The logic of the self inductance formula

nabliat
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M_{12}=\frac{N_2\phi _{12}}{I_1}

i need to remmeber it

but i can't see the logic of
the formula
??
 
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nabliat said:
M_{12}=\frac{N_2\phi _{12}}{I_1}

i need to remmeber it

but i can't see the logic of
the formula
??

I thought the formula looked something like: \frac{\mu_{0}N^2 A}{l}. A is the cross section & magnetic field inside the solenoid is B=\mu_0 NI/l, flux through one loop is BA=\mu_0 N IA/l and \Phi= \mu_0 N^2 I A/l is the total flux. The inductance is just total flux divided by current.
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
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