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)?
1 Answer
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).
In general, an inductor consists of a coil that produces a magnetic field perpendicular to that coil, regardless of how the inductor is oriented with respect to the rest of the circuit, so it will produce magnetic flux through a surface enclosed by a contour along the inductor, and since you can often assume that all of a circuit's inductance is in an inductor, you can take the line integral just around the coil of the inductor.
