Derive an expression for the mutual inductance of two long co-axial solenoids of same length wound one over the other,

Obtain the expression for the the mutual inductance of two long co-axial solenoids S_{1} and S_{2} wound one over the other, each of length L and radii r_{1} and r_{2} and n_{1} and n_{2} number of turns per unit length, when a current I is set up in the outer solenoid S_{2}.

#### Solution

Consider two long solenoids *S*_{1} and *S*_{2} of same length (*l* ) such that solenoid *S*_{2} surrounds solenoid *S*_{1} completely.

Let:

*n*_{1} = Number of turns per unit length of *S*_{1}

*n*_{2} = Number of turns per unit length of *S*_{2}

*I*_{1} = Current passed through solenoid *S*_{1}

Φ_{21} = Flux linked with *S*2 due to current flowing through *S*_{1}

Φ_{21} ∝ *I*_{1}

Φ_{21 }= *M*_{21}*I*_{1}

_{Where}

*M*_{21} = Coefficient of mutual induction of the two solenoids

When current is passed through solenoid *S*_{1}, an *emf* is induced in solenoid *S*_{2}.

Magnetic field produced inside solenoid *S*_{1} on passing current through it is given by

*B*_{1} = μ_{0}*n*_{1}*I*_{1}

Magnetic flux linked with each turn of solenoid *S*_{2} will be equal to *B*_{1} times the area of cross-section of solenoid *S*_{1}

Magnetic flux linked with each turn of the solenoid *S*_{2} = *B*_{1}*A*

*Therefore, total magnetic flux linked with the solenoid S _{2} is given by *

Φ_{21} = *B*_{1}*A* × *n*_{2}*l* = μ_{0}*n*_{1}*I*_{1} × *A*× *n*_{2}*l*

*Φ _{21} = μ_{0}n_{1}n_{2}AI_{1}*

*∴ M _{21} = μ_{0}n_{1}n_{2}Al*

Similarly, the mutual inductance between the two solenoids, when current is passed through solenoid S_{2} and induced emf is produced in solenoid S_{1}, is given by

*M*_{12} = μ_{0}*n*_{1}*n*_{2}*Al*

∴ *M*_{12} = *M*_{21} = *M* (say)

Hence, coefficient of mutual induction between the two long solenoids is given by

**M=μ _{0}n_{1}n_{2}Al**