Calculate the self inductance of a solenoid of length 0.30 m containing 1000 turns and having area of cross section `36 cm^2`.

Explanation

Given l =0.3m , `N=10^3` turns, `A=36 xx10^(-4)m^2, (di)/(dt)=-100 A S^(-1)`
using `L= (mu_0N^2A)/l`
We get, `L=(4xx3.142xx10^(-7)xx(10^3)^2(36xx10^(-4)))/(0.3)`
`L= 1.508 xx10^(-2)H`