There could be some small back EMF effect when the slug moves. However, I seriously doubt that is what is causing the high voltage on closing. There may be other effects:
- Relay contacts bounce. That means the solenoid will be disconnected multiple times even during a overall "on" operation. These short disconnects that happen after some current has built up could cause high voltage for a short time.
- Ringing. There is inevitable capacitance in the system accross the coil. When the coil is switched on, it is like energizing a tank circuit. In ideal conditions, this could ring up to twice the input voltage especially with contact bounce. In practise, the DC resistance of a solenoid is usually substantial enough to damp the system well enough, and the R and L of the solenoid dominate.
- It's not really there. The scope may be showing you things that aren't really happening at transients and especially with poor probe grounding.
I don't know what exactly is happening, except that I'm quite skeptical the EMF is really going to 200 V. I also don't like the PTC fuse being in series for testing these things. Try shorting it out and see what that does. Also try putting a reverse diode immediately accross the solenoid, not at the other end of a wire or on the other side of the PTC fuse. This should be a fast diode.
Let me reduce the complexity of your question by asking you to forget about power - the power you talk about is irrlevent to the question - 3 volts at 100 amps doesn't tell you anything about the magnetics - it just defines the resistance of the coil at 0.03 ohms.
Next is the permanent magnet producing a flux density of 0.5 teslas. You are questioning what voltage will be induced in the solenoid by this magnet and you are quite rightly equating B (flux density), solenoid area and time in order to calculate the induced voltage.
Well, it's a bit more complex because the flux density from a permanent magnet will not be constant and the induced voltage in the solenoid will depend on the distribution of flux from the magnet. The solenoid's area you say is 0.15 sq m or a diameter of about 0.44 metres and it is very unlikely that a magnet that can be moved around this solenoid will be producing a reasonably constant B across the whole area of the solenoid but, putting that to one side let's assume it is.
The voltage induced in a 100 turn solenoid will be 75 volts.
Keeping the current at 100A in the solenoid can still be achieved using a fairly straightforward (but powerful) current source and as the magnet moves around the solenoid you will see the terminal voltage of the current source produce 75V too. The trouble is that a real current source will not like being reversed biased because the induced voltage is alternating each time the magnet moves around the solenoid.
In fact the current flowing through the coil is irrelevant - if no dc current flowed through the coil and the magnet was moved around you'd still see an induced voltage of 75 volts.
A practical idea for a circuit that keeps the current constant is to use filtering components that block the alternating 75V at the current source but, the low speed and high current requirement make this practically impossible to achieve.
Maybe you'd like to explain your idea a little more.
Best Answer
Personally, I have never understood the "induced-emf" line of thinking. To me, it is easier, and much easier not to get it wrong, from the circuit and equations point of view.
I am simplifying the solenoid to be just an inductor. The equation for Circuit 1 is: $$ V_L = V_1 = L \frac{di}{dt} $$ It tells us right away that the voltage across L must be V1 and never goes to zero. The positive change of current never stops and i goes to infinity. Obviously, this is a bad model.
The obvious improvement to the model (unless the battery is underpowered) is to include the ohmic resistance that you want to neglect. That is Circuit 2. The equation: $$ V_1 = V_L + V_R = L \frac{di}{dt} + R i $$ Initially, i is zero. All of V1 is across L. i increases.
As i increases, voltage drops across R. Voltage across L decreases, rate of i increase slows down.
Eventually, when i reaches the level where R x i = V1. Voltage across L becomes 0.
If you solve for \$V_L\$ and \$i\$, you get something like: $$ V_L = V_1e^{-\frac{R}{L}t} $$ $$ i = \frac{V_1}{R} (1 - e^{-\frac{R}{L}t}) $$ \$V_L\$ is not the voltage across the solenoid, we need to include the voltage across R for that, the total would then be \$V_1\$.