Is magnetic force conservative in an elastic collision?

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SUMMARY

The discussion centers on whether magnetic force is conservative during an elastic collision between two carts equipped with permanent magnets. The participants conclude that, despite the presence of a moving magnet creating a time-varying magnetic field, the magnetic force can be considered conservative for practical purposes. This is because the magnetic field acts uniformly on the carts, which are not charged particles, thus allowing energy conservation to hold true in this scenario. Minor losses such as eddy currents may occur, but they can generally be ignored in a realistic setup.

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  • Understanding of elastic collisions and kinetic energy transformation
  • Familiarity with magnetic forces and the Lorentz force equation
  • Knowledge of Faraday's law of electromagnetic induction
  • Basic concepts of magnetic fields and their behavior in motion
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Physics students, educators, and researchers interested in electromagnetism, energy conservation, and the dynamics of magnetic forces in mechanical systems.

LmdL
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Hi,
Suppose I perform a basic experiment of an elastic collision between 2 carts. I do not want the carts to touch each other so I adjust a permanent magnet to each cart and give each one a starting velocity. Carts get close to each other, kinetic energy is transformed into "potential magnetic energy" and back, and carts are getting repelled.
In this case, is magnetic force is conservative? That is, can I say, that energy is conserved before and after the collision or equivalently that work of a magnetic force in this case is zero?

Attempt to see it from physics:
Curl of a conservative force is zero, but in case of magnetic force:
Lorentz force:
F=qE+qv\times B
Curl of force:
\nabla \times F=q \nabla \times E+q \nabla \times \left ( v\times B \right )
From Faraday's law:
\nabla \times E =-\frac{\partial B}{\partial t}
Therefore:
\nabla \times F=-q \frac{\partial B}{\partial t}+q \left ( v\left ( \nabla \cdot B \right ) - B \left ( \nabla \cdot v \right )\right )
Since \nabla \cdot B = 0 always, and \nabla \cdot v = \frac{\partial }{\partial t}\left ( \nabla \cdot r \right )=0 we are left with:
\nabla \times F=-q \frac{\partial B}{\partial t}
In my experiment I don't have magnetic field that varies in time, or does a moving magnet actually produce a magnetic field that varies in time?

Thanks in advance.
 
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LmdL said:
In this case, is magnetic force is conservative? That is, can I say, that energy is conserved before and after the collision or equivalently that work of a magnetic force in this case is zero?
Yes - to a very good approximation. You can have some losses (eddy currents, change of magnetization, ...), but for a realistic setup you can probably ignore them.

LmdL said:
In my experiment I don't have magnetic field that varies in time, or does a moving magnet actually produce a magnetic field that varies in time?
Sure, the magnetic field for a fixed position changes in time as the car moves by.
 
So, I don't understand. I actually have a moving magnet and therefore a magnetic field that varies in time. Therefore curl of a magnetic force in this case is not zero. So why energy is conserved?
 
LmdL said:
I actually have a moving magnet and therefore a magnetic field that varies in time.
Right.
LmdL said:
Therefore curl of a magnetic force in this case is not zero.
So what? There is no charged particle accelerating in this field.
LmdL said:
So why energy is conserved?
To a good approximation, your cars are not charged and the magnetic field acts on the whole magnetic part of the car in the same way.
 
Ahhh, I got it. So the reason is not because dB/dt is not zero, but because q is negligible small.
Thank you!
 

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