A current carrying conductor of length `l` is bent into two loops one by one. First loop has one turn of wire and the second loop has two turns of wire. Compare the magnetic fields at the centre of the loops

Options

A. `B' = 4B`
B. `4B' = B`
C. `2B' = B`
D. `B' = 2B`

Answer

To solve the problem of comparing the magnetic fields at the center of two loops formed by bending a current-carrying conductor of length `L`, we will follow these steps:

### Step 1: Understanding the Magnetic Field Formula
The magnetic field \( B \) at the center of a circular loop carrying current \( I \) is given by the formula:
\[ B = \frac{n \mu_0 I}{2a} \]
where:
- \( n \) = number of turns in the loop
- \( \mu_0 \) = permeability of free space
- \( I \) = current through the wire
- \( a \) = radius of the loop

### Step 2: Analyze the First Loop
For the first loop:
- It has 1 turn, so \( n = 1 \).
- The length of the wire used for this loop is \( L \), which corresponds to the circumference of the loop:
\[ 2\pi a = L \implies a = \frac{L}{2\pi} \]
Now, substituting \( n \) and \( a \) into the magnetic field formula:
\[ B_1 = \frac{1 \cdot \mu_0 I}{2 \cdot \frac{L}{2\pi}} = \frac{\mu_0 I \cdot \pi}{L} \]

### Step 3: Analyze the Second Loop
For the second loop:
- It has 2 turns, so \( n = 2 \).
- The length of the wire used for this loop is still \( L \), which corresponds to the circumference of the two loops combined:
\[ 2\pi a' = \frac{L}{2} \implies a' = \frac{L}{4\pi} \]
Now, substituting \( n \) and \( a' \) into the magnetic field formula:
\[ B_2 = \frac{2 \cdot \mu_0 I}{2 \cdot \frac{L}{4\pi}} = \frac{2 \cdot \mu_0 I \cdot 2\pi}{L} = \frac{4\mu_0 I \cdot \pi}{L} \]

### Step 4: Compare the Magnetic Fields
Now we have:
- \( B_1 = \frac{\mu_0 I \cdot \pi}{L} \)
- \( B_2 = \frac{4\mu_0 I \cdot \pi}{L} \)

To compare \( B_1 \) and \( B_2 \):
\[ \frac{B_1}{B_2} = \frac{\frac{\mu_0 I \cdot \pi}{L}}{\frac{4\mu_0 I \cdot \pi}{L}} = \frac{1}{4} \]
This implies:
\[ B_2 = 4B_1 \]

### Conclusion
The magnetic field at the center of the second loop (with two turns) is four times greater than that of the first loop (with one turn).