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Lorentz force? & Magnetic fields!

  1. Jul 12, 2012 #1
    Greetings everyone(New guy here!).

    I wanted to know something about "Lorentz force". A magnetic force causes a loop carrying current to rotate and torque is created due to those forces. Are magnetic fields doing work in this cause?

    Based upon this law:

    F = q(E + v + B)

    F = I x L x B (Current flowing thorugh a wire.)

    I believe its obvious that without the magnetic fields no work can be done. Thus the"motor effect" is caused due to those forces.

    Is the magnetic field perpendicular to the loop wire?

    + In general are magnetic fields conservative or nonconservative?

    Thanks all!

    You're newly fresh member Miyz!
    Last edited: Jul 12, 2012
  2. jcsd
  3. Jul 12, 2012 #2
    Not sure what exactly you're referring to with a torque on a wire loop.

    This equation is incorrect, you need a cross product between the velocity and the magnetic field:

    F = q(E + v x B)

    F,E,v and B are all vectors. Magnetic force is not conservative, this can be seen from the Biot-Savart law where a moving charge generates a circulation of the magnetic field around it, and if there is a circulation of the field, there is a curl, ie. curl(B) is nonzero. Now, for a field to be conservative, there has to be a scalar potential, say "theta", such that B = -grad(theta). But the curl of a gradient is always zero so there is no theta which can lead to a non-zero circulation.

    Put simply, an electron can go in a circle and continuously gain momentum from the magnetic field. It will reach its starting point with a load more energy than it began with, which is not possible for a conservative field, eg., gravity, where if you travel around then finish where you started, your gravitational potential energy is unchanged.
  4. Jul 12, 2012 #3
    Torque generated by the total magnetic force for a wired loop that experienced a force from the magnetic field"the motar effect?". A torque is generated because of the magnetic forces.

    Ops! Honest mistake there.

    How is that possibile?! Very interesting fact never knew that before! Due to the magnetic force the particle would constantly gain momentum!! How?!
  5. Jul 12, 2012 #4
    I'm not sure what MikeyW is talking about. A charged particle entering a uniform magnetic field processes in a circle. The direction of its momentum changes, but the magnitude (and hence, kinetic energy) does not.
  6. Jul 12, 2012 #5
    What law is this based on? I'd like to look this up.
  7. Jul 12, 2012 #6
    Maybe I've confused myself a bit. The net work about a closed loop is zero, but the circulation of the field itself is nonzero. Surely that makes it non-conservative?

    Probably best to ignore the "put simply" paragraph for now!
  8. Jul 12, 2012 #7
    I'm confused here...
  9. Jul 12, 2012 #8
    MikeyW, you're thinking of an induced *electric* field. In that case the fact the circulation of the field is nonzero implies a net amount of energy gained in a loop due to Stokes' theorem. This reasoning does not apply to the magnetic field: Stokes' theorem indeed gives that the loop integral of the magnetic field is non-zero, but this does not imply energy transfer since the particle moving in a circle does not feel this magnetic field (the particle only feels components of the magnetic field perpendicular to its velocity). So indeed, you confused yourself a bit there.
  10. Jul 12, 2012 #9
    Based on wikipedia: "some authors classify the magnetic force as conservative, while others do not. The magnetic force is an unusual case;"

    Unusual case huh...
  11. Jul 12, 2012 #10
    Miyz: that's because you can have two conflicting definitions of "conservative force"

    1) The net work done (on a test particle) over a closed loop is zero.

    2) The line integral for a closed loop is zero.

    The magnetic field satisfies the first, not the second (see my previous post on this issue).

    Why two definitions? Well most of the time, i.e. in the cases of force that can *do* work (unlike the magnetic force), they're equivalent, and these are the cases that "matter" (so according to that view (1) should be thé definition). On the other hand, sometimes we're not particularly interested in energy transfer but more in knowing whether or not the force can be derived from a potential. In that case (2) should be thé definition, since there is a theorem that states that a field has a potential if and only if the line integral across any closed loop is zero.
  12. Jul 12, 2012 #11
    The big thing is that most forces only depend on position, and this makes it easy to define conservative forces--if the force admits a scalar potential (curl is zero) and that scalar potential is single-valued (saying work done is path-independent), then the force is conservative.

    The magnetic field doesn't admit a scalar potential. It also doesn't do work on free particles, and its direction depends on the velocity of the particle. Honestly, it is so different I think saying it's conservative or non-conservative has no meaning because one associates non-conservative forces with forces that are still dependent only on position.
  13. Jul 12, 2012 #12
    What about the case of the "Motor effect"? Do things differ in that cause?

    Or its the same exact thing?

    Based on the magnetic force law on a current carrying wire: F = I x L x B
    (This law keep tell me that under that condition maybe things could differ since were generating a force on a wire carrying current)
  14. Jul 12, 2012 #13


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    The magnetic part of the Lorentz force, which completely reads
    [tex]\vec{F}=q \left (\vec{E}+\frac{\vec{v}}{c} \times \vec{B} \right )[/tex]
    (written in Heaviside-Lorentz or Gaussian units, for which the force law reads the same) is in no case a conservative force since it is not derivable from a potential, i.e., it is not given as the gradient of a scalar potential.

    However, the magnetic part never does work on the particle since the corresponding power
    [tex]P_m=\vec{v} \cdot \vec{F}_m=\vec{v} \cdot \left (\frac{\vec{v}}{c} \times \vec B \right )=0.[/tex]
    The electric part is a conservative force for the case of static fields since then
    [tex]\vec{\nabla} \times \vec{E}=0,[/tex]
    which means that there is a scalar potential of the electric field, such that (at least locally)
    [tex]\vec{E}=-\vec{\nabla} \Phi.[/tex]
  15. Jul 13, 2012 #14
    F = I x L x B Or F = IL x B

    Is the MAIN force applied on a current carrying wire... In this cause can and will do work! Example: Motors/Generators... If B were to be 0 the WHOLE Magnetic force on the wire is not be present to create that "twist" for torque to be present... We can't create motors/generatros without magnetic fields can we? No, without the magnetic field no work can be transfered... Thats my main point!

    Again: F = IL x B

    If B = 0 , F = 0 , Work = 0.

    I think magnetic field & force under certain circumstance can preform work, there has to be a source of force/energy present for work to be done. Plain old simple magnetic field may apply a force on a particle or a charged particle but no work will be ever done.
  16. Jul 13, 2012 #15
    Work done is a dot product of force and displacement of point under consideration.
    In a motor, the magnetic force does work, because it causes the motion of the loop ( coil sides) in its own direction.
    The action of magnetic field on a charge is different ( where the charge moves in circular motion) . Here the magnetic force is perpendicular to direction of motion, so no work in done.
    In a practical motor, design is always aimed at maximizing the torque. Hence the magnetic field is made perpendicular to loop wire...in cases it exactly isn't, there's always a component of magnetic field perpendicular to it.
  17. Jul 13, 2012 #16
    In one way, a magnetic field can not do work, but conceptually it's not always that easy, because a magnetic field can actually cause a particle to speed up.

    Take for example a magnet fixed somewhere in outer space, not moving, and an electron is moving toward it. The magnetic field, as explained above, does no work. However, when the electron bumps into the magnet (and bounces away), the magnet moves, so we get a time-dependent magnetic field, which induces an electric field, and this electric field which does work on the particle.

    So sure, it's the electric field that does the work, but where does the energy come from that speeds up the particle? Well, from the magnetic field of the magnet (the magnet weakens a bit in this interaction, i.e. it loses energy). So in a physical sense, the magnetic field has done work on the particle.

    So to be clear, fundamentally it's never the magnetic field which directly delivers the work, but in a more physical sense one sometimes has to say that the magnetic field has delivered its energy to a particle.

    Just my two cents.

    Perhaps better said, the magnetic field is like a high lord with his peasants: although never doing any work, he does get it done.
    Last edited: Jul 13, 2012
  18. Jul 13, 2012 #17
    hahaha, well said!

    True, Magnetic fields are really interesting but they have a natural field that we can study more and more and I honestly feel that its much underrated and less attention and effort is given towards it.

    I really started to get interested in magnetic fields/force when I worked on motors. Electric currents are the main ingredient for this "chain reaction" to occur and the second one is magnetic fields! Without magnetic field no work would be done in a motor... Hence this law again: F = IL x B

    The torque generated in a motor is lead BY the reaction of both magnetic fields repelling/attracting each other that just fueled my with curiosities!
  19. Jul 14, 2012 #18
    What do you all think of that?
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