A short solenoid of length 4 cm, radius 2 cm and 100 turns is placed inside and on the axis of a long solenoid of length 80 cm and 1500 turns. A current of 3 A flows through the short solenoid. The mutual inductance of two solenoid is

Options

A. 0.12 H
B. `5.3xx10^(5)H`
C. `3.52xx10^(-3)H`
D. `8.3xx10^(-5)H`

Answer

To find the mutual inductance \( M \) of the two solenoids, we can use the formula:

\[ M = \mu_0 n_1 n_2 \pi r_1^2 L \]

where:
- \( \mu_0 \) is the permeability of free space, approximately \( 4\pi \times 10^{-7} \, \text{H/m} \)
- \( n_1 \) 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
- \( r_1 \) is the radius of the short solenoid
- \( L \) is the length of the short solenoid

### Step 1: Calculate \( n_1 \)
Given:
- Number of turns in the short solenoid = 100
- Length of the short solenoid = 4 cm = 0.04 m

\[ n_1 = \frac{\text{Number of turns}}{\text{Length}} = \frac{100}{0.04} = 2500 \, \text{turns/m} \]

**Hint:** To find the number of turns per unit length, divide the total number of turns by the length in meters.

### Step 2: Calculate \( n_2 \)
Given:
- Number of turns in the long solenoid = 1500
- Length of the long solenoid = 80 cm = 0.8 m

\[ n_2 = \frac{1500}{0.8} = 1875 \, \text{turns/m} \]

**Hint:** Use the same method as in Step 1 to find the number of turns per unit length for the long solenoid.

### Step 3: Convert the radius \( r_1 \) to meters
Given:
- Radius of the short solenoid = 2 cm = 0.02 m

**Hint:** Always convert centimeters to meters when using SI units.

### Step 4: Substitute values into the mutual inductance formula
Now we have:
- \( \mu_0 = 4\pi \times 10^{-7} \, \text{H/m} \)
- \( n_1 = 2500 \, \text{turns/m} \)
- \( n_2 = 1875 \, \text{turns/m} \)
- \( r_1 = 0.02 \, \text{m} \)
- \( L = 0.04 \, \text{m} \)

Substituting these values into the formula:

\[ M = (4\pi \times 10^{-7}) \times (2500) \times (1875) \times \pi \times (0.02)^2 \times (0.04) \]

### Step 5: Calculate the mutual inductance \( M \)
Calculating step-by-step:

1. Calculate \( (0.02)^2 = 0.0004 \)
2. Calculate \( 4\pi \times 10^{-7} \approx 1.2566 \times 10^{-6} \)
3. Now substitute and calculate:

\[ M \approx (1.2566 \times 10^{-6}) \times (2500) \times (1875) \times \pi \times (0.0004) \times (0.04) \]

Calculating this gives:

\[ M \approx 2.90 \times 10^{-4} \, \text{H} \]

### Final Answer
The mutual inductance \( M \) of the two solenoids is approximately \( 2.90 \times 10^{-4} \, \text{H} \).

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