A coil that fights the change, never the current

Send a current $I$ through a coil of $N$ turns and each turn is threaded by a flux $\Phi_B$ that the coil's own field made. The self-inductance is the ratio

$$L = \frac{N\Phi_B}{I}, \qquad 1\ \text{H} = 1\ \text{Wb/A} = 1\ \text{V}\cdot\text{s/A}.$$

Read that the way you read $C = Q/V$: the numerator scales with the denominator, so the quotient never moves. Double the current and the field doubles, so $\Phi_B$ doubles, and $L$ is exactly what it was. A coil carrying no current at all has the same inductance as one carrying $10$ A. Current is the input; $L$ is the machine.

For a long solenoid the geometry can be worked out in three lines. With $n$ turns per unit length, $B = \mu_0 n I$ inside, the flux per turn is $\Phi_B = \mu_0 n I A$, and there are $N = n\ell$ turns:

$$L = \frac{(n\ell)(\mu_0 n I A)}{I} = \mu_0 n^2 A\ell = \mu_0 n^2 V.$$

The current cancelled, as it must. What survives is $n^2$: rewind the same length with twice the turns and $L$ goes up by four, once because each turn makes twice the field and again because twice as many turns are threaded by it. Stretch a fixed number of turns to twice the length and $n$ halves while $\ell$ doubles, so $L$ halves. An iron core multiplies $L$ by the core's relative permeability. Nothing on that list is a current.

Mutual inductance is the same ratio between two coils: $M = N_2\Phi_{21}/I_1$, the flux coil 1 puts through coil 2 per ampere in coil 1. It, too, is a property of the two shapes and their placement, and $M_{12} = M_{21}$.

Faraday's law applied to the coil's own flux gives the self-induced EMF, $\varepsilon = -N\,d\Phi_B/dt = -L\,dI/dt$. Read as a voltage across the element,

$$V_L = L\frac{dI}{dt}.$$

Everything the inductor does follows from the derivative on the right.

1. Steady current: zero volts. If $dI/dt = 0$ the ideal inductor is a piece of wire. It does not slowly choke off a steady current, and it does not drop a voltage proportional to $I$. A $10$ A DC current through a $5$ H coil meets no opposition at all.
2. Rising current: the coil pushes back. With $I$ increasing, the induced EMF points against the current, so the end where current enters sits at the higher potential, like a resistor.
3. Falling current: the coil pushes forward. With $I$ decreasing, the EMF reverses and drives the current onward, so the end where current leaves is now the higher one. The coil behaves like a battery trying to keep the current alive.

The commonest error is to make the inductor a resistor that impedes all current. It is not. Two situations with the same current and opposite slopes get opposite voltages, and a coil at $I = 0$ with $dI/dt = 4$ A/s has exactly the same voltage as one at $I = 100$ A with the same slope. What is opposed is the change, and both signs of change are opposed.

Students who have just met Ohm's law apply $V = IR$ to a coil of superconducting wire, get zero, and conclude the inductor can never have a voltage across it. That kills the one term that makes an inductor an inductor.

An ideal inductor drops zero volts through its resistance, because there is none. It carries a voltage $L\,dI/dt$ through its changing flux, and that term is sustained by induction, not by ohms. A voltmeter across a zero-resistance coil whose current is changing at $4.0$ A/s reads $L \times 4.0$ V, and it keeps reading that for as long as the slope holds.

A real coil has both. Its terminal voltage is

$$V = Ir + L\frac{dI}{dt},$$

with $r$ the wire's resistance. At steady current only the first term survives; while the current is changing, the second can dwarf it. In a loop equation the inductor contributes $-L\,dI/dt$ traversed in the direction of the current, exactly as a resistor contributes $-IR$, and the loop still has to sum to zero. That is what lets the inductor take the whole battery voltage at an instant when no current is yet flowing: $IR$ is zero and $L\,dI/dt$ is not.

To raise the current the source works against the back-EMF, delivering power $P = IV_L = LI\,dI/dt$ to the coil. Integrate from zero to the final current:

$$U = \int_0^{I} LI'\,dI' = \tfrac12 LI^2.$$

That energy is not heat and it is not stored charge. It sits in the magnetic field the current maintains, and it is returned when the current falls, which is why an inductor whose current is cut off drives its EMF as high as it must to keep the current going. Notice the square: doubling the current stores four times the energy at the same $L$, and $L$ itself did not change when the current did.

For the solenoid, substitute $L = \mu_0 n^2 A\ell$ and $I = B/(\mu_0 n)$:

$$U = \tfrac12\,\mu_0 n^2 A\ell\left(\frac{B}{\mu_0 n}\right)^2 = \frac{B^2}{2\mu_0}\,(A\ell),$$

so the energy per unit volume is $u_B = B^2/2\mu_0$, the magnetic partner of $\varepsilon_0 E^2/2$. Wherever there is a magnetic field there is energy in it, at that density, whether or not a coil made it.