electromagnetic force required to levitate an object


by bks008
Tags: electromagnetic, force, levitate, object, required
bks008
bks008 is offline
#1
Feb6-12, 07:39 PM
P: 2
I'm trying to use some home-made electromagnets to lift a small object. I am a college student, but this is a personal project, so while I may reference something from a textbook, its not a homework problem.

I know that I can get some iron bar, coil it with wire, run some current through it and see if it works, but I want to know the math behind it. I know I'm trying to make a solenoid with a core, and the formula my book gave me for the B field is B=μo*i*n where μo is the permeability of free space, i is the current in amps, and n is the number of coils. My book leads me to believe that B is in Teslas, but when I try out different numbers in Mathcad I get resulting units of Tesla-meters. I found on wikipedia B=[μo*i*n]/L where L is the length of the solenoid, and this gives me the units I'm looking for. Can someone point me into the right direction here?

Next question is, how do I relate the magnetic field of my electromagnet to the mass of the object I want to lift? I know I need to set the electromagnetic force equal to m*g*h, but I'm not sure where to go from there.

Am I even on the right track? If I'm not, I'd appreciate all the help I can get. I've searched and searched online trying to find this stuff but I haven't found anything overly helpful.
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rohans
rohans is offline
#2
Feb7-12, 10:27 AM
P: 7
hey the n (for the B field is B=μo*i*n where μo is the permeability) you mentioned first time is no. of turns per unit length
the n you mentioned the second time is no. of turns N

n=N/L
L- length of the solenoid



As to the basic question , what you are trying to do is impossible because magnetic field of a solenoid OUTSIDE it is ZERO (practically negligible)
bks008
bks008 is offline
#3
Feb7-12, 11:36 AM
P: 2
Quote Quote by rohans View Post
hey the n (for the B field is B=μo*i*n where μo is the permeability) you mentioned first time is no. of turns per unit length
the n you mentioned the second time is no. of turns N

n=N/L
L- length of the solenoid



As to the basic question , what you are trying to do is impossible because magnetic field of a solenoid OUTSIDE it is ZERO (practically negligible)

Thanks for the help with the first part. I'm embarrassed I didn't see that before.

As for the second part, if I add a ferromagnetic core (loosely quoting wikipedia), then the same equation holds true as long as I take into account the permeability of the core.


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