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.
Choosing an inductor value for a buck regulator comes directly from V = \$\frac{\text{L di} }{\text{dt}}\$ . Where V is the voltage across the inductor, and i is the current through it. First, you want to design for the case where the inductor is in continuous conduction mode (CCM). This means that energy in the inductor doesn't run out during the switching cycle. So, there are two states, one where the switch is on, and another where the switch is off (and the rectifier is on). Voltage across the inductor during each state is essentially a constant (although it is a different value for each state). Anyway since the voltage is a constant, the inductor equation can be linearized (and rearranged to give L).
L = \$\frac{V \text{$\Delta $t}}{\text{$\Delta $I}}\$ this is the basis for the equation you saw in the app-note.
\$\text {$\Delta $I}\$ is something you define, not determine.
You will want to maintain CCM operation, so define \$\text {$\Delta $I}\$ as some small fraction of inductor current (I). A good choice is 10% of I. So, for your case \$\text {$\Delta $I}\$ would be 0.24A. This will also define the ripple current in the output capacitors, and less ripple current means less ripple voltage on the output.
Now you can choose an optimal value of L using \$V_{\text{in}}\$ and \$V_o\$ (and hence the duty cycle D = \$\frac {V_o} {V_ {\text {in}}}\$). But you can also make a quick over estimate for the inductance where you don't consider \$V_{\text{in}}\$ using L ~ \$\frac{10 V_o}{I_o F_{\text{sw}}}\$ (for more on this look here How to choose a inductor for a buck regulator circuit? ). An over estimate can be worthwhile, especially if you are early in development or uncertain exactly how much the output current will be (output current tends to end up higher than expected usually).
Since you are looking at Linear Tech you should (as Anindo Ghosh pointed out) also look at using their CAD support.
Best Answer
The more rapidly you remove the terminals, the more an arc will want to form.
Ideal inductance is defined by:
$$ v(t)= L\frac{\mathrm di}{\mathrm dt} $$
If you remove the terminals "instantly", then the current must stop "instantly". That means the di/dt term will approach infinity, and consequently the voltage term will, also.
The faster you stop the current, the higher the voltage generated. That energy has to go somewhere, and the harder you try to stop it, the more it will try to get out. If considering this situation in a purely theoretical, ideal situation where you posit the energy can't go anywhere, it's simply not possible. There is no real mathematical solution.
In practice, there are places the energy can go besides arcing. If you could somehow magically eliminate arcing in an otherwise real circuit, you'd have to consider:
In practice several of these will be in play, in addition to arcing. Depending on the particular design of the circuit some may be more significant than others.