Question

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

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

Answer 1

Answer: (A)

125 mH

Answer 2

We know that, $L=\frac{\mu_{0} \mu_{r} N^{2} A}{l}$ Here, $\frac{L_{2}}{L_{1}}=\frac{\mu_{0} \mu_{r} N_{2}^{2} \wedge}{\mu_{0} N_{1}^{2} A}$ $\frac{L_{2}}{L_{1}}=\mu_{r}\left(\frac{N_{2}}{N_{1}}\right)^{2}$ $L_{2}=L_{1} \mu_{r}\left(\frac{N_{2}}{N_{1}}\right)^{2}$ $L_{2}=1 \times 10^{3} \times 500 \times\left(\frac{50}{100}\right)^{2}$ $L_{2}=125 \times 10^{-3} H$ $L_{2}=125 \,mH$