Inductors & magnet - no vibration
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.)