# A misunderstanding concerning self induction of an inductor

If there is an inductor in a certain electrical circuit, from Faraday's Law we know that $$\oint \vec{E} \cdot \vec{dl} = -\frac{d\phi_{magnetic}}{dt}$$ I see everywhere that when doing this line integral over a circuit with an inductor, the change in magnetic flux is $$-L\dot{I}$$, but shouldn't the magnetic flux be measured through the surface area enclosed by the line integral? Why would the inductor produce a magnetic field in that direction, doesn't it only produce a magnetic field parallel to the surface area enclosed by the circuit (like a solenoid). I'm confused, but what does the magnetic field produced by the inductor have to do with the magnetic flux through the surface area enclosed by the line integral (the circuit conducting lines)?

Remember that the inductor is part of the circuit, so you also need to consider the flux through the inductor. Also, since inductance is defined by $$L = \frac{\varepsilon}{\dot I}$$, it follows immediately that a circuit with inductance $$L$$ and derivative of current $$\dot I$$ is going to have a magnetic flux of $$L\dot I$$ through it (within some sign convention).