The concept that we need here to give a proper solution is mutual inductance.
The mutual inductance is given by the expression
[tex]M=\frac{N\Phi}{I}[/tex]
Where,
I = current
N = Number of turns
[tex]\Phi =[/tex]Flux through the solenoid.
Part A) Then we have in our values that,
[tex]I=6.6A[/tex]
[tex]\Phi= 3.50*10^{-2}Wb[/tex]
[tex]N=450[/tex]
Replacing in the equation,
[tex]M = \frac{450*350*10^{-2}}{6.60}[/tex]
[tex]M = 2.39H[/tex]
Part B) Here is required the Flux, then using the same expression we have that
[tex]\Phi = \frac{IM'}{N}[/tex]
We conserve the same value for the Inductance but now we have a current of 2.6, then
[tex]\Phi = \frac{2.6*2.39}{690}[/tex]
[tex]\Phi = 9*10^{-3}Wb[/tex]
Therefore the flux in Solenoid 1 is [tex]9*10^{-3}Wb[/tex]
