Define mutual inductance between two long coaxial solenoids. Find out the expression for the mutual of inner solenoid of length 1 having the radius `r_(1)` and the number of turns `n_(1)` per unit length due to the second other solenoid of same length and `n_(2)` number of turns per unit length.

Explanation

Let l be the length of each solenoid `S_(1) " and " S_(2). N_(1) " and " N_(2)` be the total number of turn os `S_(1) " and " S_(2)`.
Magnetic field in `S_(1)=B_(1)=(mu_(0) N_(1)N_(1))/(l)`
Magnetic flux linked with each turn of `S_(2)=B_(1)xx` area of each turn `=B_(1) pi r_(1)^(2)`
Total magnetic flux linked with `S_(2)=B_(1) pi r_(1)^(2)N_(2)`
`therefore phi_(2)((mu_(0)N_(l)I_(l))/(l)) pi r_(1)^(2) N_(2)=(mu_(0)N_(1)N_(2)pi r_(1)^(2) I_(l))/(l)`
But magnetic flux linked with `S_(2)` is due to `I_(1)`
`therefore phi_(2) prop I_(1) " or" phi_(2)=M I_(1)`
Where M is the mutual inductance of `S_(2) " and " S_(1)`
`therefore M I_(1)=(mu_(0)N_(1)N_(2)pi r_(1)^(2)I_(1))/(l) therefore M= mu_(0)n_(1)n_(2)pi r_(1)^(2)I`.