A wire of length l is bent into a circular loop of radius R and carries a current I. The magnetic field at the centre of the loop is B. The same wire is now bent into a double loop of equal radii. If both loops carry the same current I and it is in the same direction, the magnetic field at the centre of the double loop will be

Options

A. Zero
B. 2 B
C. 4 B
D. 8 B

Explanation

To solve the problem, we need to analyze the magnetic field produced by a wire bent into loops and how it changes when the number of loops is altered.

### Step-by-Step Solution:

1. **Understanding the Initial Loop**:
- A wire of length \( L \) is bent into a single circular loop of radius \( R \).
- The circumference of the loop is given by \( 2\pi R \).
- Since the wire is fully used to create the loop, we have:
\[ L = 2\pi R \]
- The magnetic field \( B \) at the center of a single loop carrying current \( I \) is given by:
\[ B = \frac{\mu_0 I}{2R} \]

2. **Finding the Radius of the Loop**:
- From the equation \( L = 2\pi R \), we can express \( R \) in terms of \( L \):
\[ R = \frac{L}{2\pi} \]

3. **Substituting for Magnetic Field**:
- Now substituting \( R \) into the magnetic field formula:
\[ B = \frac{\mu_0 I}{2 \left(\frac{L}{2\pi}\right)} = \frac{\mu_0 I \cdot 2\pi}{2L} = \frac{\mu_0 I \pi}{L} \]

4. **Creating a Double Loop**:
- The same wire is now bent into two loops of equal radius. Each loop will have a radius \( R' \).
- The total length of wire used for two loops is still \( L \), so:
\[ 2 \times 2\pi R' = L \implies R' = \frac{L}{4\pi} \]

5. **Calculating the Magnetic Field for the Double Loop**:
- The magnetic field at the center of one loop carrying current \( I \) is:
\[ B' = \frac{\mu_0 I}{2R'} \]
- Substituting for \( R' \):
\[ B' = \frac{\mu_0 I}{2 \left(\frac{L}{4\pi}\right)} = \frac{\mu_0 I \cdot 4\pi}{2L} = \frac{2\mu_0 I \pi}{L} \]

6. **Considering Both Loops**:
- Since both loops carry the same current \( I \) in the same direction, the total magnetic field at the center of the double loop is:
\[ B_{total} = 2B' = 2 \left(\frac{2\mu_0 I \pi}{L}\right) = \frac{4\mu_0 I \pi}{L} \]

7. **Relating Back to the Original Magnetic Field**:
- We previously found \( B = \frac{\mu_0 I \pi}{L} \).
- Therefore, the magnetic field at the center of the double loop becomes:
\[ B_{total} = 4B \]

### Final Answer:
The magnetic field at the center of the double loop is \( 4B \).

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