Notice ya got no replies yet.
Vpeak = Inductance * peak coil current divided by time it take to open the circuit. It is all the rate of change of current and inductance.
peak coil current is Vdc/coil resistance
time to open depends on how fast you pull the contacts apart. FET turn off can be as fast at 40 nanoseconds.
Let us say your coil current is only 100 mA, and typical coil inductance is 10mH
Vpeak = 1E-2*1E-1/40E-9 = 25kV.
So, ya obviously NOT opening the circuit with a fast FET, more like taking a 1/4 usec for the current to drop to zero as arc ... typical af pulling apart contacts fast by hand.
Put a backwards diode across the terminals, voltage will clamp at a few volts, the current decays very slowly, in this example case,
L/R = 10mH/270 ohms = about 1/3 millisecond. The coil current hold the relay closed about that long also.
Carry the L/R to extremes, a superconducting coil and low R diode, the relay may take many seconds to open after voltage removed.
Short the superconduting coil with a superconductor an relay will /latch/ closed.
etc etc.
So Can I assume that the 100mv peak will go up proportionately with the applied voltage?
Thus, yes, since the current goes up proportionally with coil voltage at dc. Assuming you remove the applied voltage at the same speed. Note that in the old points ignition in a car, a capacitors is used across the points to 'quench' the points arcing That arcing slows the opening time to milliseconds. Transistorized ignition systems have fast opening times, as fast as the 40 nanoseonds previously mentioned.