step-by-Step Solution:

1. **Understand the Initial Setup**:
- A long wire carries a steady current \( I \).
- The wire is initially bent into a single circular loop (one turn) and the magnetic field at the center of this loop is given as \( B \).

2. **Determine the Length of the Wire**:
- Let the length of the wire be \( L \).
- For a single turn circular loop, the circumference is equal to the length of the wire:
\[ 2\pi R_1 = L \]
- From this, we can express the radius of the loop as:
\[ R_1 = \frac{L}{2\pi} \]

3. **Calculate the Magnetic Field for One Turn**:
- The magnetic field at the center of a circular loop carrying current \( I \) is given by the formula:
\[ B = \frac{\mu_0 I}{2R_1} \]
- Substituting \( R_1 \) into the equation:
\[ B = \frac{\mu_0 I}{2 \left(\frac{L}{2\pi}\right)} = \frac{\mu_0 I \cdot \pi}{L} \]

4. **Consider the Case with n Turns**:
- Now, the wire is bent into a circular loop of \( n \) turns.
- The total length of the wire remains \( L \), and the circumference of the \( n \) turns is:
\[ n \cdot 2\pi R_2 = L \]
- From this, we can express the radius of the new loop as:
\[ R_2 = \frac{L}{2\pi n} \]

5. **Calculate the Magnetic Field for n Turns**:
- The magnetic field at the center of this new loop with \( n \) turns is:
\[ B' = \frac{\mu_0 I}{2R_2} \cdot n \]
- Substituting \( R_2 \):
\[ B' = \frac{\mu_0 I}{2 \left(\frac{L}{2\pi n}\right)} \cdot n = \frac{\mu_0 I \cdot \pi n}{L} \]

6. **Relate the New Magnetic Field to the Original**:
- From the earlier calculation, we know:
\[ B = \frac{\mu_0 I \cdot \pi}{L} \]
- Thus, we can relate \( B' \) to \( B \):
\[ B' = n^2 \cdot B \]

### Final Result:
The magnetic field at the center of the coil when the wire is bent into \( n \) turns is:
\[ B' = n^2 B \]