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The correct answer depends upon the quality of the magnet and the quality of the iron that it is attracted to. In general terms, the holding force will be approximately 280 lbs. of force for each square inch of surface area that is between the magnet and the iron. The force is calculated as follows. The mks formula for force is F=B²A/2µ where F is in newtons, B in Tesla, A in sq. meters, and µ is the permeability of air which is 4·pi·1E-7. Then, considering the fact that good quality iron saturates at about B=2.2 Tesla, and 1 sq.in. is = 6.45E-4 sq.meters and plugging this into the eqution gives a force of about 1242 newtons or 279 lbs. If the horse shoe magnet is holding the iron at each end of the magnet then the force will double if lifting is exactly balanced at each end. The above assumes very good surface contact between magnet and iron. Practical holding forces may be lower which can be approximated by lowering the value of B to something like 1.7 or so. dbm 7/16/09

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