A length lof wire carries a steady current i. It is bent first to form a circular plane coil of one turn . The same length is now bent more sharply to give three loops of smaller radius . The magnetic field at the centre caused by the same current is

Options

A. One - third of itss first value
B. Unaltered
C. Three times of its initial value
D. Nine times of its initial value

Answer

To solve the problem, we need to calculate the magnetic field at the center of a circular coil formed from a wire of length \( L \) that carries a steady current \( i \). We will analyze two scenarios: first, when the wire is bent into one circular loop, and second, when it is bent into three smaller loops.

### Step-by-step Solution:

1. **Determine the Length of the Wire**:
Given that the length of the wire is \( L \).

2. **First Case: One Circular Loop**:
- The circumference of the circular loop is given by:
\[ C = 2\pi r \]
- Since the entire length of the wire is used to form this loop, we have:
\[ L = 2\pi r \]
- From this, we can solve for the radius \( r \):
\[ r = \frac{L}{2\pi} \]

3. **Magnetic Field for One Loop**:
- The magnetic field \( B \) at the center of a circular loop carrying current \( i \) is given by:
\[ B = \frac{\mu_0 i}{2r} \]
- Substituting the value of \( r \):
\[ B = \frac{\mu_0 i}{2 \left(\frac{L}{2\pi}\right)} = \frac{\mu_0 i \cdot 2\pi}{2L} = \frac{\mu_0 \pi i}{L} \]

4. **Second Case: Three Smaller Loops**:
- In this case, the same length of wire \( L \) is bent into three loops. The circumference of each smaller loop is:
\[ C' = 2\pi r' \]
- Since there are three loops, the total length of the wire used is:
\[ L = 3 \cdot 2\pi r' \implies L = 6\pi r' \]
- Solving for the radius \( r' \):
\[ r' = \frac{L}{6\pi} \]

5. **Magnetic Field for Three Loops**:
- The magnetic field \( B' \) at the center of three loops is given by:
\[ B' = \frac{\mu_0 i}{2r'} \cdot n \]
- Here, \( n = 3 \) (the number of loops):
\[ B' = \frac{\mu_0 i}{2 \left(\frac{L}{6\pi}\right)} \cdot 3 = \frac{\mu_0 i \cdot 3 \cdot 6\pi}{2L} = \frac{9\mu_0 \pi i}{L} \]

6. **Comparing the Magnetic Fields**:
- Now, we compare \( B' \) with \( B \):
\[ \frac{B'}{B} = \frac{\frac{9\mu_0 \pi i}{L}}{\frac{\mu_0 \pi i}{L}} = 9 \]
- Therefore, the magnetic field at the center of the three loops is:
\[ B' = 9B \]

### Final Answer:
The magnetic field at the center caused by the three loops of smaller radius is **9 times** the magnetic field at the center of the single loop.