College physics is too many decades in the past. I hope someone can help me estimate which of the two arrangements below will have the stronger external field as defined below.
Magnets: Neodymium N50 discs. For simplicity assume that the
8x3mm and 12x2mm types individually have the same surface pull strengths. Note that the two pairs do *not* have the same centre-to-centre distance. Instead they have the same outer edge-to-edge distance. That is, the 12x2mm types are closer to each other.
What's needed is which pair will exert the stronger pull on a magnetic object placed at a point midway between a pair several millimeters (say 6mm) above their upper surfaces. Absolute values in magnetic units are not needed.
Do you mean by placing, say, a copper wire between the two magnets and moving either the magnets or the wire, which one would create the highest voltage ? Everyghing else being equal except for the flux density from the two different magnets ?..... As in Faraday's law, V = N * dI/dT ?
No, this is about static mechanical pull force on a stationary magnetic object. What you brought up is dynamic induction of emf.
The pull force or strength of attraction exerted by a Neodymium magnet at a pole surface is well known, given the specific grade and dimensions. This pull force diminishes rapidly with distance.
The situation becomes more complex when two magnets interact and the attracted object is not perpendicular to either magnet. In the diagram below, the combined fields of the two magnets exert a pulling force on a magnetic object - say a small piece of iron - at point X. My question is about the relative strengths of that pull in the two arrangements.
In that case, the lower pair wins no matter what, doesn't it? As shown, it has the smaller gap and wins. If the magnets are moved farther apart, they still have the larger area and still win. Is that what you mean?
My thinking was that, since the larger magnets are closer to each other, their fluxes would link up more easily - sort of a partial magnetic short-circuit and their combined field would weaken more rapidly with distance and exert a weaker force on an object some distance away. Of course this is just some common sense deduction and could be way off.
To start, that's not the best assumption. The internal FIELD, and the field at the pole face, might be the same, but that is NOT SIMPLY RELATED TO PULL FORCE.
Approximately, a distant pair of magnets will be characterized by their dipole moments, but this situation has magnet size which is not small compared to the separation, so that kind of simplifying assumption won't work well. That means 'center-to-center distance' has to be, regretfully, a fiddly little detail rather than a dominant parameter.
What WILL work, is a complete formula for the B field, direction and magnitude, in all of space around a magnetized short cylinder. Add the two such, and then integrate over all space the squared ampitude of the B field. (ignore the space inside the cylinders, because that's not gonna change field according to position)
The pull between the two magnets is then
F = mu_zero * d/d{X30mm) [volume_integral_of_B_squared_d^3(V)]
Sometimes symmetry helps, but your cylinders aren't on a common axis, so the only useful symmetries are that the integral of B_squared on the upper half-space is the same as on the lower half-space., and the 'near' half-space is the mirror-image of the 'far' half space. A numerical integration calculation then is only enumerating little volume elements of a quadrant.
The derivative can be computed by redoing the integral for a 30mm span and for a 30.01mm span; it's not really infinitesimal, but it's small enough for all practical purposes.
This kind of calculation (boundary values, field solutions, odd shapes) is the reason for array processors and supercomputers.
Sadly, AFAIK no such closed-form formula exists. There's a formula for the magnitude of the B field along the z axis of symmetry of a short cylinder but I've never seen anything else.
Yep. It seems like it should be a fairly straightforward problem, but I don't think it is.
Check out Jackson _Classical Electrodynamics_ equation 5.38 in my old first edition... it's not closed-form, but neither is sine or cosine, and that case (a current loop) is just one short integral away from modeling a solenoid (stack of infinitesimal current loops == cylinder magnet).
If you take 300mm 'radius', in 1mm^3 discrete chunks, and use symmetry (so it's only a quadrant to calculate) it's only gonna take a few dozen million calculations of that formula with the three components and the K and E elliptic integrals... this laptop would handle that in background, in seconds.
Writing the code, and debugging, isn't completely trivial though.
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