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Lorentz magnetic force

  1. Aug 12, 2010 #1
    Hi.

    My question is how the Lorentz magnetic force [itex]F=q\vec{v}\times\vec{B}[/itex] is derived? Is it due to magnetic moment of charged particles? If the moving charged particle does not have magnetic moment (due to spin, so the torque on the particle is zero) how the magnetic force changes its direction of motion? The potential energy of that particle is changing too? If so, than the work is done
    on that particle?
     
    Last edited: Aug 12, 2010
  2. jcsd
  3. Aug 12, 2010 #2

    gabbagabbahey

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    No, the (Magnetic) Lorentz Force Law is empirically derived (from experiments with currents and moving charges) and, as the occurence of [itex]q[/itex] (the particle's charge) in the formula suggests, it describes the force on a moving point charge. It is not due to the particle's magnetic dipole moment, but rather its charge (electric monopole moment) and its motion (relative to the inertial observer that measures the force) through a magnetic field. The force on a dipole with magnetic moment [itex]\textbf{m}[/itex] can be derived from the Lorentz Force law , and is given by [itex]\textbf{F}=\mathbf{\nabla}(\textbf{m}\cdot\textbf{B})[/itex] (give or take a minus sign). The electron behaves like a point particle with charge [itex]e[/itex], and also has an intrinsic dipole moment [itex]\mathbf{\mu}_e[/itex], so it will be subject to both a Lorentz Force [itex]\textbf{F}=e\textbf{v}\times\textbf{B}[/itex] due to its motion (velocity) and a Lorentz force [itex]\textbf{F}=\mathbf{\nabla}(\mathbf{\mu}_e\cdot\textbf{B})[/itex].
     
  4. Aug 12, 2010 #3
    Ok i think i understand it now, thank you very much!

    But what about the second question(potential energy)? In case where the particle don't have intrinsic dipole moment the potential energy is changing (so work is non zero)? When it has intrinsic dipole moment it is obvious that the potential energy of the particle is changing so the work done by magnetic field on that particle is non zero.
     
  5. Aug 12, 2010 #4

    Born2bwire

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    The work done is always zero. This is because the force from the magnetic fields is always instantaneously directed to be normal to the velocity. This means that the force is always directed normal to the path of motion and thus the dot product of F and dx along the path o motion is zero (thus the work done is zero). The magnetic fields are enacting a change in momentum, but not a change in energy. We can extract energy out of the magnetic fields and transfer it to charged particles, but this is always done using an electric field as a mediator. The reasons why an electric field comes about are usually rather complicated though.

    As for the dipole moment, classically a dipole moment is a loop current. Classical electrodynamics does not allow for intrinsic dipole moments to arise without a current as a cause. It is only in quantum mechanics that we ascribe dipole moments to particles where we cannot use a physical current loop to adequately describe the generation of this moment. That is to say, we talk about the electron having a spin and this spin generates a magnetic dipole moment. However, we cannot satisfactorily say that this is due to a physical spin of a shperical shell of charge as the resulting properties of this spin fail to satisfy relativity. But again, spin and its moment belong to quantum, not classical, physics.

    If we take the dipole moment to be a loop current, or just an orbiting charge, then we enter those complicated situations that I mentioned above. A moving charge will see a different set of fields in its own frame. For example, a moving frame can transform a magnetic field into a superposition of magnetic and electric fields. These electric fields can do work on the particle in its frame.
     
  6. Aug 13, 2010 #5
    Thank you guys, you helped me a lot!
     
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