Question

The self-inductance of an air core solenoid of 100 turns is $1mH$. 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

• A. 125 mH
• B. 24 mH
• C. 60 mH
• D. 30 mH
• E. 45 mH

Answer 1

The correct option is A 125 mH

Inducatnce = $\mu N^2\frac{Area}{length}$

$\therefore \frac{L_2}{L_1}=\frac{{\mu }_r\times {50}^2\times 500}{{100}^2}$

$\therefore \frac{L_2}{1}=\frac{{50}^2\times 500}{{100}^2}$

$\therefore L_2=125mH$