Define co-efficient of mutual induction and find an expression for it.

Explanation

Coelicient of mutuat induction: It is defined as the magnetic flux inked with one coil when ynitrcurrent flows through the neighbouring coil.
Mufial tnductton of two long solenolds:
Let we have two long solenoids `S_(1) and S_(2)`
I = Length of each solenoid.
Suppose solenoid `S_(1)` is completely surrounded by the solenoid `S_(2)`
`n_(1)=` Number of turns per unit length of solenoid `S_(1)`
`n_(2)` Number of turns per unit length of solenoid `S_(2)`.
Let `I_(1)=` Current flowing through solenoid `S_(1)`.
Due to the flow of curent through solenoid `S_(1)`, magnetic field is produced inside it and magaetic flux will be linked with `S_(2)`.
`phi_(21)=` Magnetic flux linked with S, due to flow of current through `S_(1)`
We know magnetic flux produced in a coil is directly proportional to current flowing throughit.
So `phi_(21) prop I_(1)`
`phi_(2) =M_(21) I_(1)`
where `M_(21)` = Coefficient of mutual of the two solenoids when current is passed through coil `S_(1)` and induced emfis produced in solenoid `S_(2)`
Let `B_(1)`= Magnetic field produced inside the solenoid `S_(1)`
and `B_(1)=mu_(0) n_(1) I_(1)`
If A = Area of cross-section of solenoid `S_(1)`
Then magn tic flux linked with each turn ofthe solenoid `S_(2)= B_(1) A`
Therefore, total magnetic flux linked with the solenoid `S_(1)`,
`phi_(21)=B_(1) A xx n_(2) I=mu_(0) n_(1) I_(1) xx A xx n_(2)I`
or `phi_(21) =mu_(0) n_(1) n_(2) A 1 I_(1)" ...(i)"`
From equations (i) and (ii), we have
`M_(21) =mu_(0) n_(1)n_(2) A I" ...(iii)"`
Now, let us find the mutual inductance between thé two solenoids, when current is passed through solenoid `S_(2)` and induced emfis produced in solenoid `S_(1)` If a current `I_(2)` is passed through the solenoid `S_(2)` and magnetic flux `phi_(12)` is linked with solenoid `S_(1)`, then

`phi_(21) prop I_(2)`
or `phi_(12) =M_(12) I_(2)" ...(iv)"`
where `M_(12)` is coefficient of mutual inductance of the two solenoids when current is passed through solenoid `S_(2)` and induced emfis produced in solenoid `S_(1)`.
Now, magnetic field produced inside the solenoid `S_(2)` on passing current through it,
`B_(2) = mu_(0) n_(2)I_(2)`
The magnetic flux linked with each turn of the solenoid `S_(1)=B_(2) A`
Therefore, total magnetic flux linked with the solenoid `S_(1)`,
`phi_(12) =B_(2) A xx n_(1) I=mu_(0) n_(2) I_(2) xx A n_(1) 1`
or `phi_(12)=mu_(0) n_(1) n_(2) A l I_(2)" ....(v)"`
From equations (iv) and (v), we have
`M_(12) =mu_(0) n_(1) n_(2) A I" ...(vi)"`
From equations (ii) and (vi), it follows that in case of two long solenoids, the coefficient of mutual induction remains.the same, irrespective of whether the current is passed through the inner solenoid or the outer solenoid. Therefore,
`M_(21) =M_(12)= M` (say)
Hence, coefficient of mutual induction between two long solenoids,
`M=mu_(0) n_(1) n_(2) Al`