Force on a ferrous object inside a non-uniform magnetic field

In summary, the conversation discusses how to compute the force applied on ferromagnetic material inside a non-uniform magnetic field. The suggested method involves magnetizing the material and using Lorentz law to compute the induced currents. The use of programs like FEMM is also mentioned and an example of a simple problem is given. The conversation also mentions using the magnetic potential to compute the force and discusses the possibility of omitting elastic forces on the iron cylinder.
  • #1
EnEE
16
3
Hello everyone!

I want to know how to compute the force applied on ferrous inside a non-uniform magnetic field, to make it easier, let's assume that the field direction is constant, and the magnitude decreases linearly.

Now, I know that ferromagnetic material inside a magnetic field will become magnetized, and once he become magnetized, I can compute the induced currents on the surface and inside the matter(curl of M, and Mxn, where M is the magnetization), and therefore I can use Lorentz law to compute the force.
Is that right to do this?

I saw that usually engineers using programs like FEMM to solve such problems, and I try to figure out how to solve it analytically

simple examples:
http://uzzors2k.4hv.org/index.php?page=magneticlevitation - the force applied should be similar to this:
https://en.wikipedia.org/wiki/Force_between_magnets#Force_between_two_cylindrical_magnets

simple problem:
assume we have an air-core solenoid produces a magnetic field(instead of the iron-core in the example), I want to know what is the minimum number of turns or the minimum current I need to pull a given iron cylinder (given distance from the solenoid, and given mass, and for simplify the problem I can assume that the cylinder is right in the symetric axis of the solenoid)
 
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  • #2
OK, few thougths...
1.my first thougth about induced currents and lorentz law will not work here
2.If I will use the magnetic potential: μm=0.5⋅μ⋅H⋅H
I can compute the force by F=-∇μm

H is function of z, so F is also function of z. If I omit ellastic forces on the iron cylinder(can I? maybe tension?)
I will get: F=-∫∇μm between z1 and z2(z2-z1 is the length of the cylinder)

what do you think?
 

What is force on a ferrous object inside a non-uniform magnetic field?

The force on a ferrous object inside a non-uniform magnetic field is the result of the interaction between the magnetic field and the magnetic properties of the object. This force can cause the object to move or experience torque depending on its orientation and the direction of the magnetic field.

How is force on a ferrous object inside a non-uniform magnetic field calculated?

The force on a ferrous object inside a non-uniform magnetic field can be calculated using the formula F = B∇(m⋅H), where B is the magnetic field, ∇ is the gradient operator, m is the magnetic moment of the object, and H is the magnetic field intensity.

What factors affect the force on a ferrous object inside a non-uniform magnetic field?

The force on a ferrous object inside a non-uniform magnetic field is affected by the strength and direction of the magnetic field, the magnetic properties of the object, and its orientation in relation to the magnetic field.

What happens if a ferrous object is placed in a non-uniform magnetic field?

If a ferrous object is placed in a non-uniform magnetic field, it will experience a force and may also rotate depending on its orientation. The object will tend to align itself with the direction of the magnetic field in order to minimize the force acting on it.

How can the force on a ferrous object inside a non-uniform magnetic field be manipulated?

The force on a ferrous object inside a non-uniform magnetic field can be manipulated by changing the strength or direction of the magnetic field, altering the magnetic properties of the object, or changing its orientation in relation to the magnetic field. This can be done by using various devices such as electromagnets or by physically moving the object.

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