A solenoid of lenght 30 cm with 10 turns per centimetre and area of cross-section `40cm^(2)` completely surrounds another co-axial solenoid of same length, area of cross-section `20cm^(2)` with 40 turns per centimetre. The mutual inductance of the system is

Options

A. 10 H
B. 8 H
C. 3 mH
D. 30 mH

Explanation

To solve the problem of finding the mutual inductance of the coaxial solenoids, we will follow these steps:

### Step 1: Convert Given Values
1. **Length of solenoids (L)**: Given as 30 cm, convert to meters:
\[ L = 30 \, \text{cm} = 30 \times 10^{-2} \, \text{m} = 0.30 \, \text{m} \]

2. **Turns per unit length (n1 and n2)**:
- For the outer solenoid (n1): Given as 10 turns/cm, convert to turns/m:
\[ n_1 = 10 \, \text{turns/cm} = 10 \times 100 = 1000 \, \text{turns/m} \]
- For the inner solenoid (n2): Given as 40 turns/cm, convert to turns/m:
\[ n_2 = 40 \, \text{turns/cm} = 40 \times 100 = 4000 \, \text{turns/m} \]

3. **Area of cross-section (A)**:
- For the inner solenoid: Given as 20 cm², convert to m²:
\[ A = 20 \, \text{cm}^2 = 20 \times 10^{-4} \, \text{m}^2 = 2 \times 10^{-3} \, \text{m}^2 \]

### Step 2: Use the Formula for Mutual Inductance
The formula for mutual inductance (M) between two coaxial solenoids is given by:
\[ M = \mu_0 n_1 n_2 A L \]
where:
- \(\mu_0 = 4\pi \times 10^{-7} \, \text{H/m}\) (permeability of free space)
- \(n_1\) and \(n_2\) are the number of turns per unit length for the outer and inner solenoids respectively.
- \(A\) is the area of the cross-section of the inner solenoid.
- \(L\) is the length of the solenoids.

### Step 3: Substitute the Values into the Formula
Substituting the values we have:
\[ M = (4\pi \times 10^{-7}) \times (1000) \times (4000) \times (2 \times 10^{-3}) \times (0.30) \]

### Step 4: Calculate the Mutual Inductance
1. Calculate the product:
\[ M = (4\pi \times 10^{-7}) \times (1000) \times (4000) \times (2 \times 10^{-3}) \times (0.30) \]
\[ = (4\pi \times 10^{-7}) \times (1000 \times 4000 \times 2 \times 0.30 \times 10^{-3}) \]
\[ = (4\pi \times 10^{-7}) \times (2400000 \times 10^{-3}) \]
\[ = (4\pi \times 10^{-7}) \times (2400) \]
\[ = 4 \times 3.14 \times 2400 \times 10^{-7} \]
\[ = 30144 \times 10^{-7} \, \text{H} \]
\[ = 3.0144 \times 10^{-3} \, \text{H} \approx 3 \times 10^{-3} \, \text{H} \]

### Final Answer
The mutual inductance of the system is:
\[ M \approx 3 \, \text{mH} \, (3 \times 10^{-3} \, \text{H}) \]

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