The self-inductance of an air core solenoid of 100 turns iz 1 mH. The self-inductance of another solenoid of 50 turns (with the same length and cross sectional area) with a core having relative permeability 500 is

Options

A. 125 mH
B. 24 mH
C. 60 mH
D. 30 mH

Explanation

To solve the problem, we need to find the self-inductance of a solenoid with 50 turns and a core having relative permeability of 500, given that the self-inductance of an air-core solenoid with 100 turns is 1 mH.

### Step-by-Step Solution:

1. **Understanding the Given Data:**
- Self-inductance of the air-core solenoid (L1) = 1 mH = \(1 \times 10^{-3}\) H
- Number of turns of the first solenoid (N1) = 100 turns
- Number of turns of the second solenoid (N2) = 50 turns
- Relative permeability of the core (μr) = 500

2. **Formula for Self-Inductance of a Solenoid:**
The self-inductance \(L\) of a solenoid can be expressed as:
\[ L = \mu_0 \cdot n^2 \cdot A \cdot l \]
where:
- \(L\) = self-inductance
- \(\mu_0\) = permeability of free space
- \(n\) = number of turns per unit length
- \(A\) = cross-sectional area
- \(l\) = length of the solenoid

3. **Finding the Permeability of Free Space (\(\mu_0\)):**
From the first solenoid:
\[ L_1 = \mu_0 \cdot n_1^2 \cdot A \cdot l \]
Rearranging gives:
\[ \mu_0 \cdot n_1^2 \cdot A \cdot l = 1 \times 10^{-3} \text{ H} \]

4. **Calculating the Inductance for the Second Solenoid:**
The self-inductance for the second solenoid (L2) with a core of relative permeability \(\mu_r\) is given by:
\[ L_2 = \mu_r \cdot L_1 \cdot \left(\frac{N_2^2}{N_1^2}\right) \]
Plugging in the values:
\[ L_2 = 500 \cdot (1 \times 10^{-3}) \cdot \left(\frac{50^2}{100^2}\right) \]

5. **Calculating the Ratio of Turns:**
\[ \frac{50^2}{100^2} = \frac{2500}{10000} = \frac{1}{4} \]

6. **Substituting the Values:**
\[ L_2 = 500 \cdot (1 \times 10^{-3}) \cdot \frac{1}{4} \]
\[ L_2 = 500 \cdot 0.00025 = 0.125 \text{ H} = 125 \text{ mH} \]

7. **Final Answer:**
The self-inductance of the second solenoid is \(125 \text{ mH}\).