I was trying to make a device which can vibrate - so I took inductor, some 220uH 22uH, put 12V pulsed current through it (~10Hz), and put neodymium magnet - I was expecting to feel 10Hz vibration of a magnet in my hands when holding it 1mm from the inductor - but I haven't feel anything.
I tried 2 different magnets, result is the same.
What I did wrong? Is that due to the fact that my inductor is 'cored'?
What do you want to vibrate, how much does it weigh? Is a vibration motor from an old mobile phone a possible solution? It can make a 100g phone vibrate a move while laying on a table.
If you want to do it with the coil and magnet, the energy is way too low (see calculation below). Neodymium makes the highest flux permanent magnets, and maybe this isn't the problem here, and neither is the coil core, but think about a speaker. Pulsing 12V in an 8\$\Omega\$ speaker can be felt and heard. So what's the difference? To start with the number of turns. You need many turns on a coil to have a resistance of several ohms. In your coil this may be limited (you want low resistance in a coil!) so the field will be less strong. But more important: magnetic coupling between magnet and coil. In a speaker this is optimal: the coil's core = the magnet. You can't have better than that. You may have to place your coil very close to the magnet to have some vibration, but this will of course limit the vibration amplitude.
edit (after your inductance value correction)
Just for fun I decided to also make some calculations. I picked a 22\$\mu\$H coil similar-looking to the one in your question for reference. At 10Hz the inductance will be much lower than the coil's resistance, so we first look up the latter. Datasheet says 0.13\$\Omega\$. Then the reactance: \$X_L = 2 \pi f L = 2 \pi \times 10 \times 22\times10^{-6} = 1.4m\Omega\$. That's only 1% of the resistance, so we're going to ignore it. \$ I = \frac{12V}{0.13 \Omega } = 92A \$. A lot less than Russell's value, but still too much for your power supply (unless you use a welder). Assuming your power supply voltage collapses you'll hardly being inserting any energy in the coil.
note: this calculation assumes a sine wave current. In reality it will be more of a square wave, and then Russell's way of calculating is better. (At least in theory. In practice Russell ignores a factor which is \$10^2\$ larger than his calculated value, while Steven ignores a factor \$10^2\$ smaller than his calculated value.)
The inductance and frequency are almost certainly totally unsuited to the drive level provided. Unless you are providing 1000's of amps of drive (at which stage you'd need welding goggles to view the inductor) then you need a different arrangement.
The current through an ideal inductor will increase linearly with time when a constant voltage is applied.
Here
L = 220 uH = 2.2E-4
t = 1/2 of 10 Hz cycle = 0.05 = 5E-2second
V = 12 Volt
Most available power supplies are not capable of this sort of current :-).
For operation at around 10 Hz you need a MUCH larger inductance value. The inductor may not be much larger physically but it will have a very large number of turns comparatively. For a physically very small inductor the wire will become very fine, resistance will be high and resistance will start to play a very significant part.
To use the existing inductor at sensible power and current levels you would need to increase the frequency substantially and the resultant pulses would not be able to b sensed as vibration. For example, to reduce peak current to ~=2.7A would require operation at 10 kHz.
To start to experiment in this area the easiest way is probably to dismantle a small relay. This will usually have a laminated core, DC operates currents in the 10's of mA's to 100s of mAs range and would produce "pull in" forces that are noticeable.
Note that you can use a spring loaded "pole piece" to get mechanical vibration. Your magnet will successively provide attraction and repulsion on subsequent cycles. As the magnet will also be attracted to the core material you may still need to utilise a spring if you want approximately symmetrical attraction and repulsion forces.
* ADDED COMMENTS *
There are a number of comments on other answers. I opine that the basic core shape is OK but that the material and amp-turns are wrong. So -
Start with pulling almost any small relay apart as per my suggestion. They have done the work of optimising core and using VERY fine wire and many turns and ... .
Once you have a feel for it try your own. The core you are using MAY work with many many turns of the finest wire you can find BUT ideally you want a core with MUCH higher "permeability". This will greatly increase your inductance for a given ampere-turn product. A core of soft iron wires would work very well. Metal (iron) powder also.
The inductors can be anything of 3-5 kinds.
Your inductor is probably the last kind. So there is no much field around inductor.
Edit: After debugging. The cause is that inductor is too low on inductance for experiment to provide noticable force.