You don't say what your application is, what current and frequency.
If you wind them yourself an RLC meter will tell you what the inductance is. This will depend on the frequency, so make sure you set the meter for the right measuring range. (Mike rightly notes that RLC meters don't always support HF measurements.)
If you don't have an RLC meter, you can measure it by applying your AC signal to the coil in series with a resistor. On a scope you can measure the phase shift between voltage and current, from which you can derive the inductance/resistance ratio.
For air coils there are online calculators which give you the inductance as a function of inner and outer diameter, length, and the number of turns. The formula is
\$L (\mu H) = \dfrac{0.315 (N A)^2}{6 A + 9 B + 10 C} \$
Where N = number of turns, A = average coil radius, B = coil length and C = coil thickness, all in cm.
I've used coils from Coilcraft, but IIRC they're not cheap.
Vishay is another manufacturer.
Sumida is very inexpensive. I always had to have their SMD power catalog close at hand.
But actually there's dozens of them. Look for specific values at DigiKey, Mouser or Farnell.
Yes, a inductor sortof resists current changes, just like a capacitor resists voltage changes. In fact, inductors and capacitors are current/voltage mirrors of each other. The way I like to think of inductors in circuits is that they give inertia to current. They don't of course, but it seems a useful conceptualization technique.
In the schematic without the diode, if everything starts out at 0 and the switch is closed, the current will be a exponential decay toward Vs/R. Initially all the voltage is accross the inductor, and in the steady state there is 0 voltage accross it.
The interesting stuff happens when the switch is opened. At any one instance, the inductor will maintain its current constant. This includes the instance the switch is opened. Without the diode, there is no obvious path for the current. The inductor voltage will increase to whatever maintains the current thru it.
A mechanical switch works by touching together two conductors. When the switch opens, the conductors move away from each other. This can't happen instantly, so when the switch first tries to stop the current thru it, the contacts will be very close together. It won't take much voltage to cause arc over. Once the arc is started, the gas between the contacts becomes a plasma, which has high conductivity. The arc can therefore continue for a while as the contacts move farther apart. During this time, the voltage accross the switch isn't zero, so the inductor current decreases. As the contacts move further apart, the arc voltage increases, decreasing the inductor current more rapidly.
Eventually the current is low enough that it can't sustain the arc and the switch finally opens for real. At that point there is little energy left in the inductor. The only place for that current to go is onto the inevitable parasitic capacitance accross the inductor and other parts of the circuit. Every two conductors in the universe have some non-zero capacitance between them. This capacitance is small, and therefore the voltage will rise quickly. This also decreases the current in the inductor rapidly. Eventually a peak is reached where the voltage on the capacitance actually starts to push the inductor current the other way. In a perfect system, all the energy on the capacitance would be transferred to the inductor as current, but this time in the opposite direction. Then it would charge up the capacitance again in the opposite direction, and the whole cycle would repeat indefinitely. In the real world there is some loss, so each swing back and forth will be a little lower in amplitude as energy is lost as it is being sloshed back and forth between the inductor and the capacitance. Voltage plotted as a function of time (as a oscilloscope does) will show a sine wave with amplitude decaying exponentially towards Vs.
Best Answer
No it's SRF (not SFR) i.e. self-resonant frequency and it's the frequency that the inductor naturally resonates at due to self capacitance of the windings.
For RF stuff Q factor is important to note especially when they tell you at what frequency the Q factor was measured - if it is at a high frequency and the value is OK (20 upwards is a rule of thumb I guess) then you can be sure that at that frequency the eddy current losses in the core are reasonably low.
Ideally, for critical uses (such as oscillators) having a graph of Q factor versus frequency is quite useful because, for a given value of inductance you'd expect Q to rise proportional to frequency - at some high frequency this graph will begin to flatten and this usually tells you what the max frequency of use is.