Solenoid Relative Force

For devices like this, force is proportional to the square of the voltage, minus nonlinearities and losses. So the force you'll get at 40 VDC is (40*40)/(30*30) or about 76% more than 30 VDC, assuming you're already in the zone that overcomes starting friction etc. There's also a return spring -- that has a static force to overcome, so the equation may become: (40*40*kL - kS)/(30*30*kL - kS) kL is the "force of inductor based on squared voltage constant" and kS is "force of spring constant." You can see from this that the higher the kS, the more relative force you'll actually get out at 40 volts than at 30 volts.

That being said, check if your benchtop power supply can be bridged; mine does 30V single, but bridged it can go to 60V.

Force from a solenoid = \$(N\cdot I)^2 \dfrac{μ_0\cdot A}{(2 g^2)}\$

Where:

• μ0 = 4π×10-7
• F is the force in Newtons
• N is the number of turns of the coil
• I is the current in Amps in the coil
• A is the area in length units squared (cross sectional area of the coil)
• g is the length of the gap between the coil and the iron.

If you want to check your solenoids pull force and you know the number of turns and have a good estimate of the "gap" and the cross sectional area of the winding then use the formula.

However, it's notable that force is proportional to current-squared and if the 12V you used dropped down to (say) 6V under the load of the solenoid, the pull force would quarter because the current would have halved. It's also worth noting that the force is inversely proportional to the gap-squared - double the gap and the force quarters.