The related concept to solve this exercise is given in the expressions that the magnetic field has both as a function of the number of loops, current and length, as well as inductance and permeability. The first expression could be given as,
The magnetic field H is given as,
[tex]H = \frac{nI}{l}[/tex]
Here,
n = Number of turns of the coil
I = Current that flows in the coil
l = Length of the coil
From the above equation, the number of turns of the coil is,
[tex]n = \frac{Hl}{I}[/tex]
The magnetic field is again given by,
[tex]H = \frac{B}{\mu_t}[/tex]
Where the minimum inductance produced by the solenoid coil is B.
We have to obtain n, that
[tex]n = \dfrac{\frac{B}{\mu_t}l}{I}[/tex]
Replacing with our values we have that,
[tex]n = \dfrac{\frac{1.1Wb/m^2 }{200000}(2m)}{4mA}[/tex]
[tex]n = \dfrac{(\frac{1.1Wb/m^2 }{200000})(\frac{10^4 guass}{1Wb/m^2})(2m)}{4mA(\frac{10^{-3}A}{1mA})}[/tex]
[tex]n = 27.5 \approx 28[/tex]
Therefore the number of turn required is 28Truns
