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Wait, I thought magnetism was nonconservative

  1. Sep 4, 2014 #1
    i ran a simulation that calculates the trajectory of a particle in the presence of a magnetic field from maxwells equations.
    it also calculates the work (line integral of force) and the kinetic energy gained(kinetic energy at the end of simulation minus kinetic energy at the beginning)
    i was pretty surprised to see that the values are the same. i know its not a programming error.
    here the magnetic field is constant(maybe thats why?)
    B=3,5,10
    and there is an electrostatic force.
    E=-100[x,y,z]/R^3
    (i know these differ by permissivity or permittivity of free space or whatever but it shouldnt matter, besides i get the same result with just a straight magnetic force)
    initial conditions x,y,z=[3,1,2] and dx/dt,dy/dt,dz/dt=[1,4,3]
     

    Attached Files:

  2. jcsd
  3. Sep 4, 2014 #2

    Drakkith

    User Avatar

    Staff: Mentor

    To my understanding, a static, constant magnetic field does no work on a charged particle, so there should be no gain of kinetic energy.
     
  4. Sep 4, 2014 #3

    vanhees71

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    Science Advisor
    2016 Award

    Now this discussion starts again, and I try again: A magnetic field does no work on a particle.

    A particle moving in an electromagnetic field obeys the equation of motion (for simplicity I use the non-relativistic expression)
    [tex]m \ddot{x}=\vec{F}(t,\vec{x},\vec{v})=q \left (\vec{E}+\frac{\vec{v}}{c} \times \vec{B} \right ).[/tex]
    The work done is
    [tex]\frac{m}{2} [\vec{v}^2(t_2)-\vec{v}^2(t_1)] = \int_{t_1}^{t_2} \mathrm{d} t \; \vec{v}(t) \cdot \vec{F}[t,\vec{x}(t),\vec{v}(t)]=\int_{t_1}^{t_2} \mathrm{d} t \; q \vec{v}(t) \cdot \vec{E}[t,\vec{x}(t)] .[/tex]
    Here, [itex]\vec{x}(t)[/itex] is the trajectory of the particle, i.e., a solution to the equation of motion. Then it doesn't matter, whether the force is conservative or not. It's only conservative, if the electric field is static, and the magnetic field never contributes to the work done (no matter whether the magnetic field is stationary or not or whether it's homogeneous or not).
     
  5. Sep 4, 2014 #4
    ahh ok that makes sense the force is perpendicular to velocity so all the work is done by the electrostatic force.
    What about other conserved quantities? ie angular momentum
     
  6. Sep 4, 2014 #5

    Dale

    Staff: Mentor

    These quantities are guaranteed to be the same, even for a nonconservative force like friction. The only requirement is that the object must be rigid.

    The proof is called the work energy theorem.
     
  7. Sep 4, 2014 #6
    i did some research and realized the integral of torque (position cross force) equals the change in angular momentum. its really kinda interesting when you think about it.
    b
    ∫Force[itex]\bullet[/itex]Velocity *dt=kinetic energy gained from a to b.
    a

    and
    b
    ∫Position[itex]\times[/itex]Force *dt=angular momentum gained from a to b
    a

    It makes me wonder what if you try crossing force and velocity?
    or doing the dot product of position and force?
    Do these have physical meanings also?

    (sorry if im being annoying with all the weird questions)
     
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