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 determine 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 loop carrying current \( I \) is given by the formula:
\[ B = \frac{\mu_0 n I}{2R} \]
where:
- \( B \) is the magnetic field,
- \( \mu_0 \) is the permeability of free space,
- \( n \) is the number of turns,
- \( I \) is the current, and
- \( R \) is the radius of the loop.

2. **Calculating \( B_1 \)**:
For the first case, where the wire is bent into one turn:
- Number of turns \( n_1 = 1 \)
- Let the radius of the loop be \( R_1 = R \).
Thus, the magnetic field \( B_1 \) is:
\[ B_1 = \frac{\mu_0 \cdot 1 \cdot I}{2R} = \frac{\mu_0 I}{2R} \]

3. **Calculating \( B_2 \)**:
For the second case, where the wire is bent into four turns:
- Number of turns \( n_2 = 4 \)
- The length of the wire remains the same, so the radius of the new loop \( R_2 \) is related to \( R \) by the inverse of the number of turns:
\[ R_2 = \frac{R}{4} \]
Thus, the magnetic field \( B_2 \) is:
\[ B_2 = \frac{\mu_0 \cdot 4 \cdot I}{2R_2} = \frac{\mu_0 \cdot 4 \cdot I}{2 \cdot \frac{R}{4}} = \frac{\mu_0 \cdot 4 \cdot I \cdot 4}{2R} = \frac{16 \mu_0 I}{2R} = \frac{8 \mu_0 I}{R} \]

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

Thus, the ratio of \( B_1 \) to \( B_2 \) is:
\[ \frac{B_1}{B_2} = \frac{1}{16} \]

### Final Answer:
The ratio of \( B_1 \) and \( B_2 \) is \( \frac{1}{16} \).

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