When we're finding out the mutual induction of coaxial solenoids, and particularly of the outer solenoid; why are we taking the area of the inner solenoid in N2 Φ2 =N2 A1 B1? and why not A2? After all it should be of the solenoid S2. But they're saying we aren't taking this A2, area of cross section of outer solenoid, because there's no magnetic field in the region b/w the two solenoids. Then why why find the mutual inductance for the outer solenoid at all? I mean then, the outer solenoid shouldn't be having any back EMF at all since there's no change of flux through it!!!!!
Please help.

Question

When we're finding out the mutual induction of coaxial solenoids, and particularly of the outer solenoid; why are we taking the area of the inner solenoid in N2 Φ2 =N2 A1 B1? and why not A2? After all it should be of the solenoid S2. But they're saying we aren't taking this A2, area of cross section of outer solenoid, because there's no magnetic field in the region b/w the two solenoids. Then why why find the mutual inductance for the outer solenoid at all? I mean then, the outer solenoid shouldn't be having any back EMF at all since there's no change of flux through it!!!!!
Please help.

dustin_whitelock · Answer

Here's the figure for clarification

dustin_whitelock · Answer

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vincent-lyon.fr · Answer

Because B created by S1 is not uniform.
Flux through one loop of S2 is $B_1.\pi \,r_1^2+0.(\pi \,r_2^2-\pi \,r_1^2)$

dustin_whitelock · Answer

How did you obtain this: B1.πr21+0.(πr22−πr21)?
I only know this formula flux= vector field . vector area
Where did this second part of your formula come from?

vincent-lyon.fr · Answer

The actual formula is :
flux= Sum (vector field . vector area)

Here, one part of the total area has B1 flowing through it and the other part has no field acting at all.