Relating Teslas to Force

Question

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.

Answer 1

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.

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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.

Answer 2

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.