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The translational force on a piece of iron in a magnetic field

by deep838
Tags: field, force, iron, magnetic, piece, translational
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deep838
#1
Feb9-13, 12:09 PM
P: 105
Ok so if there be a piece of iron in a uniform magnetc field B, what will be the force on the iron?
You can assume any variable you want, but i want to find the magnetic force that causes the iron pieces to move towards the magnet in terms of any variables, like it corss-sectional area, length, position etc.
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mfb
#2
Feb9-13, 02:46 PM
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P: 11,589
The net force is 0. There might be a torque, depending on the orientation of the iron and its magnetic field (if present) and the external magnetic field.
You need a field which is not uniform to move magnets.
deep838
#3
Feb9-13, 10:04 PM
P: 105
Quote Quote by mfb View Post
The net force is 0. There might be a torque, depending on the orientation of the iron and its magnetic field (if present) and the external magnetic field.
You need a field which is not uniform to move magnets.
Oops... sorry! i forgot that! what i really had in mind was the force exerted on the iron by a bar magnet.... in that case the field is not unifrom.... sorry for that bit of wrong info! Anyway, i plan to first find the pole strength that develops on either faces of the iron and then use F=sB, s=pole strength... So how do i get my s?

mfb
#4
Feb10-13, 07:13 AM
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P: 11,589
The translational force on a piece of iron in a magnetic field

The iron bar will get a magnetic field similar to the external field, and this multiplied with the gradient of the external field should give some reasonable estimate for the force.
http://en.wikipedia.org/wiki/Force_between_magnets
deep838
#5
Feb11-13, 02:58 AM
P: 105
Quote Quote by mfb View Post
The iron bar will get a magnetic field similar to the external field[/url]
Isn't the field inside the iron bar B=μ(I+H) where H= magnetizing filed and I= intensity of magnetization?
But then what to do? I need to find the developed pole strengths first, right?
How do I do that?
mfb
#6
Feb11-13, 12:15 PM
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P: 11,589
A magnet does not consist of two separate poles. If you know the magnetic field (and assume it is the same everywhere in the magnet), it is fine.
deep838
#7
Feb11-13, 09:19 PM
P: 105
Quote Quote by mfb View Post
A magnet does not consist of two separate poles. If you know the magnetic field (and assume it is the same everywhere in the magnet), it is fine.
Tell me one thing.... to obtain the force on a magnetic material why should'nt I multiply the inensity B with the developed pole strength in the material (s)?
Have I got a misconception about these things? Please help me. I am absolutely at a loss now!
mfb
#8
Feb12-13, 09:36 AM
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P: 11,589
I think it is a detour. It is possible, but you just multiply with things (like the volume) and divide by them again afterwards.
deep838
#9
Feb12-13, 09:51 AM
P: 105
Ok... One more thing, does the developed pole strength at each end of the iron bar change with its position? I think the pole strength will keep varying as a function of the field the iron bar is in. Am I right?

By using the properties of the material, I got an equation for the pole strength. Its given by,
s=χBA/μ , where χ is the susceptibility of iron, B is the field it is kept in, A is its cross-sectional area and μ is the permeability of the surrounding medium. Is this correct?
mfb
#10
Feb12-13, 02:54 PM
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P: 11,589
One more thing, does the developed pole strength at each end of the iron bar change with its position? I think the pole strength will keep varying as a function of the field the iron bar is in. Am I right?
Right.

s=χBA/μ , where χ is the susceptibility of iron, B is the field it is kept in, A is its cross-sectional area and μ is the permeability of the surrounding medium. Is this correct?
I would expect the length somewhere, too, but it could be true.
deep838
#11
Feb12-13, 09:50 PM
P: 105
Quote Quote by mfb View Post
I would expect the length somewhere, too, but it could be true.
Well surprisingly enough, the length canceled when i equated the magnetizing field in terms of the moment and the volume. Here it is:

I=M/V=M/A.l=s.l.M^/Al=M^s/A
Then again, χ=I/H=Iμ/B... and then the above expression of the pole strength


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