A wire of certain length carries a current l.It is bent to form a circle of one turn and the magnetic field at the centre is B1. If it is bent to form a coil of four turns, then magnetic field at centre is `B_2`.The ratio of `B_1` and `B_2 `is

Options

A. 1 : 16 `
B. 1 : 4`
C. 64 : 1 `
D. 2 : 1 `

Answer

To solve the problem, we need to find the ratio of the magnetic fields \( B_1 \) and \( B_2 \) produced at the center of a circular wire when it is bent into one turn and then into four turns.

### Step-by-Step Solution:

1. **Understanding the Magnetic Field Formula**:
The magnetic field at the center of a circular coil is given by the formula:
\[ B = \frac{\mu_0}{4\pi} \cdot \frac{2\pi N I}{R} \]
where:
- \( B \) is the magnetic field,
- \( \mu_0 \) is the permeability of free space,
- \( N \) is the number of turns,
- \( I \) is the current,
- \( R \) is the radius of the coil.

2. **Calculating \( B_1 \)**:
For the first case, where the wire is bent into one turn (\( N_1 = 1 \)):
\[ B_1 = \frac{\mu_0}{4\pi} \cdot \frac{2\pi \cdot 1 \cdot I}{R} = \frac{\mu_0 I}{2R} \]

3. **Calculating \( B_2 \)**:
For the second case, where the wire is bent into four turns (\( N_2 = 4 \)):
The radius of the new coil will be \( R_2 = \frac{R}{4} \) (since the length of the wire remains constant).
Thus, we can express \( B_2 \) as:
\[ B_2 = \frac{\mu_0}{4\pi} \cdot \frac{2\pi \cdot 4 \cdot I}{R/4} = \frac{\mu_0}{4\pi} \cdot \frac{8\pi I}{R/4} = \frac{\mu_0 \cdot 32 I}{4R} = \frac{8\mu_0 I}{R} \]

4. **Finding the Ratio \( \frac{B_1}{B_2} \)**:
Now, we can find the ratio of \( B_1 \) to \( B_2 \):
\[ \frac{B_1}{B_2} = \frac{\frac{\mu_0 I}{2R}}{\frac{8\mu_0 I}{R}} = \frac{\mu_0 I}{2R} \cdot \frac{R}{8\mu_0 I} = \frac{1}{16} \]

5. **Final Result**:
Therefore, the ratio of \( B_1 \) to \( B_2 \) is:
\[ \frac{B_1}{B_2} = 1 : 16 \]