# Is magnetic force conservative in an elastic collision?

1. Dec 4, 2014

### LmdL

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:
$$\nabla \times F=q \nabla \times E+q \nabla \times \left ( v\times B \right )$$
$$\nabla \times E =-\frac{\partial B}{\partial t}$$
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?

2. Dec 4, 2014

### Staff: Mentor

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.

Sure, the magnetic field for a fixed position changes in time as the car moves by.

3. Dec 4, 2014

### LmdL

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?

4. Dec 4, 2014

### Staff: Mentor

Right.
So what? There is no charged particle accelerating in this field.
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.

5. Dec 4, 2014

### LmdL

Ahhh, I got it. So the reason is not because dB/dt is not zero, but because q is negligible small.
Thank you!