Choosing between two magnet arrangements

[Pjdd]

College physics was too long ago and I haven't kept myself updated in magnetics. So I'm asking for help from those with a more current knowledge. Precise numerical values are not necessary. Informed opinions are welcome, but no wild guesses please.

I want to make a simple magnetic stirrer and am trying to work out which of two alternative arrangements will be more effective. That is, which one - A or B in the diagram - will have the stronger magnetic loop. I can't just get the result by experiment because I don't have the materials yet.

According to the online calculator at kjmagnetics.com, the two magnet sizes - 8x3.2mm and 12x2mm - have equal surface pulling powers alone. But I reasoned that, since the 12x2mm pair are closer together, much of their fields would link up without reaching the stir bar - forming a partial magnetic short circuit - and thus exert a weaker pull on the bar. Am I on the right track or is there a flaw in my logic? Thanks in advance for your inputs.

All magnets are Neodymium N50 grade. The magnets in both schemes have the same outer edge-to-edge (not center-to-center) distance of 30mm. Hence the two 12x2mm types are closer to each other.

The gap between the stir bar and the upper surfaces (not the bottom) of the magnets are the same (6mm) in both arrangements.

To help with visual judgment, the drawing is made to scale within +/-1 pixel.

 

[FvM]

What do you mean with "12x2"? Rod with 2 mm diameter? Or quadratic block? How's the magnet orientation?

 

[Pjdd]

Oops! Sorry about the omissions. The magnets are discs (short cylinders) axially magnetized with their axes vertical in the diagram. 8 and 12 are the diameters in mm while 3.2 and 2 are the lengths or thicknesses. Each pair has the poles in opposite directions.

I did mark the pole orientations at first but I must have erased them at some point and forgot to put them back. Here's the corrected diagram.

 

[c_mitra]

I did mark the pole orientations at first but I must have erased them at some point and forgot to put them back. Here's the corrected diagram.

The basic idea is that the magnetic stirrer is basically (yes!) a simple motor. The rotating magnetic field is produced by the rotating magnets in the figure- the stronger the field the better.

The stir bar is the motor rotor. It is a simple bar magnet- the stronger the better.

You can replace the rotating magnets (shown in your picture) with flat coils that are driven with AC (then you will not have to rotate the disc)- they will produce a rotating magnetic field.

 

[Pjdd]

True, and I can think of other ways to generate the required rotating magnetic field. But the object here is to use readily available materials that need nothing more than some screws and a bit of glue to assemble. The rotating magnets are to be fixed onto a common computer cabinet fan (not my own original idea; I got it from the internet).

Any idea about my question? That is, about the relative efficiencies of the magnetic coupling in the two alternative arrangements?

 

[c_mitra]

Any idea about my question? That is, about the relative efficiencies of the magnetic coupling in the two alternative arrangements?

Your question is rather tricky. Let me try to elaborate:

I assume by "equal pulling power" you mean same or similar pole strength. In that case, the thicker magnet will have larger magnetic moment (pole strength X distance between the poles)

The bar magnet provides a low reluctance path; the amount of flux that will NOT go through the bar magnet is not easy to calculate.

The mounting surface (30 mm iron sheet) should have some holes to guide the flux path. Without doing calculations, I would go with the thicker magnets, because it will have greater moment. Greater moment has greater reach.

With the iron sheet below shorting the N-S poles (of the two disk magnets) the arrangement is like a horseshoe magnet. In that case, the magnetic moment will depend on the distance between the two disks- the smaller dia magnets are better.

Overall, I feel that the smaller dia magnets will be better for your experiments.

 

[Pjdd]

Thanks for your interest. Your thoughts agree with mine, but since my recollection of magnetics is rather fuzzy, I was going by intuition and common sense. I couldn't be sure and was looking for someone to confirm or contradict it.

 

[FvM]

Overall, I feel that the smaller dia magnets will be better for your experiments.

The larger discs have a higher volume and thus store more energy. But the larger center distance of small magnets may compensate the effect by generating more torque. I don't think that the question can be clearly answered without either an empirical test or a 3D simulation.

 

[c_mitra]

The larger discs have a higher volume and thus store more energy

The energy consideration is irrelevant because this energy is not transferred; the magnetic field simply acts as a slippy coupling (and the dissipated energy comes from the mechanical energy of rotation of the disks).

Once we consider the magnetic dipole moment, the volume of the magnet does not matter. This dipole moment produces a magnetic field with a 1/r^3 law (approx; valid only at large distance). At this point, you need not bother how the magnetic field is being produced. (strictly speaking, you need to consider the quadruple and octupole moments too because the distances are not large)

effect by generating more torque...

I agree that the original question was badly stated; in the diagram shown, the stir bar will follow the rotating disk perfectly and slippage will increase with the viscosity of the solution being stirred. But this energy is coming from the rotation of the disks containing the two magnets.

At high rotations, the stir bar may fail to move at all; some energy will be dissipated as eddy current losses. But the magnetic coupling will not be affected.

I may be wrong, as usual.

 

[Pjdd]

.... slippage will increase with the viscosity of the solution being stirred. ....... At high rotations, the stir bar may fail to move at all;

I anticipated that and it's why I'm making a PWM speed control to run it at optimum speed for each operation.

 

[FvM]

The energy consideration is irrelevant because this energy is not transferred; the magnetic field simply acts as a slippy coupling (and the dissipated energy comes from the mechanical energy of rotation of the disks).

Thanks for explaining obvious things. At the end, the external field produced by magnets with similar configuration is related to the contained energy, though.