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Assume I have tiny magnet particles in vacuum, which are not considered as a points, although they are very tiny. Magnet particles are situated in a uniform magnetic field.

How can I can describe the movement of the particles?


My try

Lets consider 2D case. Paricles are $4\mu m*2\mu m$ rectangles. I want to model mechanics by the next way:

Consider object as array of vertices that are connected with a thread. Force can be applied only to vertex. Because of vertices are conected, when one moves, it pulls the rest(image below will help to understand)

enter image description here

If it would be gravity I should define the mass of each vertex, and the gravity field among it. In case of magnetism I am confusing of what should I correspond with each vertices. Note that the vertices are not elementary particles, they are just some imaginary parts of the microscoping body(or macroscoping, nevermind), so the Lorentz force can't be applied.

Variants to set to vertices:

  • Magnetic inductance $\vec{B}$. Unknown
  • Magnetic susceptibility value of the whole body $\chi$. Known for me, $\chi = 3*10^{-6}$
  • Assume each vertex is a current loop, to use Amper Law, so I need the current $I$. Unknown.

So basicly, for 2D case, can somebody who maybe already have been doing the modeling similar to this, figure me in which way should I move?


Update 1

Found this answer https://physics.stackexchange.com/a/20265/255554. So the force at a point can be found from the product of magetic moment $\vec{m}$ and magnetic inductance $\vec{B}$ at this point:

$$\vec{F}(x) = \vec{m}(x)\times \vec{B}(x)$$

To find magnetic moment I have to find, again, the current:

$$m=IS\vec{n}$$

,where $I$ is current of the loop, $S$ is space of the loop, $\vec{n}$ is normal vector to loop plane.

Can megnetic moment can be expressed from magnetic susceptibility value?


Update 2

Okay, the relationship between $\chi$ and $\vec{m}$ can be found:

$$\vec{m}=\vec{M}V$$ , where $M$ - magnetization, $V$ - volume $$\vec{M}=\chi \vec{H}$$ , where $\vec{H}$ - external magnetic field strength $$\vec{m}= \chi \vec{H} V$$

Although, earlier I considered a 2D case, the volume is known, and equals $95 \mu m^3$, but this is the volume of the whole body, and I need only a small area of vertex, so let it be equal $95/100=0.95 \mu m^3$, and $\vec{H}=100 \frac{A}{m}$, then:

$$\vec{m}=3*10^{-6}\space *\space 100\frac{A}{m}\space *\space 9.5*10^{-19} m^3=2.85*10^{-22} A*m^2$$

Magnetic moment is found, now wondering what I should do with $\vec{B}$. Initially, I wanted to set external magnetic field as a magnetic induction field, but now I have to deal with magnetic field strength.

Will it be correct if I define magnetic induction in each point as

$$\vec{B}=\mu_0(\vec{H}+\vec{M})$$?

If yes, I think that’s all

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    $\begingroup$ $m \times B$ is the torque on a dipole not the force $\endgroup$
    – hyportnex
    Commented Mar 2, 2020 at 16:24
  • $\begingroup$ A magnetic dipole does not experience a net force in a uniform magnetic field. $\endgroup$
    – R.W. Bird
    Commented Mar 2, 2020 at 18:53
  • $\begingroup$ @R.W.Bird how much should field not be uniformed? I mean, if it “almost” uniform? $\endgroup$
    – Artur
    Commented Mar 2, 2020 at 18:58
  • $\begingroup$ The net force on a dipole is proportional to the divergence of the field. A magnetic field spreads most noticeably in the vicinity of the north pole of a magnet. $\endgroup$
    – R.W. Bird
    Commented Mar 3, 2020 at 14:35
  • $\begingroup$ @R.W.Bird but the divergence of the magnetic field is zero. And the last sentence, You meant north and south, not exactly north, yes ? $\endgroup$
    – Artur
    Commented Mar 3, 2020 at 18:44

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