How is a Push Solenoid Constructed Differently from a Pull Solenoid?

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SUMMARY

The construction of push and pull solenoids differs primarily in the attachment of the plunger. A pull solenoid features a hook on the left side of the plunger, while a push solenoid has a pushpin on the right side. Both types utilize a copper wire wound around a tube, and the plunger moves from left to right when activated. The force exerted by the plunger is determined by the magnetic energy within the coil, represented by the equation F = dEmagn / dx.

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I was reading http://www.rossdecco.com/pdf/how_a_dc_solenoid_works.pdf about push and pull solenoids. I don't quite understand what differentiates the two in terms of construction. What I have is a copper wire wound around a tube and the ends are attached to a battery, like in this I believe in this video it is a pull solenoid. How is the push constructed differently from the pull exactly? What can I do to construct that same thing seen in the video into a push type? Is it just a matter of length of the plunger?
 
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The only difference is whether a hook is attached to the left side of the plunger, or if a pushpin is attached to the right side.

The plunger itself is always moving from left to right when activated, because the plunger will take up more volume within the coil, when doing so. The "passive" position is skew to the center of the coil.

By seeking toward the center of the coil, the plunger will weaken the energy in the magnetic energy, Emagn , as much as possible. Placing an x-axis along the center axis of the coil, the force yielded by the plunger will be:

F = dEmagn / dx
 
Last edited:
Hesch said:
The only difference is whether a hook is attached to the left side of the plunger, or if a pushpin is attached to the right side.

The plunger itself is always moving from left to right when activated, because the plunger will take up more volume within the coil, when doing so. The "passive" position is skew to the center of the coil.

By seeking toward the center of the coil, the plunger will weaken the energy in the magnetic energy, Emagn , as much as possible. Placing an x-axis along the center axis of the coil, the force yielded by the plunger will be:

F = dEmagn / dx
Hm, that's helpful to understand. Though, a better question to ask now is what would the load be in both cases (an example of what would be)?
 

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