The coefficient of self-inductance of a solenoid is `0.18 mH`. If a crude of soft iron of relative permeability `900` is inserted, then the coefficient of self-inductance will become nearly

Options

A. `5.4 mH`
B. `162 mH`
C. `0.006 mH`
D. `0.0002 MH`

Explanation

To solve the problem of finding the new coefficient of self-inductance when a soft iron core is inserted into a solenoid, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Initial Self-Inductance**:
The initial self-inductance of the solenoid is given as \( L_0 = 0.18 \, \text{mH} = 0.18 \times 10^{-3} \, \text{H} \).

2. **Identify the Relative Permeability**:
The relative permeability of the soft iron core is given as \( \mu_r = 900 \).

3. **Use the Formula for Self-Inductance**:
The self-inductance of a solenoid can be expressed as:
\[ L = \frac{n^2 \mu A l}{l} \]
where \( n \) is the number of turns per unit length, \( A \) is the cross-sectional area, \( l \) is the length of the solenoid, and \( \mu \) is the permeability of the material inside the solenoid.

4. **Relate Permeability to Relative Permeability**:
The permeability \( \mu \) of the material is given by:
\[ \mu = \mu_0 \mu_r \]
where \( \mu_0 \) is the permeability of free space (approximately \( 4\pi \times 10^{-7} \, \text{H/m} \)).

5. **Calculate the New Self-Inductance**:
When the soft iron core is inserted, the new self-inductance \( L_1 \) can be calculated as:
\[ L_1 = L_0 \mu_r \]
Substituting the values:
\[ L_1 = 0.18 \times 10^{-3} \, \text{H} \times 900 \]

6. **Perform the Calculation**:
\[ L_1 = 0.18 \times 900 \times 10^{-3} = 162 \times 10^{-3} \, \text{H} = 162 \, \text{mH} \]

### Final Answer:
The new coefficient of self-inductance when the soft iron core is inserted is approximately **162 mH**.