2

Is the Lorentz Force acting on a wire, that has current $I$ in a magnetic field $B$ conservative? Or non-conservative? I understand that all the fundamental forces are conservative, am I correct?

Qmechanic
  • 201,751
Pupil
  • 1,110

2 Answers2

5

The Lorentz force $$ \vec F = \frac{d}{dt}\vec p_\text{charge} = q ( \vec E + \vec v \times \vec B ) $$ is in general not conservative, because it is possible to construct a system of two charges where the forces are not equal and opposite. Conservation of energy and momentum in electrodynamics requires you to also consider the energy and momenta associated with the fields.

Keep in mind that the Lorentz force on a moving charge $\vec F = q\vec v \times \vec B$, never does work. If magnetic field causes a current-carrying wire to accelerate, the energy comes from either the power supply controlling the magnetic field $\vec B$ or the power supply maintaining the current $\vec I$, and the work done must be computed using the induced electric field $\vec \nabla \times \vec E = -\partial \vec B / \partial t$. This is a very different creature from, say, a mass oscillating on a lossless spring, where the kinetic energy of the mass and the potential energy of the spring convert into one another.

rob
  • 89,569
  • what about the case of a wire that carries a lot of current? – Pupil May 20 '14 at 13:02
  • An infinite, uniform wire in an infinite, uniform field will feel a constant force $\vec F = \vec I \times \vec B$. Compare to other constant forces. – rob May 20 '14 at 13:22
  • Why would it matter if it's constant or not? In comparison to a conservative force? So in any case where the Lorentz force is applied its always non-conservative? – Pupil May 20 '14 at 22:36
  • @Key Well, $\vec F = m\vec g$ near Earth's surface is a conservative, constant force. I suppose that if you had an infinite straight wire carrying in an infinite uniform field you could swing it on a pendulum. But the example seems quite contrived. I'm not sure it's a well-posed problem. – rob May 21 '14 at 15:42
  • rob, is there ever a case where the Lorentz force is indeed conservative? – Pupil May 22 '14 at 12:29
  • @Key I think that a force that cannot do work cannot be conservative. – rob May 22 '14 at 14:26
  • I know that the lorentz force does no work on a charge, but what about the case of a current carrying wire? Work is done by the lorentz force, motion us gained... Due to the lorentz force? Thus a non-conservative force did indeed do work? – Pupil May 23 '14 at 00:45
  • Although you stated the energy comes from the power source that maintains the magnetic field or the current, however, the force is due to the lorentz and even if we wanted to calculate the force we used the basic formula : $F =$ $IL$ $x$ $B$ – Pupil May 23 '14 at 00:46
  • As I said in my answer, only electric fields can do work. A moving current-carrying wire generates a time-dependent magnetic field, which in turn generates an electric field (with nonzero curl). – rob May 23 '14 at 01:07
  • I deleted another post since you are right. However, I I wanted to be sure of the following: 1) The force is always equal to $F$ = $IL$ $x$ $B$, that can be used to calculated the work done. 2) The work done is directly due to the electric field, that is induced by the change in magnetic field, that force is a non-conservative forces. Am I right so far? I feel that I understood something – Pupil May 23 '14 at 10:35
  • 1
    @Key I think that you are right. – rob May 23 '14 at 13:04
  • @Key Another possible answer is in the title of the "duplicate" of your question: a magnetic field can be represented by a scalar potential, like a conservative force can, but only in the absence of currents. – rob May 23 '14 at 13:07
  • Rob, Something confused me earlier, we are discussing about an example that is without a power supply, what if there is a power supply? And the magnetic field in the Lorentz force is constant, the work done is by the electric force that is induced by the power supply, and the e-force is conservative? Its not magnetically induced, even when the conductor gains KE due to the work done by the force... this could change our conclusion. – Pupil May 26 '14 at 04:11
  • @Key I think you may be having an XY problem. "Conservative force" is not a pedagogically useful way to consider interactions with magnetic fields, except possibly in extremely contrived circumstances. Coming up with other contrivances won't make it sensible to say "magnetic forces are conservative." Do you have a deeper question lurking? – rob May 26 '14 at 04:25
  • Well, in this situation that i used multiple times where a wire is placed in a magnetic field and it experiences the Lorentz force,the work done is by the electric force, usually the magnetic field induces the electric field to do work, and the electric force applied then is non-conservative. However, when the magnetic field is created by a power supply, certainly the electric field is so as well, thus.. work is done by a conservative electric force. This is what I mean, nothing else. – Pupil May 26 '14 at 10:06
  • Aside from that... I have no other questions in mind. – Pupil May 26 '14 at 21:28
1

Is the Lorentz Force acting on a wire, that has current I in a magnetic field B conservative? Or non-conservative?

As the wiki page you linked to explains, conservative force is a force that acts on particle and work done by this force when the particle moves from A to B does not depend on the path, just on the points A,B.

Wire is not particle, so the original meaning does not apply. If you want to use the word conservative for wire, you'll need to define what it means.

I understand that all the fundamental forces are conservative, am I correct?

Electrostatic and Newtonian gravitational forces are conservative. If the particles in the source of the force move, electric force will not be electrostatic and thus won't be conservative.

Ján Lalinský
  • 37,229