A steady current is flowing in a circular coil of radius R, made up of a thin conducting wire. The magnetic field at the centre of the loop is `B_L`. Now, a circular loop of radius `R//n` is made from the same wire without changing its length, by unfounding and refolding the loop, and the same current is passed through it. If new magnetic field at the centre of the coil is `B_C`, then the ratio `B_L//B_C` is

Options

A. `1:n^2`
B. `n^(1//2)`
C. `n:1`
D. none of these

Explanation

To solve the problem, we need to find the ratio of the magnetic fields \( B_L \) and \( B_C \) produced by two different circular coils made from the same wire. Let's break down the solution step by step.

### Step 1: Determine the magnetic field \( B_L \) at the center of the original coil
The magnetic field at the center of a circular coil of radius \( R \) carrying a current \( I \) is given by the formula:
\[ B_L = \frac{\mu_0 I}{2 R} \]
where \( \mu_0 \) is the permeability of free space.

### Step 2: Determine the length of the wire used in the original coil
The circumference of the original coil is:
\[ L = 2 \pi R \]
This length \( L \) is the length of the wire used to make the coil.

### Step 3: Determine the radius of the new coil
The new coil is made from the same wire but has a radius of \( \frac{R}{n} \).

### Step 4: Calculate the number of turns in the new coil
The circumference of the new coil is:
\[ L = 2 \pi \left(\frac{R}{n}\right) = \frac{2 \pi R}{n} \]
The number of turns \( n \) that can be made with the same length of wire is:
\[ \text{Number of turns} = \frac{\text{Length of wire}}{\text{Circumference of new coil}} = \frac{2 \pi R}{\frac{2 \pi R}{n}} = n \]

### Step 5: Determine the magnetic field \( B_C \) at the center of the new coil
The magnetic field at the center of a circular coil with \( n \) turns and radius \( \frac{R}{n} \) is given by:
\[ B_C = n \cdot \frac{\mu_0 I}{2 \left(\frac{R}{n}\right)} = n \cdot \frac{\mu_0 I n}{2 R} = \frac{n^2 \mu_0 I}{2 R} \]

### Step 6: Calculate the ratio \( \frac{B_L}{B_C} \)
Now we can find the ratio of the two magnetic fields:
\[ \frac{B_L}{B_C} = \frac{\frac{\mu_0 I}{2 R}}{\frac{n^2 \mu_0 I}{2 R}} = \frac{1}{n^2} \]

### Final Result
Thus, the ratio \( \frac{B_L}{B_C} \) is:
\[ \frac{B_L}{B_C} = \frac{1}{n^2} \]