Obtain an expression for the mutual inductance of two long solenoids.

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Solution

Explanation
The Magnetic field produced at any point inside the solenoid
$S_{1}$ due to the current $l_{1}$ is $B_{1} = \frac{μ_{0} N_{1} I_{1}}{l}$
Flux linked with each turen of $S_{2} = B_{1} A$
Total Magnetic flux with linked with solenoid having $N_{2}$ turns is
$ϕ_{2} = B_{1} A N_{2} = (\frac{μ_{0} N_{1} I_{1}}{l}) A N_{2}$
up to
$ϕ_{2} = M l_{1}$
$M = \frac{μ_{0} N_{1} N_{2} A}{l}$

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Obtain an expression for the mutual inductance of two long solenoids.

Explanation The Magnetic field produced at any point inside the solenoid S1 due to the current l1 is B1=μ0N1I1lFlux linked with each turen of S2=B1ATotal Magnetic flux with linked with solenoid having N2 turns isϕ2=B1AN2=(μ0N1I1l)AN2up toϕ2=Ml1M=μ0N1N2Al