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I'm attempting to design a solenoid that pushes on a permanent magnet. I have a specific amount of force I need to apply to the magnet. I cannot, however, find anything on relating Teslas to direct force.

I'm aware that the Tesla is one Newton/Amp Meter. I get that this indicates the amount of deflection caused in a wire by a magnet based on its length and amperage. But I'm attempting to deal with another magnet. It has no amps.

What I need is to determine the force (in Newtons or Pounds) that the field will exert. Can I just ignore the amp part of the Newton and treat the Tesla as Newtons per Meter? That honestly seems like I'm missing something.

Brandon Thompson
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2 Answers2

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A Tesla is a unit of the magnetic field. The units of a force field are Newtons per unit relevant charge.

For example, the units of a gravity field are Newtons per unit mass $m$.

The units of an electric field are Newtons per unit electric charge $q$.

Similarly, the units of a magnetic field are Newtons per unit magnetic charge $q_m$.

Magnetic charges are sort of a fiction made up by physicists to be able to do math$^1$. They don't sit around existing intrinsically the way that masses and electric charges do. Rather, a magnetic charge exists when an electric charge is moving.

For a current carrying wire, Amperes tell us the electric charge flux - how many electric charges are moving through a given thin slice of material at any moment. The meters tell us how many thin slices of material there are. By multiplying them together we get the total flow rate of electric charges, which is our magnetic charge.

A permanent magnet can be modeled as an array of a huge number of tiny, atom-sized current loops, all lined up in the same direction with their axis of rotation on the magnet's north-south axis. Current flowing in a loop gets you a dipole$^2$ - a positive charge and a negative charge separated by a distance, and if we add them all up, we can represent the magnet as a single large dipole, with one large positive magnetic charge on the south pole and one large negative magnetic charge on the north pole. So, even though we wouldn't describe a magnet as having a current flowing through it, the huge number of tiny currents each contribute a tiny amount of amperes across a microscopic fraction of a meter, and the dipoles' charges are correctly expressed in A*m.

1: (once you get to quantum mechanics, all the charges turn into fictions and we just deal with field interactions, anyway, so magnetism isn't any different from a certain point of view)

2: if we're being completely correct, our straight current-carrying wire is best represented as a loop of infinite radius. Magnetic charges never exist by themselves, and all circuits have to loop back on themselves eventually, or the circuit is broken and current won't flow.

g s
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    Re “magnetic charges always exist as a dipole”: this has a technical meaning you may not have intended, which means it’s not really true. You’re right that there aren’t current distributions which have a magnetic monopole moment. But it’s possible to construct magnetic fields whose dipole moment is zero, but whose quadrupole, octupole, etc. moments are nonzero. – rob Jul 06 '21 at 13:39
  • Thanks. I'll alter my answer to be more correct. – g s Jul 06 '21 at 16:48
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Dipole-dipole forces are nonlinear and messy, much messier than monopole-monopole electric forces. A dipole in a uniform field experiences a torque, but not a net force; it’s non-uniformities in the field which cause dipoles to attract each other.

In your problem, if you are able to mechanically constrain the permanent magnet so that its magnetic moment $\vec \mu$ is parallel to the axis of your solenoid, then the force will be

$$ \vec F = (\vec\mu \cdot \vec\nabla)\vec B = \mu_z \frac{\partial}{\partial z}B_z \hat z $$

See a derivation and caveats here.

The magnetic moment $\mu$ for a loop with current $I$ and area $A$ is $\mu = IA$. This magnet supplier claims the moment for a permanent magnet is

$$ \mu = \frac{B_r V }{ \mu_0 } $$

where $B_r$ is the remnant field (typically about 1 tesla for neodymium magnets), $V$ is the volume of the magnet, and $\mu_0 \approx 4\pi\times 10^{-7}\rm\, T\, m / A$ is the permeability of free space. That formula is new to me, but the units are right, so it’s probably okay.

Your favorite E&M textbook will tell you how to compute $\partial B/ \partial z$ along the axis of your solenoid.

Beware that it’s more robust to have your solenoid attract (“pull on”) your permanent magnet than to repel (“push on”) it. In the short term, interacting dipoles in the repulsive orientation can reduce their energy by flipping around into the attracting orientation. In the long term, as your “permanent” magnet loses its intrinsic magnetization, your solenoid field can repolarize it; the repolarization will go in the attracting orientation.

rob
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  • First, let me thank you for the breadth of your answer which, honestly was a bit beyond my knowledge. I've never had a reason to learn vector math. However, I did find the equations I was going to use for this here:

    What I need to be able to do is convert Teslas to a unit of Force. If your equation does that, could you provide a key for it?

     – Brandon Thompson Jul 06 '21 at 15:46
  • Your points about the basic design are well taken as well. But I wanted to harness the drop in force specifically. The solenoid will be on for only .009 seconds at a time, up to 5 or 6 times a second, and will never actually contact the permanent magnet. If this is still an issue, I may have to exam the possibility of using stabilizing magnets to maintain its field and orientation (if that's even possible) or design the unit to easily swap the permanent magnets. Thank you again for your reply. – Brandon Thompson Jul 06 '21 at 15:51
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    The first equation gives the net force on the whole dipole, assuming the size of the dipole is small relative to the divergence of the magnetic field and the magnets are positioned such that they won't exert a torque on one another. You could use it to find the force between two magnets at long distances, although it would start to give incorrect results when the magnets were close together. – g s Jul 06 '21 at 16:59
  • Can I assume that force is in Newtons? – Brandon Thompson Jul 08 '21 at 04:06
  • @BrandonThompson If you use $\mu$ in $\rm A,m^2$ and $\partial B/\partial z$ in $\rm T/m$, then the force is in newtons. – rob Jul 08 '21 at 11:20