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What is the equation to find the expected force of electromagnet? What do the variables mean and how do you find each variable?

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    $\begingroup$ It you look in the "Related" sidebar you'll see many related (or duplicate) questions including Pull Force of an electromagnet. This is a very popular question, but it is also one that does not yield a tidy formula or equation. Sorry, but that's the long and short of it. $\endgroup$ – dmckee May 1 '16 at 1:14
  • $\begingroup$ You need numerics for this. For starters try FEMM femm.info/wiki/HomePage for simple 2.5d magnet designs. $\endgroup$ – CuriousOne May 1 '16 at 1:25
  • $\begingroup$ Doesn't the force depend on what the force is exerted on? $\endgroup$ – don_Gunner94 May 1 '16 at 2:07
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    $\begingroup$ @don_Gunner94: Yes, it does, and ferromagnetic materials at the magnetization that is useful for mechanical applications tend to be non-linear, that's why it is almost inevitable to use numerics. $\endgroup$ – CuriousOne May 1 '16 at 2:12
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    $\begingroup$ @don_Gunner94 You don't? I mean, you can wave your hands at it, but frankly the most that will accomplish is giving them a false sense of comprehension. The subject simply requires a certain level of knowledge to encompass. $\endgroup$ – dmckee May 1 '16 at 4:37
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I agree that this question may be a duplicate, but previous questions seemed to be focused on particular geometries and situations. Here's my two cents..

The Lorentz force is

$J \times B$

You might combine Maxwell's equations (while neglecting the timescales for charges to move to conductor surfaces ($\frac{\partial \mathbf{E}}{\partial t}$)) to get the induction equation

$\frac{\partial \mathbf{B}}{\partial t} + \nabla \times \left(\sigma^{-1} \left[ \nabla \times \frac{\mathbf{B}}{\mu} \right] \right) = \nabla \times (\mathbf{u} \times \mathbf{B}), \qquad \nabla \bullet \mathbf{B} = 0.$

Assuming there's no moving conductor, we may write this as

$\frac{\partial \mathbf{B}}{\partial t} = - \nabla \times \left(\sigma^{-1} \left[ \nabla \times \frac{\mathbf{B}}{\mu} + \mathbf{J}_{external} \right] \right).$

Where $\mathbf{J}_{external}$ is the externally applied current. The magnetic field can be solved from the induction equation, and current ($\mathbf{J}$) may be found from ampere's law $\mathbf{J} = \nabla \times \frac{\mathbf{B}}{\mu}$

In summary, the induction equation can help give you some intuition as to how the magnetic field behaves, this gives you information about the magnetic field and current (via ampere's law).

Hope this helps.

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