We start from the Maxwell's Equation
$$ \mathbf{\nabla} \times \mathbf{B} = \mu \mathbf{J} + \overbrace{\mu \epsilon \dfrac{\partial \mathbf{E}}{\partial t}}^0. $$
We take surface integration of both sides, for the surface (\$s\$) inside the mean path (\$c\$) of the core.
$$ \int_s \left( \mathbf{\nabla} \times \mathbf{B} \right) \cdot d\mathbf{s} = \mu \int_s \mathbf{J} \cdot d\mathbf{s} $$
We use the Stroke's Theorem to rewrite the left hand side; where \$c\$ is in the same direction with the magnetic flux \$\Phi\$.
$$ \oint_c \mathbf{B} \cdot d \mathbf{\ell} = \mu N I $$
(The integral on the left hand side results \$NI\$, because there are \$N\$ different wires on the winding.)
The magnetic field density inside these kind of cores are considered to be uniform. So, we can write
$$ B \ell_c \overset\sim= \mu NI \implies B = \dfrac{\mu NI}{\ell_c}; $$
where \$\ell_c\$ is the mean path lenght of the core.
We can find magnetic flux from the magnetic flux density we have found by using the cross sectional area of the core \$A_c\$.
$$ \Phi = BA_c = \dfrac{\mu NIA_c}{\ell_c} $$
By definition, inductance is the amount of magnetic flux generated per applied current, that is
$$ L \overset\triangle= \dfrac{\Phi}{I}. $$
So, we find inductance of the system as
$$ \boxed{ L = \dfrac{\Phi}{I} = \dfrac{\dfrac{\mu NIA_c}{\ell_c}}{I} = \dfrac{\mu NA_c}{\ell_c} }. $$
But, all other sources (example) give inductance of an inductor like this as
$$ L = \dfrac{\mu N^2A_c}{\ell_c}. $$
What is the mistake I did in my derivation? Please explain in detail.
Answer
You calculate the core flux with the equotion above, and the inductance takes the sum of all fluxes through each turn. The flux through each turn is the same and equal to core flux. The core flux is proportional to N, and the per-turn sum of flux is proportional to \$N^2\$.
Another way to express this dependancy is to say: because of magnetic coupling between turns.
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