# Is self-inductance dependent on geometry? Is $\Phi$ net flux or just flux due to cirucit?

Self-inductance of a circuit is defined as $$L=\frac{\Phi}{I}.$$ But it is not clear what $$\Phi$$ (read phi) is. Is it net magnetic flux or flux due to the magnetic field created due to current in the circuit?

Now suppose we have a loop, as shown in the first fig, through which some non-zero current $$I$$ is passing. A magnetic field will be created and flux through any surface bounded at the circuit will be non-zero, in this case. Hence the inductance of the coil is $$L=\frac{\Phi}{I}\not=0.$$

Now suppose if a machine creates such a magnetic field so that net magnetic flux (due to the field of the coil and the machine) through the same circuit be zero. Then if $$\Phi$$ is net magnetic flux the inductance of the coil will be $$L=0.$$

But it can be found in all books that self-inductance is dependent on geometry. But here the geometry of the circuit and even the current through it is not changed but still, its inductance changed but it shouldn't. This happens because I took $$\Phi$$ to be net magnetic flux, but if I take it to be flux due to the magnetic field of the circuit only then no problem arises. But I can't be sure of it, since it is not explicitly mentioned anywhere I checked. So am I correct?

The flux $$\Phi=\int \vec B \cdot d\vec A$$ is thus not zero.
As for geometry do you not think that a loop with one turn will have the different total flux passing though it as a solenoid of the same area but having $$100$$ turns with the same current passing through it?