# Lorentz Force in Magnetic Circuit Air Gap

1. Apr 25, 2010

### f=qe+qvXb

[URL]http://en.wikipedia.org/wiki/File:Magnetischer_Kreis.svg[/URL]

The attached image shows a magnetic circuit with a small air gap. The magnetic material on each side of the gap is attracted to the other side. I've seen the magnitude of the force calculated based on principle of virtual work from an energy gradient.

My "problem" is I can't see the how the force arises in terms of the Lorentz equation. The amphere currents or electron spins are in the plane of the face of the magnetic material. The field is mostly perpendicular to the face. So in terms of the Lorentz equations how is the B field and currents oriented to get a force in a direction perpendicular to the face?

My first though is because the B field is in reality not uniform across the gap. However the virtual work method (which seems the be very popular for this problem) starts off by assuming the field is uniform across the gap.

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2. Apr 25, 2010

### gabbagabbahey

I suppose if you wanted to use the Lorentz Force equation directly, you could model the magnetized material as a bunch of tiny current loops, apply the Lorentz force equation to each current loop, add up the effects and get the same result. However, this is essentially how the equation $\textbf{F}=\mathbf{\nabla}(\textbf{m}\cdot\textbf{B})$ is derived in the first place.

You have to remember that each slab of magnetic material contains many accelerating charges that produce both time-varying magnetic fields and hence Faraday induced electric fields. The net magnetic field may be approximately uniform in the gap, but that's not what's really directly causing the force between the slabs. It's the Faraday induced electric fields that cause the force of attraction. If you do the calculation carefully enough, you will end up with the same net force between the slabs using the Lorentz Force Law.

3. Apr 25, 2010

### Phrak

I don't see how either charge or Faraday's law enters into this where the magnets are static. For an ideal, magnet were the dipoles are aligned and equally distributed, unless I'm mistaken, each magnet may be modeled as a solenoid. Populated with current loops in place of dipoles, the currents will cancel everywhere but the surface of the magnet.

Last edited: Apr 25, 2010
4. Apr 25, 2010

### gabbagabbahey

A current loop contains accelerating charges (the direction of their velocity necessarily changes in order for their motion to form a loop). Those accelerating charges produce Faraday fields, which result in an electric force on all the other charges. If it wasn't for these induced fields, there would be no attractive/repulsive force between current loops. Magnetic forces do no work. In order to account for any electrodynamic force that does work, like the force of attraction between the two slabs, you always have to look at the electric fields present if you plan on using the Lorentz Force law directly.

5. Apr 26, 2010

### Phrak

I don't quite understand your explanation. Two parallel wires will exert a force on one another where we won't have acceleration of charges as in the loop. Second, what sort of charge displacements did you have in mind where a Coulomb force or F=Eq would apply?

6. Apr 26, 2010

### gabbagabbahey

I feel we may be getting off topic here, but to answer your first question; two parallel wires running a steady current will attract/repel. However, in order to maintain a steady current work must be done on the charges in the wire and that work is mediated by an electric force, usually from the battery that is maintaining the current. For a more in depth explanation, see Griffiths Introduction to Electrodynamics example 5.3 (3rd edition).

7. Apr 26, 2010

### Phrak

I don't have a Griffiths, but if you consider this too far off topic, that's OK too.

8. Apr 26, 2010

### f=qe+qvXb

[/STRIKE]

First, thanks to you both for taking time to reply. I really appreciate it.

gabbagabbahey I’m going to need to study your answer more to understand it. This is a magnetostatics situation. Actually the faces of the slab could be faces of a permanent magnet. If there is an E field causing the attraction, then if the faces came together, shouldn’t there be current in the direction perpendicular to the faces?

I also don’t have Griffiths but will look for a copy to study the example you referenced.

In looking at the equation $\textbf{F}=\mathbf{\nabla}(\textbf{m}\cdot\textbf{B})$ something came to mind that I had not considered. While the B field is constant, the moment does make a step function change at the boundary of air and the slab. I’ve only studied force calculation with a constant moment and non-uniform fields. I will try looking a constant fields and non-uniform moments to get a normal force at least for fun. Although at first glance I don't see it creating a force in the correct direction either.