A length of 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 double loop of smaller radius. If the same current I is passed, the ratio of the magnitude of magnetic field at the centre with its first value is:

Explanation

To solve the problem, we need to analyze the magnetic field produced by a wire bent into a circular coil and then into a double loop of smaller radius. We will derive the expressions for the magnetic field in both cases and find their ratio.

### Step 1: Determine the length of the wire
Let the total length of the wire be \( L \).

### Step 2: Magnetic field for the single turn circular coil
For a circular coil of radius \( R \) with one turn, the magnetic field at the center is given by the formula:
\[ B_1 = \frac{\mu_0 I}{2R} \]
where \( \mu_0 \) is the permeability of free space and \( I \) is the current flowing through the wire.

### Step 3: Relate the radius to the length of the wire
The circumference of the circular coil is equal to the length of the wire:
\[ L = 2\pi R \implies R = \frac{L}{2\pi} \]

### Step 4: Substitute \( R \) into the magnetic field formula
Substituting \( R \) into the expression for \( B_1 \):
\[ B_1 = \frac{\mu_0 I}{2 \left(\frac{L}{2\pi}\right)} = \frac{\mu_0 I \cdot 2\pi}{2L} = \frac{\mu_0 I \pi}{L} \]

### Step 5: Magnetic field for the double loop of smaller radius
Now, when the wire is bent into a double loop of smaller radius \( r \), the total number of turns \( n = 2 \). The length of the wire will still be \( L \), and the circumference of the double loop is:
\[ L = 2 \cdot 2\pi r \implies L = 4\pi r \implies r = \frac{L}{4\pi} \]

### Step 6: Calculate the magnetic field for the double loop
The magnetic field at the center of the double loop is given by:
\[ B_2 = \frac{\mu_0 n I}{2r} = \frac{\mu_0 (2) I}{2r} = \frac{\mu_0 I}{r} \]

### Step 7: Substitute \( r \) into the magnetic field formula
Substituting \( r \) into the expression for \( B_2 \):
\[ B_2 = \frac{\mu_0 I}{\left(\frac{L}{4\pi}\right)} = \frac{\mu_0 I \cdot 4\pi}{L} \]

### Step 8: Find the ratio of the magnetic fields
Now, we find the ratio of the magnetic field at the center of the double loop to that of the single loop:
\[ \text{Ratio} = \frac{B_2}{B_1} = \frac{\frac{\mu_0 I \cdot 4\pi}{L}}{\frac{\mu_0 I \pi}{L}} = \frac{4\pi}{\pi} = 4 \]

### Final Answer
The ratio of the magnitude of the magnetic field at the center of the double loop to that of the single turn coil is:
\[ \text{Ratio} = 4 \]