A solenoid having a core of cross - section `4xx10^(-4) m^2` , half air and half iron `(mu_r=500)` is 0.22m long. If the number of turns in it is 1000, then calculate the self inductance.

Explanation

For air -cored solenoid `= (mu_0A_1N^2)/l=L_1`
For iron cored solenoid = `(mu_0A_2N^2mu_r)/l=L_2`
but `L=L_1+L_2,A_1=A_2`
`therefore L=(mu_0AN^2)/l(1+mu_r)`
`= (4xx3.142xx10^(-7)xx4xx10^(-4)xx(10^(3))^2(1+500))/(0.22)`
L= 0.57 H