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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.
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.