The magnetic field at the centre of a circular coil carrying current I ampere is B. It the coil is bent into smaller circular coil of n turns then its magnetic field at the centre is B, the ratio between B' and B is

Options

A. `1:1`
B. `n:1`
C. `n^(2):1`
D. `(n+1):1`

Answer

To solve the problem, we need to find the ratio of the magnetic field \( B' \) at the center of a smaller circular coil with \( n \) turns to the magnetic field \( B \) at the center of the original circular coil carrying current \( I \).

### Step-by-Step Solution:

1. **Magnetic Field of the Original Coil**:
The magnetic field \( B \) at the center of a circular coil carrying current \( I \) and having radius \( R \) is given by the formula:
\[ B = \frac{\mu_0 I}{2R} \]
where \( \mu_0 \) is the permeability of free space.

2. **Length of the Original Coil**:
The total length \( L \) of the wire used to make the coil is:
\[ L = 2\pi R \]
From this, we can express \( R \) in terms of \( L \):
\[ R = \frac{L}{2\pi} \]

3. **Substituting \( R \) into the Magnetic Field Formula**:
Substituting \( R \) into the magnetic field formula, we get:
\[ B = \frac{\mu_0 I}{2 \left(\frac{L}{2\pi}\right)} = \frac{\mu_0 I \pi}{L} \]
We will refer to this as Equation (1).

4. **Magnetic Field of the Smaller Coil with \( n \) Turns**:
When the original coil is bent into \( n \) smaller coils, the new radius \( R' \) of each smaller coil can be expressed as:
\[ L = 2\pi R' \implies R' = \frac{L}{2\pi n} \]

5. **Magnetic Field of the Smaller Coil**:
The magnetic field \( B' \) at the center of the smaller coil with \( n \) turns is given by:
\[ B' = \frac{n \mu_0 I}{2R'} \]
Substituting \( R' \) into this formula gives:
\[ B' = \frac{n \mu_0 I}{2 \left(\frac{L}{2\pi n}\right)} = \frac{n \mu_0 I \cdot 2\pi n}{2L} = \frac{n^2 \mu_0 I \pi}{L} \]
We will refer to this as Equation (2).

6. **Finding the Ratio \( \frac{B'}{B} \)**:
Now, we can find the ratio of the magnetic fields:
\[ \frac{B'}{B} = \frac{\frac{n^2 \mu_0 I \pi}{L}}{\frac{\mu_0 I \pi}{L}} = n^2 \]

7. **Final Result**:
Thus, the ratio of the magnetic field at the center of the smaller coil to that of the original coil is:
\[ \frac{B'}{B} = n^2 \]

### Conclusion:
The ratio between \( B' \) and \( B \) is \( n^2 \).