gilshultz:
With this configuration your on voltage will be 12 Volts when you turn it off your transient voltage should be less then 0.5 with a reasonable sized MOSFET.
Care to explain what you mean by that?
gilshultz:
The nice thing about this solution there is no polarity to worry about.
There will be if the magnet already includes a diode!
As most engineers know when a diode is conducting the voltage across it is the Vf (Voltage Forward) While standard silicon diodes have a forward voltage drop of about 0.6 V and germanium diodes 0.2 V, Schottky diodes' voltage drop at forward biases of around 1 mA is in the range of 0.15 V to 0.46 V (see the 1N5817 and 1N5711), which makes them useful in voltage clamping applications and prevention of transistor failure. With MOSFETs you need to look at the avalanche break down voltage. Now with that knowledge we also know that when we have an inductor charged and disconnect it the polarity will reverse and the voltage will rise until limited by some external means, even insulation and leakage count. That in my example will cause the MOSFET to avalanche limiting the voltage to about 0.5 including wire drop. Hopefully this helps. MOVs will work for a while but are slithered metal and work by being consumed so they will eventually quit working. To simplify my conclusion: the energy is stored in the inductor, not the switch. This is why SMPSs work, the inductor is charged and then discharged into the load by switching devices. Remove the inductor and they will not work. You can do that with capacitors but you get very bulky and not much power and poor efficiency. The radiated or conducted interference is caused by the power to the inductor being switched off. The faster it is switched the bigger the transient. Most of what is here can be found in basic and intermediate electronics books and what we were taught in our schooling in our electrical and electronic courses. We see many examples of students looking for an answer without wanting to to do the work, that is there loss. This response is to help you get started in solving your problem, not solve it for you.
Good Luck & Have Fun!
Gil
control logic- long wire - driver(FET) - diode - solenoid.
power suply- long wire - resrvoir cap - driver(FET) - diode - solenoid.
gilshultz:
With this configuration your on voltage will be 12 Volts when you turn it off your transient voltage should be less then 0.5 with a reasonable sized MOSFET.
OK, but I am still waiting for an explanation of that statement. 
Where do you visualise this "transient voltage", what has a "reasonable sized MOSFET" got to do with the matter and what is an "avalanche rated MOSFET" supposed to do, and in your reply #19 were you describing using a diode across the magnet or what?
I've asked @gilshultz the same before - at least you got something of a reply 
Some explanation here (video with transcript). Indeed MOSFETs can absorb some amount of energy from a switching coil, the problem is that unless you know exactly what the inductance is and the current involved you can not calculate the energy involved and whether the MOSFET can handle it.
Also when applying PWM it's becoming a different story, as the current is switched hundreds of times a second.
While an interesting parameter, I don't see it as a replacement for a flyback diode.
Paul, I'm an experienced EE and I have some issues with the Magical Thinking critique.
I do agree the switch would experience a transient in current. However the large voltage spike which causes much of the radiated emissions comes from the high voltage generated by the inductor, therefore I expect it it still good practice to close that current loop by placing a diode or MOV close to the inductor.
I also agree that properly bypassing the supply switching device is important.
What I can't get my mind around is why you suggest placing the diode or other protection device close to the switch. While there is a very fast current change, I only see a small capacitor at the switching node necessary to clip the voltage spikes which may show up as a result of the di/dt.
After reading several other authoritative sources, I can't find anyone else who validates the idea of keeping the diode close the supply.
Please do educate, justify, and elaborate.
Side note: Maglocks typically do have MOVs (not diodes) and are polarized mostly as many have status LEDs or other circuitry in them which is polarized.
gabriel_escape_room_techs:
Please do educate, justify, and elaborate.
My previous explanation is point by point. I suggest reading it a few times over again. I am happy to discuss any point where you feel I have made an error in the logic. 
gabriel_escape_room_techs:
Side note: Maglocks typically do have MOVs (not diodes) and are polarized mostly as many have status LEDs or other circuitry in them which is polarized.
The ones I cite from China use diodes. Yes, I looked! This is consistent with being polarised and containing the status LED. If nothing else, I think diodes are cheaper than MOVs. 
Don't MOVs have a very limited life span of just a handful of transients, as the material gets used up in the process?
On further consideration, I fancy I might add a little to the previous discussion.
gabriel_escape_room_techs:
I do agree the switch would experience a transient in current. However the large voltage spike which causes much of the radiated emissions comes from the high voltage generated by the inductor, therefore I expect it it still good practice to close that current loop by placing a diode or MOV close to the inductor.
On Gabriel's part, this demonstrates the exact "magical thinking" I described as it ignores pretty much everything I explained!
The inductor does indeed generate a voltage whose value is determined purely by the switching device (and the diode). But if we neglect the behaviour of the transmission line in between the switching device and the inductor, this voltage - and its transients - will be exactly the same at each end of the transmission line, so there is no difference in regard to radiation of voltage transients.
So referring to "close that current loop" is clearly muddled thinking. Were we talking about voltage or current? Two entirely different things. If you "close the current loop" with a diode at the inductor, then the inductor sees a smoothly (exponentially) declining current but the transmission line sees an almost instantaneous transient in the current and it is exactly such transients - not the modest decline in the circuit containing the inductor - which radiate interference.
So now let's take a look at that transmission line.
I won't consider resistance; the simple model of a transmission line is a series inductor and parallel capacitance. The capacitance can perhaps be considered at each end. Well, the transmission line inductance is in series with the primary inductor. If you put the diode at the switching device end, then the inductance acts with the primary inductance and as described, resists sudden changes in the current which is to say, minimises inductive transient radiation.
If you place the diode at the primary inductance end however, you have now created in the transmission line, a second inductor in series with the primary which will add to the voltage transient at the switching device and enhance inductive transient radiation. It may not in itself contain enough energy to damage the unprotected switching device.
Capacitance at either end of the transmission line will indeed serve to slow the voltage transient but will conversely cause a transient when the inductor is switched on. Capacitance toward the inductor end will tend to cause radiation from the transmission line while capacitance at the switching device end will increase more the current surge seen by the switching device.
Note the general principle that it is the transmission line which radiates interference due to switching transients and the current transients occur in that part of the transmission line which is on the switching device side of the diode, so placing that diode near the primary inductor causes all of the transmission line to be such a potential radiator while placing it at the switching device limits this to the loop originally described, formed by the power supply (or the local bypass capacitor), the switching element and the diode.