The self-iriductance 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

Options

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

Answer

A

Explanation

We know that,
`L = (mu_(0) mu_(r) N^(2) A)/( l)`
Here, `(L_2)/(L_1) = (mu_(0) mu_(r) N_(2)^(2) A)/( mu_(0) N_(1)^(2) A)`
`(L_2)/(L_1) = mu_(r) ((N_2)/(N_1))^2`
`L_2 = L_1 mu_(r) ((N_2)/(N_1))^2`
`L_2 = 1 xx 10^(-3) xx 500 xx ((50)/(100))^2`
`L_2 = 125 xx 10^(-3)`H
`L_2 = 125` mH