A wire of certain length carries a steady current. It is first bent to form a circular wire of one turn. The same wire is bent to form a circular coil of three turns. The ratio of magnetic induction at the centre of the coil in the two cases is ________

Answer

To solve the problem, we need to find the ratio of the magnetic induction (magnetic field) at the center of a circular coil formed from a wire of length \( L \) when it is bent into one turn and then into three turns.

### Step-by-Step Solution:

1. **Understanding the Problem**:
- We have a wire of length \( L \) carrying a steady current \( I \).
- The wire is first bent to form a circular coil with **1 turn** and then with **3 turns**.
- We need to find the ratio of the magnetic induction at the center of the coil in both cases.

2. **Magnetic Field for 1 Turn**:
- When the wire is bent into **1 turn**, the circumference of the circular coil is equal to the length of the wire:
\[ 2\pi r = L \implies r = \frac{L}{2\pi} \]
- The magnetic field \( B_1 \) at the center of a circular coil with radius \( r \) and carrying current \( I \) is given by:
\[ B_1 = \frac{\mu_0}{4\pi} \cdot \frac{2\pi I}{r} \]
- Substituting \( r \):
\[ B_1 = \frac{\mu_0}{4\pi} \cdot \frac{2\pi I}{\frac{L}{2\pi}} = \frac{\mu_0 I}{2L} \]

3. **Magnetic Field for 3 Turns**:
- When the wire is bent into **3 turns**, the total length of the wire is still \( L \), so the circumference for one turn is:
\[ 3 \cdot 2\pi r' = L \implies r' = \frac{L}{6\pi} \]
- The magnetic field \( B_2 \) at the center of this coil is given by:
\[ B_2 = \frac{\mu_0}{4\pi} \cdot \frac{2\pi (3I)}{r'} \]
- Substituting \( r' \):
\[ B_2 = \frac{\mu_0}{4\pi} \cdot \frac{2\pi (3I)}{\frac{L}{6\pi}} = \frac{\mu_0}{4\pi} \cdot \frac{12\pi I}{L} = \frac{3\mu_0 I}{L} \]

4. **Finding the Ratio \( \frac{B_1}{B_2} \)**:
- Now, we find the ratio of the magnetic fields:
\[ \frac{B_1}{B_2} = \frac{\frac{\mu_0 I}{2L}}{\frac{3\mu_0 I}{L}} = \frac{1}{2} \cdot \frac{L}{3} = \frac{1}{6} \]
- Therefore, the ratio of the magnetic induction at the center of the coil in the two cases is:
\[ \frac{B_1}{B_2} = \frac{1}{6} \]

### Final Answer:
The ratio of magnetic induction at the center of the coil in the two cases is \( 1:6 \).