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Can Magnetic Force Perform Work?

  1. Jun 2, 2010 #1
    Magnetic forces are no-work forces.But when a magnet suspended freely in the earth's magnetic field(from its center of gravity) rotates to settle itself in an approximately north south direction ,performing some work in the process. Who does this work?Obviously the magnetic forces which are again assumed to be no work forces!If we consider the initial and the final orientations of the magnet there is indeed a change in the electromagnetic energy density[We must consider both the fields:that due to the magnet and that due to the earth].
    Now let us consider a horizontal conductor[a thin rod for our example] moving in a vertical magnetic field,perpendicular to its own length.As the electrons move axially, work performed on unit charge is BLV[which is indeed the induced emf]
    Here,
    B: Magnetic field
    L: Length of the conductor
    V:Speed of the moving conductor

    Now the axial force on an electron is BeV. Again the axial motion invites a magnetic force perpendicular to the length of the conductor.This particular force gets continuously canceled by the surface forces. The total magnetic force is indeed a no-work force.But if a part of this "total magnetic force" gets canceled by some other force for example the surface forces the remaining part can indeed perform work!By this mechanism work done on unit charge is indeed BLV. So by this mechanism we can derive work from the magnetic forces.[You may consider my postings at sci.physics.foundations "Why don't we do that?" and "Power from Motional Emf----A simple Illustration"[Dated 1st March] ]
    Can a reasoning of this type explain the work done by the magnetic forces in the example of the rotating magnet?
    We can get work out of the magnetic forces . Would it be reasonable to modify Poynting's theorem in the view of this fact to make it more realistic? It is to be noted that the derivation of the Poynting theorem assumes that the magnetic forces are not capable of doing any work.
     
  2. jcsd
  3. Jun 2, 2010 #2
    Could you review this explanation?

    Why should the electrons move axially?
    Why should the conductor move?

    Please note we have just had a long discussion thread here about the difference between the simple engineers method, which allows magnetic fields to do work and the more sophisticated classical physics method which does not.

    https://www.physicsforums.com/showthread.php?t=229056&highlight=lorenz+force
     
  4. Jun 4, 2010 #3
    I have used this explanation in a similar discussion in General Physics. I am mentioning it here for its relevance to the general nature of the problem.

    Magnetic force[from the Lorentz force equation] is given by:

    F=q[V[tex]\times[/tex]B]

    Elementary work done :

    dW= q[V[tex]\times[/tex]B].dr

    =q[V[tex]\times[/tex]B].Vdt

    =0

    Let us break up the velocity into two parts

    V=V1 + V2

    We make sure that V1 , V2 and B do not lie in the same plane

    Work performed is given by:

    dW=q[[V1+V2][tex]\times[/tex]B].[V1+V2]dt

    =q[V1[tex]\times[/tex]B].V1dt + q[V1[tex]\times[/tex]B].V2dt + q[V2[tex]\times[/tex]B].V1dt + q[V2[tex]\times[/tex]B].V2dt
    =0

    q[V1[tex]\times[/tex]B].V1dt=0

    q[V2[tex]\times[/tex]B].V2dt=0

    q[V1[tex]\times[/tex]B].V2dt [tex]\neq[/tex]0

    q[V2[tex]\times[/tex]B].V1dt[tex]\neq[/tex]0

    dW= q[V1[tex]\times[/tex]B].V2dt + q[V2[tex]\times[/tex]B].V1dt
    =0
    dW=dW1+ dW2
    =0
    But
    dW1 and dW2 are in general individually not equal to zero.

    A technologist may try to get the benefits of the individual works though the total work performed is zero!

    If somehow we could produce a non-magnetic force=-q [V2[tex]\times[/tex]B]
    then

    Work performed by this force:

    = -q [V2[tex]\times[/tex]B][V1 + V2]dt
    = -q [V2[tex]\times[/tex]B].V1dt
    dW2 gets canceled by the work done by this newly introduced force and we are left with dW1 which is of non zero value!
     
    Last edited: Jun 4, 2010
  5. Jun 6, 2010 #4
    A magnetic force F can do work if the force is a result of a change in the stored magnetic energy W.

    W = ½∫B·H dVvolume

    Fx = +∂W/∂x


    Work =∫Fx dx

    Bob S
     
  6. Jun 6, 2010 #5
    Bob ,I believe, you have made a mistake in your interpretation of the work done.When a magnetic field is set up the field value rises from zero to some non-zero value. This creates an electric field which is capable of doing work.This matter has been explicitly discussed in "Introduction to Electrodynamics" by David J. Griffiths in Chapter 7[Electrodynamics]section 7.2.3 Energy in Magnetic Fields. The potential energy created is basically of electrical origin!
     
    Last edited: Jun 6, 2010
  7. Jun 7, 2010 #6
    The sign in my previous post

    Fx = +∂W/∂x where W = ½∫B·H dVvolume

    is correct. See Smythe Static and Dynamic Electricity third edition Sections 7.18 and 8.02, and the discussion in the previous thread

    https://www.physicsforums.com/showthread.php?t=406509&highlight=smythe

    It is true that when the source of a magnetic field is due to electrical currents, the external source of current must do work to maintain the magnetic fields. Smythe explicitly discusses the difference between the energy stored in electric and magnetic fields.

    Bob S
     
  8. Jun 7, 2010 #7
    Smythe - that's a deuced pricey book!
     
  9. Jun 7, 2010 #8

    gabbagabbahey

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    I disagree with this interpretation. All of classical electrodynamics can be derived from the Lorentz force Law and Maxwell's equations. The Lorentz force law says that magnetic fields never directly do any work, thus the statement "magnetic field do no work".

    There are however many situations where it appears that a magnetic force does work. For example, if we look at the work done when a magnetic dipole is placed in a non-uniform external magnetic field, [itex]W=\Delta(\textbf{m}\cdot\textbf{B})[/itex], it certainly looks like [itex]\textbf{B}[/itex] is directly responsible for this work. However, [itex]\textbf{F}=\mathbf{\nabla}(\textbf{m}\cdot\textbf{B})[/itex] is a composite force law, which can be derived directly from the Lorentz force law, which says magnetic forces do no work.
     
  10. Jun 7, 2010 #9
    When a magnetic dipole alligns itself with an external magnetic field, there is a change in magnetic potential energy. Are you saying there is no work done in this process? There must be something doing work on the dipole for its kinetic energy to increase.
     
  11. Jun 7, 2010 #10

    gabbagabbahey

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    Of course work is done in the process, but according to the fundamental force law (Lorentz force) it can't come directly from the magnetic field. It can be very tricky to find out exactly what agent is directly responsible for the work, but it is almost always an electric field , internal to the object that the work is being done on. For example, take a look at ex. 5.3 of Griffiths' Introduction to Electrodynamics (3rd edition). To see how that example relates to a dipole, consider that classically, a dipole is just a limiting case of a current loop.
     
  12. Jun 7, 2010 #11

    But all of classical electrodynamics cannot explain every physically observable elctrodynamic effect, as everybody here has pretended I haven't already pointed out.

    In fact I have made several pertainent observations that are being politely or not so politely ignored all because they conflict with the catechism.


    I particularly like the statements that flatly contradict conservation of energy.
     
  13. Jun 7, 2010 #12

    gabbagabbahey

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    What's your point? We don't currently have any theory that can explain every physically observable effect. How is that relevant to a discussion about whether or not, in classical physics---which has a wide range of applicability--- magnetic fields can do work?

    I'm not even sure where you've pointed to any electrodynamic observation that can't be explained classically.

    Again, I'm not sure which observations you are referring to.


    And which statements do you feel contradict conservation of energy, and why?
     
  14. Jun 7, 2010 #13
    Tripped myself up here and posted in the wrong thread. Sorry.
     
  15. Jun 7, 2010 #14
    The person who hung the magnet.
     
  16. Jun 8, 2010 #15
    I dont think that is a sufficient answer, because that would apply to all force fields - not just magnetic.
     
  17. Jun 9, 2010 #16
    Hi.
    Magnetic force applied to magnetic moment m is grad(mB). In case m is coming from classical motion of charged particles, e.g. currents of electrons, grad(mB) is interpreted as Lorentz force thus does no work. In case m is originated from quantum features, e.g. electron spin and orbit angular momentum, we cannot interpret grad(mB) as Lorentz force thus it does work.
    Now I am reaching this idea.
    Regards.
     
  18. Jun 9, 2010 #17

    gabbagabbahey

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    Classically, we just treat intrinsic dipole moments (due to spin) on the same footing as ideal orbital dipole moments (coming from current loops) since both experience the same net force and torque in an external field.
     
  19. Jun 9, 2010 #18
    Hi,
    Classical current loops participate in energy deal so accelerate/dissipate. Spin or electron orbit cannot. It makes difference. Am I right?
    Regards.
     
  20. Jun 9, 2010 #19

    gabbagabbahey

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    I'm not sure what you mean by this.
     
  21. Jun 9, 2010 #20
    Hi, gabbagabbahey

    Let two similar loop currents facing each other. They will attract. In approaching, currents are weakened due to induced electric field. We have to feed energy to loops if we want to keep current constant.

    Let two similar electron spins facing each other. They will attract. In approaching, spin magnetic moments are not weakened by quantum feature. We do not have to feed energy to keep magnetic moment constant.

    Regards.
     
  22. Jun 9, 2010 #21

    Born2bwire

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    But spin is not classical electromagnetics.
     
  23. Jun 9, 2010 #22
    Hi, Born2bwire.
    No, it isn't. Permanent magnets, whose power comes from electron spin or orbit angular momentum, is not classical electromagnetics either.
    Please forget about Coulomb repulsion force or say electric charge is neutralized by forming atoms with plus charged nucleus as it is in metal loops or in permanent magnets.
    Regards.
     
    Last edited: Jun 9, 2010
  24. Jun 9, 2010 #23

    Born2bwire

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    Yes, and as gabbagabbahey stated above,

     
  25. Jun 9, 2010 #24

    gabbagabbahey

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    This is why I was careful to put the word "ideal" in my earlier statement. An ideal magnetic dipole is the limit of a current loop, in which the current loop's size goes to zero, but its magnetic moment [itex]\textbf{m}=I\textbf{a}[/itex] remains unchanged.

    Classically, we treat both the magnetic moment due to spin (intrinsic angular momentum) and the magnetic moment due to orbital motion of the electron's in an atom (orbital angular momentum) as ideal magnetic dipoles.
     
  26. Jun 9, 2010 #25
    Consider the following scenario:

    A cathode ray beam is being accelerated from a cathode towards a distant anode.
    Clearly an electric field exists between the cathode and anode, increasing in potential as the anode is approached.
    So the work done on an individual electron increases as it moves from cathode towards the anode.

    Now let there be a magnetic field established, perpendicular to the beam's motion.
    The Lorenz force will deflect the beam perpendicular to the magnetic field, but still perpendicular to the beam's motion.
    No work is done as the sideways displaced electron still encounters the same electric potential.

    Now let a second pair of electrodes be established such that their field will partly oppose the deflection by the magnetic field, ie be at right angles to the original field.

    Work must now be done to move the electron sideways, as well as in the direction of the anode.

    What does this extra work?
     
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