Q140 / 150 · MHT CET 2026 (20 April Shift 1) · Physics
The self inductance of an air-core inductor (solenoid) is $0.03$ mH. By introducing an iron core into inductor the self-inductance increases to $30$ mH. The relative permeability of the core used is
- A$10^{-3}$
- B$10^{-2}$
- C$10^{2}$
- Correct answerD$10^{3}$
Solution
The self-inductance of an air-core solenoid is given by $L_{0} = \mu_{0} n^{2} A l$.
When an iron core is introduced, the self-inductance becomes $L = \mu_{r} \mu_{0} n^{2} A l = \mu_{r} L_{0}$.
Given $L_{0} = 0.03 \text{ mH}$ and $L = 30 \text{ mH}$.
$\mu_{r} = \dfrac{L}{L_{0}} = \dfrac{30}{0.03} = 1000 = 10^{3}$.
Answer: $10^{3}$
When an iron core is introduced, the self-inductance becomes $L = \mu_{r} \mu_{0} n^{2} A l = \mu_{r} L_{0}$.
Given $L_{0} = 0.03 \text{ mH}$ and $L = 30 \text{ mH}$.
$\mu_{r} = \dfrac{L}{L_{0}} = \dfrac{30}{0.03} = 1000 = 10^{3}$.
Answer: $10^{3}$
Hence, the correct answer is option: 'D'.