Two circular coils are made of two identical wires of same length and carry same current. If the number of turns of the two coils are 4 and 2, then the ratio of magnetic induction at the centres will be

Options

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

Explanation

To solve the problem, we need to find the ratio of magnetic induction (magnetic field) at the centers of two circular coils made from identical wires carrying the same current. The number of turns in the coils are given as 4 and 2.

### Step-by-Step Solution:

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

2. **Identifying Variables**:
Since both coils are made of identical wires and carry the same current, we can denote:
- For Coil 1 (4 turns): \(n_1 = 4\)
- For Coil 2 (2 turns): \(n_2 = 2\)
- The current \(I\) is the same for both coils.

3. **Calculating Magnetic Induction for Each Coil**:
For Coil 1:
\[ B_1 = \frac{\mu_0 n_1 I}{2R_1} \]
For Coil 2:
\[ B_2 = \frac{\mu_0 n_2 I}{2R_2} \]

4. **Finding the Ratio of Magnetic Induction**:
The ratio of the magnetic inductions \(B_1\) and \(B_2\) can be expressed as:
\[ \frac{B_1}{B_2} = \frac{\frac{\mu_0 n_1 I}{2R_1}}{\frac{\mu_0 n_2 I}{2R_2}} = \frac{n_1 R_2}{n_2 R_1} \]
Since the wires are identical, the lengths of the wires are the same, which implies that the radius of the coils will be different. However, we can assume that the radius is proportional to the number of turns, meaning:
\[ R_1 \propto \frac{L}{n_1} \quad \text{and} \quad R_2 \propto \frac{L}{n_2} \]
Thus, we can express \(R_1\) and \(R_2\) in terms of the number of turns:
\[ R_1 = k \cdot \frac{L}{n_1} \quad \text{and} \quad R_2 = k \cdot \frac{L}{n_2} \]
where \(k\) is a constant.

5. **Substituting the Values**:
Now substituting \(R_1\) and \(R_2\) into the ratio:
\[ \frac{B_1}{B_2} = \frac{n_1 \cdot \frac{L}{n_2}}{n_2 \cdot \frac{L}{n_1}} = \frac{n_1^2}{n_2^2} \]
Substituting \(n_1 = 4\) and \(n_2 = 2\):
\[ \frac{B_1}{B_2} = \frac{4^2}{2^2} = \frac{16}{4} = 4 \]

6. **Final Result**:
Therefore, the ratio of magnetic induction at the centers of the two coils is:
\[ \frac{B_1}{B_2} = 4 \]