Question

derive an expression for mutual inductance of two long co axial solenoid

Answer 1

Answer:

let I₁ And I₂ be the current flowing through the solenoid .

S₁ & S₂ will be solenoids .

we know the magnetic feild inside the solenoids is,

B₁ = μ(N₁ / L) I₁

the magnetic flux is passing through S2 also,so magnetic flux linked with each turn will be

Ф = B₁ X A

Ф= [μ (N₁ / L) I₁ X A

the mutual inductance between 2 solenoids are ,

M= N₂ Ф/I₁

M= N₂[ μ (N₁/L)/I₁ (I₁A)

M= μN₁N₂A/ l

total flux linkage with inner solenoid are

N Ф₁ = (n₁l B₂I₁)

M₁₂I₂= ( μΠn1n2lr²)I₂