A long conducting wire having a current I flowing through it,is bent into a circular coil of `N` turns.Then it is bent into a circular coil of `n` turns.The magnetic field is calculated at the centre of coils in both the cases.The ratio of the magnetic field in first case to that of second case is

Options

A. ` n : N `
B. `N : n`
C. `N^(2):n^(2)`
D. `n^(2):N^(2)`

Answer

To solve the problem, we need to find the ratio of the magnetic fields at the center of two circular coils formed by bending a long conducting wire with a current \( I \) flowing through it.

### 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 in the coil,
- \( I \) is the current flowing through the wire,
- \( R \) is the radius of the coil.

2. **Case 1: Coil with \( N \) Turns**:
For the first coil with \( N \) turns, the magnetic field \( B_1 \) at the center is:
\[ B_1 = \frac{\mu_0 N I}{2R} \]

3. **Case 2: Coil with \( n \) Turns**:
When the wire is bent into a second coil with \( n \) turns, let’s denote the radius of this coil as \( R' \). The magnetic field \( B_2 \) at the center of this coil is:
\[ B_2 = \frac{\mu_0 n I}{2R'} \]

4. **Relating the Radii**:
Since the length of the wire remains constant, we can relate the two coils using the circumference:
\[ 2\pi R N = 2\pi R' n \]
Simplifying this gives:
\[ R N = R' n \quad \Rightarrow \quad \frac{R'}{R} = \frac{N}{n} \]

5. **Substituting for \( R' \)**:
We can express \( R' \) in terms of \( R \):
\[ R' = R \frac{n}{N} \]

6. **Substituting \( R' \) into \( B_2 \)**:
Now substitute \( R' \) into the expression for \( B_2 \):
\[ B_2 = \frac{\mu_0 n I}{2(R \frac{n}{N})} = \frac{\mu_0 n I N}{2R n} = \frac{\mu_0 N I}{2R} \]

7. **Finding the Ratio \( \frac{B_1}{B_2} \)**:
Now we can find the ratio of the magnetic fields:
\[ \frac{B_1}{B_2} = \frac{\frac{\mu_0 N I}{2R}}{\frac{\mu_0 n I}{2R}} = \frac{N}{n} \]

### Final Result:
Thus, the ratio of the magnetic field in the first case to that of the second case is:
\[ \frac{B_1}{B_2} = \frac{N}{n} \]