When the number of turns in the two circular coils closely wound are doubled (in both) their mutual inductance becomes

Options

A. becomes 4 times
B. becomes 2 times
C. becomes `(1)/(4)` times
D. remains unchanged

Explanation

To solve the problem of how the mutual inductance changes when the number of turns in two closely wound circular coils is doubled, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Formula for Mutual Inductance**:
The mutual inductance \( M \) between two coils is given by the formula:
\[ M = \mu_0 \frac{n_1 n_2 A}{L} \]
where:
- \( \mu_0 \) is the permeability of free space,
- \( n_1 \) and \( n_2 \) are the number of turns in the first and second coil respectively,
- \( A \) is the area of the coils,
- \( L \) is the length of the coils.

2. **Identify Constants**:
In this scenario, \( \mu_0 \), \( A \), and \( L \) are constants. Therefore, the mutual inductance \( M \) is directly proportional to the product of the number of turns \( n_1 \) and \( n_2 \):
\[ M \propto n_1 n_2 \]

3. **Doubling the Number of Turns**:
If the number of turns in both coils is doubled, we have:
\[ n_1' = 2n_1 \quad \text{and} \quad n_2' = 2n_2 \]

4. **Calculate the New Mutual Inductance**:
The new mutual inductance \( M' \) can be expressed as:
\[ M' = \mu_0 \frac{n_1' n_2' A}{L} = \mu_0 \frac{(2n_1)(2n_2) A}{L} \]
Simplifying this gives:
\[ M' = \mu_0 \frac{4n_1 n_2 A}{L} = 4 \left(\mu_0 \frac{n_1 n_2 A}{L}\right) = 4M \]

5. **Conclusion**:
Therefore, when the number of turns in both coils is doubled, the new mutual inductance \( M' \) becomes:
\[ M' = 4M \]

### Final Answer:
The mutual inductance becomes **4 times** the original mutual inductance.

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