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 a current-carrying conductor, we can follow these steps:

### Step 1: Define the parameters
Let:
- \( L \) = Length of the conductor
- \( I \) = Current flowing through the conductor
- \( r_1 \) = Radius of the first loop (1 turn)
- \( r_2 \) = Radius of the second loop (2 turns)

### Step 2: Calculate the radius of the loops
For the first loop with one turn:
\[ L = 2\pi r_1 \implies r_1 = \frac{L}{2\pi} \]

For the second loop with two turns:
\[ L = 2 \times 2\pi r_2 \implies L = 4\pi r_2 \implies r_2 = \frac{L}{4\pi} \]

### Step 3: Calculate the magnetic field at the center of the first loop
The magnetic field \( B \) at the center of a single loop is given by:
\[ B = \frac{\mu_0 I}{2r} \]
For the first loop:
\[ B_1 = \frac{\mu_0 I}{2r_1} = \frac{\mu_0 I}{2 \left(\frac{L}{2\pi}\right)} = \frac{\mu_0 I \cdot \pi}{L} \]

### Step 4: Calculate the magnetic field at the center of the second loop
For the second loop with two turns, the magnetic field \( B' \) at the center is:
\[ B' = \frac{\mu_0 I \cdot n}{2r} \]
where \( n \) is the number of turns. For the second loop:
\[ B' = \frac{\mu_0 I \cdot 2}{2r_2} = \frac{\mu_0 I}{r_2} = \frac{\mu_0 I}{\left(\frac{L}{4\pi}\right)} = \frac{4\mu_0 I \cdot \pi}{L} \]

### Step 5: Compare the magnetic fields
Now we can compare \( B_1 \) and \( B' \):
\[ \frac{B_1}{B'} = \frac{\frac{\mu_0 I \cdot \pi}{L}}{\frac{4\mu_0 I \cdot \pi}{L}} = \frac{1}{4} \]
Thus,
\[ B' = 4B_1 \]

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