Define mutual inductance between a pair of coils. Derive expression for the mutual inductance of two long coaxial solenoids of same length wound one over the other.

Explanation

Mutual Inductance of two coils or circutits is defind as magnetic
flux linked with the secondary coil due to the flow of unit current
in the primary coil.

Let, I = length of each solenoid .
`r_(1) , r_(2) ` = radii of two solenoids
A ` = pi r _(1)^(2)` = area of cross-section of inner solenoid `S_(1) `
`N_(1) , N_(2)` number of turns in the two solenoids,
First , we pass a time varying current `I_(1)` through `S_(2)` . The magnetic flux is `mu_(0) n_(2) I_(2)`, where
`n_(2) = N_(2)//I` = the number of turns per unit length of `S_(2)`.
Total magnetic flux linked with the inner solenoid `S_(1)` is
`Phi_(1) mu_(0) n_(2) I_(2) AN_(1)`
`therefore ` Mutual inductance of coil 1 w.r.t. coil 2 is
`M_(12) (Phi_(1))/(I_(2)) = mu_(0) n_(2) I_(2) AN_(1)= (mu_(0) N_(1) N_(2) A)/(I)`
Now, consider the flux linked with outer solenoid `S_(2)` due to the current `I_(1)` in the inner
solenoid `S_(1) ` . The field `B_(1)` due to `I_(1)` inside `S_(1) = B_(1) mu_(0)n_(1) I_(1)`
Total flux linked with the outer solenoid is
` Phi_(2) =B_(1) AN_(2) = mu_(0) n_(1) I_(1) AN_(2)= (mu_(0) n_(1) N_(2)AI_(1))/(I)`
`therefore ` Mutual inductance of coil 2 w. r . t coil 1 is
`M_(21) = (phi_(2))/(I_(2)) = (mu_(0) N_(1) N_(2) A)/(I)`
`therefore M = (phi_(2))/(I_(2)) = (mu_(0) N_(1) N_(2) A)/(I)= mu_(0) n_(1)n_(2) AI = mu_(0) n_(1) n_(2) pir^(2) I`.
If a medium of relative permeability ` mu_(r)` is present within the solenoids , then
` M = mu_(r) mu_(0) n_(1) n_(2) pi r^(2) I` .