Explain the mutual induction between two long solenoids. Obtain an expression for the mutual inductance.

Explanation

i. `S_(1)andS_(2)` are two long solenoids each length l. The solenoid `S_(2)` is wound closely over the solenoid `S_(1).`
ii. `N_(1)andN_(2)` are the number of turns in the solenoids `S_(1)andS_(2)` respectively. Both the solenoids are considered to have the same area of cross section A as they are closely wound together.

iii. `I_(1)` is the current flowing through the solenoid `S_(1).` The magnetic field `B_(1)` produced at any point inside the solenoid `S_(1)` due to the current `I_(1)` is
`B_(1)=mu_(0)(N_(l))/(l)I_(1)" "......(1)`
iv. The magnetic flux linked with each turn of `S_(2)` is equal to `B_(1)A.`
Total magnetic flux linked with solenoid `S_(2)` having `N_(2)` turns is
`phi=B_(1)AN_(2)`
Substituting for `B_(1)` from equation (1)
`phi_(2)=(mu_(0)N_(1)N_(2)AI_(1))/(l)" "....(2)`
But, `phi_(2)=MI_(1)" ".......(3)`
where M is the coefficient of mutual induction between `S_(1)andS_(2).`
From equtions (2) and (3),
`ML_(1)=(mu_(0)N_(1)N_(2)AI_(1))/(l),M=(muN_(1)N_(2)A)/(l)`
v. If the core is filled with a magnetice material of permeabilty `mu,`
`M=(muN_(1)N_(2)A)/(l)`