A solenoid of length 50cm with 20 turns per centimetre and area of cross-section 40cm? completely surrounds another coaxial solenoid of the same length, area of cross-section `25cm^(2)` with 25 turns per centimetre. Calculate the mutual inductance of the system.

Explanation

`l= 50cm = (1)/(2)m`
`N_(1)= 20 xx 50 = 1000, A_(1)= 40 xx 10^(-4) m^(2)`
`N_(2)= 25 xx 50 = 1250, A_(2)= 25 xx 10^(-4) m^(2)`
`M= (mu_(0)N_(1)N_(2)A_(2))/(1)`
`=(4pi xx 10^(-7) xx 1000 xx 1250 xx 25 xx 10^(-4))/(1//2)H`
`=7.9 xx 10^(-3) H= 7.9mH`
For calculation of M, we have to consider the cross-sectional area of the inner solenoid.