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Reaction force due to the curvature and gradient drift

  1. Nov 10, 2014 #1
    We know that a charged particle will have a drift velocity in both a curved magnetic field and when there is a transverse spatial gradient in the magnitude of the magnetic field.

    This drift velocity is added to the rotation velocity around the the field line.

    In both cases the force vector on the particle, $$\mathbf{F}$$ is perpendicular to the velocity vector.

    $$\mathbf{v_D}=\frac {c}{q} \frac{\mathbf{F} \times \mathbf{B}}{B^2}$$

    Now the question is what the "reaction force", $$-\mathbf{F}$$ is applied on, in particular in the case of gradient drift?

    Is it the "source" of the magnetic field?

    But what can we say about a radiation field that spans throughout spacetime, far from the source?

    I mean we know that there is nothing like a free electromagnetic field even though for practical purposes we treat radiation like a dynamical entity "unconnected to the source."

    Is the reaction force due to either curvature or gradient drift irrelevant here?

    Have we assumed a time-independent $$\mathbf{B}$$ for the derivation of $$\mathbf{v_D}$$?
  2. jcsd
  3. Nov 10, 2014 #2


    Staff: Mentor

    When dealing with fields Newton's third law as a law of forces becomes somewhat irrelevant. Generally we do not speak about the force on a field. However, Newton's third law is the law that encapsulates the conservation of momentum for mechanical systems, and the conservation of momentum does generalize to fields.

    So, when dealing with fields, the change in momentum of the particle is equal and opposite to the change in momentum of the fields. Generally the change in momentum of the particle is considered to be the net force on the particle, but you usually don't consider the change in momentum of the fields to be a force on the fields (although I suppose that you could).
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