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In the following system, enter image description here I created the following equation in terms of what I assumed to be the force acting on the movable object with a mass M. The reluctance of the core is neglected, so the total reluctance of the system only consists of the air gaps:

\begin{align} \ddot{x}m + kx - \tau &= 0 \\\\ \ddot{x}m + kx - \phi R(x) &= 0 \\\\ \ddot{x}m + kx - \phi \frac{2x}{\mu_{o}A_{Gap}} &= 0 \end{align}

The solution however, has the following formula for the force applied on the movable part:

\begin{align} F_{Gap} &= \frac{\phi^2}2 \frac{dR(x)}{dx} \\\\ &= \left(\frac{Ni\mu_{o}A_{Gap}}{2x} \right)^2 \frac{1}2\frac{2}{A_{Gap}\mu_o} \\\\ F_{Gap} &= \frac{(Ni)^2\mu_{o}A_{Gap}}{4x^2} \end{align}

Where was the solution's equation derived from, and why is the mmf not used instead?

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    \$\begingroup\$ en.wikipedia.org/wiki/Magnetomotive_force Since the units for MMF are not Newtons, I guess the answer to your title is no. But I would interpret "force generated by the magnetic field" as the mechanical force being applied. \$\endgroup\$ – DKNguyen Oct 13 '20 at 4:45
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    \$\begingroup\$ Unfortunately, the term 'magnetic field' is ambiguous. Some people use it to mean H field, some to mean B field. The total force on the moveable part includes k, but those equations don't. So perhaps the force being derived is only due to the magnetic component? \$\endgroup\$ – Neil_UK Oct 13 '20 at 4:48
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    \$\begingroup\$ MMF is the potential to create a magnetic field, just as EMF is the potential to create an electric field (voltage) : neither is mechanical force. Just as you divide EMF by resistance to find current, you can divide MMF by reluctance to find magnetic flux. \$\endgroup\$ – Brian Drummond Oct 13 '20 at 10:49

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