A short soleniod of radius a, number of turns per unit length `n_(1)`, and length L is kept coaxially inside a very long solenoid of radius b, number of turns per unit length `n_(2)`. What is the mutual inductance of the system?

Options

A. `mu_(0)pib^(2)n_(1)n_(2)L`
B. `mu_(0)pia^(2)n_(1)n_(2)L^(2)`
C. `mu_(0)pia^(2)n_(1)n_(2)L`
D. `mu_(0)pib^(2)n_(1)n_(2)L^(2)`

Explanation

To find the mutual inductance \( M \) of a short solenoid placed inside a long solenoid, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Configuration**:
- We have a short solenoid of radius \( a \), number of turns per unit length \( n_1 \), and length \( L \).
- This short solenoid is placed coaxially inside a very long solenoid of radius \( b \) and number of turns per unit length \( n_2 \).

2. **Formula for Mutual Inductance**:
- The mutual inductance \( M \) between two solenoids can be expressed as:
\[ M = \mu_0 \cdot A \cdot n_1 \cdot n_2 \cdot L \]
- Where:
- \( \mu_0 \) is the permeability of free space.
- \( A \) is the cross-sectional area of the short solenoid.
- \( n_1 \) is the number of turns per unit length of the short solenoid.
- \( n_2 \) is the number of turns per unit length of the long solenoid.
- \( L \) is the length of the short solenoid.

3. **Calculate the Area \( A \)**:
- The cross-sectional area \( A \) of the short solenoid is given by:
\[ A = \pi a^2 \]

4. **Substituting Values**:
- Now, substitute the area \( A \) into the mutual inductance formula:
\[ M = \mu_0 \cdot (\pi a^2) \cdot n_1 \cdot n_2 \cdot L \]

5. **Final Expression**:
- Therefore, the mutual inductance \( M \) of the system is:
\[ M = \mu_0 \pi a^2 n_1 n_2 L \]

### Conclusion:
The mutual inductance of the system is given by:
\[ M = \mu_0 \pi a^2 n_1 n_2 L \]