Cross sectional area of a solenoid is `10cm^(2)`. Half of its cross section is filled with iron `(mu_(r)=450)` and the remaining half with air `(mu_(r)=1)`. Calculate the self-inductance of the solenoid if its length is 2 m and number of turns is 3000.

Explanation

If `alpha_(1)` part of cross section of the solenoid is filled with a substance of relative permeability `mu_(r_(1))` and the remaining part of cross section `alpha_(2)` with another substance of relative permeablity `mu_(r_(2))` then self-inductance of the solenoid is,
`L=(mu_(0)n^(2)A)/(l)(mu_(r_(1))alpha_(1)+mu_(r_(2))alpha_(2))`
Here `mu_(0)=4pixx10^(-7)` H/m, number of turns, n = 3000, length of solenoid, l = 2 m, area of cross section, `A=10cm^(2)=0.001m^(2),alpha_(1)=0.5andalpha_(2)=0.5`.
`therefore` Self-inductance of the solenoid,
`L=(4pixx10^(-7)xx(3000)^(2)xx0.001)/(2)(1xx0.5+450xx0.5)H`
`=2pixx9xx10^(-4)xx(0.5+225)H`
= 1.27 H (approx).