Why dont magnetic fields made by a wire impede the current within said wire after the magnetic fields are made? In an inductor why dont the magnetic fields impede current after the magnetic fields are fully created? Because it seems counter intuitive that they wouldnt make it really hard for current to flow. Like the negative or south pole of an inductor is on the positive side of current so shouldnt that make it harder for current to flow and why is the negative pole there and not on the negative side?
Any answers would be greatly appreciated! Thank you!
 A: This is a reflection of Faraday’s law $$\nabla \times \vec E =-\frac{\partial \vec B}{\partial t}$$ when the magnetic field is “fully created” that means it is no longer changing or it is at a maximum so the time derivative is 0 and therefore there is no voltage across the inductor.
A: Imagine a circuit that has a voltage source $V$ and a resistor $R$. There may or may not be a technical inductor in the circuit, but an inductor is just resistive wire wound many times so we'll treat is as regular wire for now.
This circuit forms a geometric loop we notate as $\Gamma$. When current $I$ is flowing in this circuit a magnetic field is generated in space. The flux of the magnetic field through the circuit loop can be calculated as the integrated flux $\phi$.
We quantify the "self"-flux $\phi$ (i.e. the magnetic flux a circuit flowing current generates through its own geometry) per unit of current $I$ as the self-inductance of the circuit:
$$
L = \phi/I
$$
Faraday's law tells us
$$
\oint_{\Gamma} \boldsymbol{E}\cdot d\boldsymbol{l} = -\frac{d\phi}{dt}
$$
Combining these definitions we see
$$
\oint_{\Gamma} \boldsymbol{E}\cdot d\boldsymbol{l} = -L\frac{dI}{dt}
$$
But, on the other hand, for our simple circuit, we can calculate $\oint_{\Gamma} \boldsymbol{E}\cdot d\boldsymbol{l}$. The electric field is 0 in the conductive wires (including the wires making up any inductor) so we need only consider the contribution from the voltage source $V$ and resistor.
We have
$$
\oint_{\Gamma} \boldsymbol{E}\cdot d\boldsymbol{l} = -V + I R
$$
I take $\Gamma$ to traverse in the direction from the negative terminal of the battery towards the positive. The first term is then negative because the electric field in the battery points from positive to negative terminal and the second term is positive for the opposite reason.
Putting this altogether we get
$$
-V + IR = -L\frac{dI}{dt}
$$
Note that if the circuit has no self-inductance (or we neglect the circuit's self-inductance) we recover Kirchoff's voltage loop rule.
We can rewrite
$$
V-L\frac{dI}{dt} = IR
$$
Ok, we can now answer many questions at once.

*

*Do magnetic fields from an inductor make it harder for current to flow? Yes. If we solve $V-L\frac{dI}{dt}=IR$ we'll see the current is reduced for non-zero $L$. So yes, the magnetic field created by the self-inductance of a circuit reduces the current that flows in that circuit.

*What if a circuit doesn't have an inductor? It still creates a magnetic field? Does that reduce the current in the circuit? Yes. In my example $L$ stood for the inductance of any circuit wire geometry including one that does not include many explicit windings. A circuit with no looping will have a small inductance $L$ compared to one with loops, but the inductance can still be calculated as $L=\phi/I$. For low-frequency circuits self-inductance can be neglected unless an explicit inductor is included in the circuit. But at high frequencies the "accidental" inductances due to circuit geometry can be critical to enhance or deteriorate circuit performance.

