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Magnetic repulsion
Magnetic repulsion  
Hello Steve,

I am working on a project now that involves using electromagnets to move other nearby objects, and I need to calculate the minimum field strength required to perform this operation. I am trying to understand the first step to finding the answer, and I would really appreciate your expert opinion on my reasoning. You have given me a lot of help in the past and I am really grateful. I have never really had much science education and I am always plagued with doubts that I have missed something!

I have a plastic tube containing an electromagnet at one end. On top of the electromagnet is a plastic cylinder which contains a permanent magnet. When the electromagnet is switched off, the permanent magnet is attracted to the iron core of the electromagnet and the plastic cylinder is held in place. When the electromagnet is switched on it produces a magnetic field which repels the permanent magnet in the plastic cylinder, lifting it up. The structure of the tube prevents the plastic cylinder from moving to the side, or from moving more than a set distance away from the electromagnet. (The entire structure could be moved and tilted to any orientation, but for these calculations I have imagined the tube is vertical, which I am assuming is going to present the greatest resistance.)
Have I imagined the up and down movement of the plastic cylinder accurately?

I am trying to work out the necessary field strength the electromagnet needs to produce in order to repel the permanent magnet, and the forces that need to be overcome to do this. My thinking that the required field strength must be greater than the weight of the permanent magnet, plus the field strength/pull force of the permanent magnet.
Is this correct?

I am guessing that the extent to which the required field strength needs to be greater than the weight of the permanent magnet combined with the pull force of the permanent magnet is related to the friction between the plastic cylinder and the sides of the plastic tube - is this right?
Could you tell me how I might factor in the friction to these calculations?

Once again thanks very much for all your previous answers - they were a great help.
Eddie

Answer
Hello Eddie,

I apologize for the delay in replying. I agree with your description of the operation of your device.

Regarding your sentence "My thinking that the required field strength must be greater than the weight of the permanent magnet, plus the field strength/pull force of the permanent magnet." I would say that this way just to make things as clear as possible: "When the permanent magnet is in the fully repelled position, the force of repulsion must be greater than the weight of the permanent magnet." To accomplish this, these 3 factors must be considered, the field strength of the electromagnet, the field strength of the permanent magnet, and the distance the permanent magnet is to move when fully repelled. I'll refer to the distance the permanent magnet is to move when fully repelled with the variable name d. And yes, the attraction of the permanent magnet for the core would work against the repulsion.

A similar worst case that you need to consider is with the device rotated to upside down compared to your figure. The permanent magnet's attraction to the core when the separation distance is d also needs to exceed the weight of the permanent magnet for it to work in the upside down orientation.

Yes, any force of friction that the permanent magnet experiences will have the same affect as its weight. So the required magnetic force has to be greater than the sum of the weight and the friction (when in the orientation shown).

If you go to the following website and scroll down to the section "Force between two cylindrical magnets", you will see the complexity involved here.
http://en.wikipedia.org/wiki/Force_between_magnets
Notice the terms inside the brackets. Their x is my d. That shows that the force of attraction or repulsion will decrease at the rate that the square of the separation distance d increases. So if had it working, and decided to double the distance d that you want the permanent magnet to move, you would have to increase the magnetic strength by a factor of 4.

Let me suggest an experiment. Put 2 equal strength permanent magnets in your tube and measure the height at which the top one floats. That is d. If you know the performance data of the magnets, that will suggest the strength that the electromagnet should have.

I hope this helps,
Steve

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Steve Johnson

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I would be delighted to help with questions up through the first year of college Physics. Particularly Electricity, Electronics and Newtonian Mechanics (motion, acceleration etc.). I decline questions on relativity and Atomic Physics. I also could discuss the Space Shuttle and space flight in general.

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I have a BS in Physics and an MS in Electrical Engineering. I am retired now. My professional career was in Electrical Engineering with considerable time spent working with accelerometers, gyroscopes and flight dynamics (Physics related topics) while working on the Space Shuttle. I gave formal classroom lessons to technical co-workers periodically over a several year period.

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BS Physics, North Dakota State University
MS Electrical Engineering, North Dakota State University

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