A basic question about inductor current in a switching circuit
This phenomenon confuses most beginners due to the whole, "A complete circuit must exist for current to flow!" thing. I find it is A LOT easier to understand if you add in the parasitic impedance around the inductor. 

Closing the circuit

With an IDEAL switch, change your circuit and add a VERY LARGE impedance across the inductor, as shown below.







simulate this circuit &ndash; Schematic created using CircuitLab

That impedance exists in real life, though perhaps not with the value I chose in this example. It comprises of, amongst other things,  air resistance, PCB resistance, and capacitive coupling across the inductor.


  In a theoretical ideal circuit that impedance is infinite, but it still factors in, read on.


Now re-examine the events. 

If the switch is initially closed long enough, a steady state current, in this example 15A, exists in the inductor along with a magnetic field. At this point, since an ideal inductor has no resistance, there is zero voltage across the inductor. The current is dictated entirely by R1 which has the full supply voltage across it.

When the switch opens, the magnetic field maintains the current in the coil in the same direction till it decays.

Now the only place for that current to come from is through the parasitic impedance. That means I-Coil, 15A, is flowing up through Z1. By simple Ohm's Law you can see that there will be a LARGE voltage drop up across Z1.

That puts the top of L1 at a very negative voltage,  -15 terravolts with the indicated values.


  Again, in a theoretical ideal circuit, the impedance, as I mentioned earlier, is infinite.  That makes the peak -15 x Infinite = -Infinite Volts. See, it still factors in.
  
  In reality of course something somewhere will break down before those voltages will be reached. Either an arc will form, insulation will breakdown, or some component will fail.


Of course, since Z1 is large, the LZ time constant is very small, so the voltage spike is very short.

Reality Check

In the real world, the impedance will actually be across the switch. 

As the contacts break an arc will be formed due to the high negative voltage on the right side of the switch coupled with the initially very short contact distance. The arc acts like a switch, ionizing the air between the contacts, forming a much smaller impedance across the switch. The voltage on the right of the switch will then be at some less negative level low enough to just maintain that arc. Since LZ is less, the current will take much longer to decay.

In fact, with the wrong kind of switch, the arc resistance can be so low as to allow the voltage on the inductor to rise back up positive close to the supply voltage. At that point the arc will be maintained indefinitely, driven by the supply. Or at least till the whole thing melts or catches fire and falls apart.

The energy from the inductor will be released in the heat of the arc and cooking the switch, and may even blow your power supply.







simulate this circuit


  If the switch is actually a transistor of some type, it will, in all likelihood, be destroyed. This is why you will see folks on here continually reminding people to add fly-back diodes to any circuit driving significantly inductive loads, like relay, solenoid, motor and transformer drivers. Even the inductance of long wires to a purely resistive load can cause the same destructive effect if the switching edges are fast enough and the currents are high.


Adding A Diode

In your second example..



You have added a low resistance path, the diode, for the current to continue to flow when the left end of the inductor tries to go negative. 

That is, when the switch opens the current initially tries to pass through the parasitic impedance. This drives the left end negative as before. However, when the voltage reaches about -0.7V, the diode will turn on and carry the current and hold the left end of the inductor at close to -0.7V.

The voltage on the right side of the inductor is dictated by the voltage drop across the resistor, which initially, with 15A still flowing, is still 15V.

Before the switch opens there is zero voltage across the ideal coil. When the switch first opens, the initial voltage across the coil is -15.7V







simulate this circuit