A solenoid of length 20cm, area of cross- section `4.0 cm^2` and having 4000 turns is placed inside another solenoid of 2000 turns having a cross - sectional area `8.0cm ^2` and length `10 cm`. Find the mutual inductance between the solenoids.

Explanation

Mutual inductance
`M= mu_(0) N_(1) N_(2) pir_1^2 l`
`= 4pi xx 10^(-7) xx 4 xx 10^(3) xx 2 xx 10^(3)`
`xx 4 xx 10^(-4) xx 10 xx 10^(-2)`
`= 0.04 xx 10^(-2) H`.
a_1 = 4 cm^2, a_2 = 8 cm^2`
`n_1 = 4000//0.2 m`,
`n_2 = 2000//0.1 m`
`l_1 = 20 cm = 0.20 m`,
`l_2 = 10 cm = 0.1 m`
`B = mu_(0) n_(2) i`
[Let the current through outer solenoid be `i]`
`phi = n_(1) B.A = n_(1) n_2) mu_(0) i xx a_(1)`
`= 2000 xx (2000)/(0.1) xx 4pi xx`
`10^(-7) xx i xx 4 xx 10^(-4)`
`E = (dphi)/(dt) = 64pi xx 10^(-4) xx (dl)/(dt)`
Now `M = E/lt`
`= 6 xx 10^(-4) H = 2 xx 10^(-2) H`.