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How do magnetic fields do no work?

  1. Mar 14, 2010 #1
    I remember learning in Electrodynamics last year that magnetic fields don't actually do any work. I don't seem to remember any of the explanation behind that. Can anyone explain why magnetic fields do no work (ie. How they differ from electric and gravitational fields). Thanks!
  2. jcsd
  3. Mar 14, 2010 #2


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    Via the Lorentz force, the force created by a magnetic field acting on a moving charge is always normal to the charge's velocity. Hence, when you take the integral of the applied force over the path of the charge, it will be zero because the magnetic force is normal to the charge's path. The magnetic field instead can only directly change the momentum of the charge.

    Often times though, when you transform to the frame of reference as seen by the charge's perspective, the magnetic field may transform into an electric field. This electric field can do work on the charge. This can be seen when analyzing the case of two current carrying wires. So while the magnetic field does no work, that doesn't mean that work can't be done in these situations. It just requires a different perspective that allows the electric field to mediate energy from the magnetic field to the charge.
  4. Mar 14, 2010 #3
    Thank you, Born. It makes sense now!
  5. Mar 14, 2010 #4
    Hence it's based on our definition of work....analogous to you pushing as hard as you can against and wall and getting tired....yet no work is done....
  6. Mar 14, 2010 #5
    Indeed. I can see that since
    [tex]\vec{F}=q\vec{v} \times \vec{B} \text{ and } W = \int \vec{F} \cdot d\vec{r}[/tex]
    [tex]W = q\int (\vec{v} \times \vec{B}) \cdot d\vec{r}[/tex]

    Since [tex]\vec{v} \times \vec{B}[/tex] is perpendicular to [tex]d\vec{r}[/tex], the dot product, and therefore the work are zero.

    From the particle's perspective though, it is being influenced by an electric field instead of a magnetic one, so that
    [tex]\vec{F} = q\vec{E} \text{ and } W = q\int \vec{E} \cdot d\vec{r}[/tex]
    Since the electron will follow the electric field lines, [tex]\vec{E}[/tex] and [tex]d\vec{r}[/tex] are parallel, making
    [tex]W=qE \int dr[/tex]

    Thanks for the help guys! This has helped a lot.
  7. Mar 14, 2010 #6


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    Magnetic fields can do work on magnetic materials, such as other magnets or other objects made of magnetic material. Putting steel shot (a ferromagnetic sphere) in a magnetic field will induce in the sphere a magnetic dipole moment. If the field is non-uniform, the shot will experience a force given by


    and this force can do work.
  8. Mar 14, 2010 #7
    Yes I think you should be careful of the context of this statement.

    It may be preparing you for the the difference between real power and reactive power in ac circuits, where inductors may do no work.

    As has been pointed out magnetic fields can do work on magnetic materials, including other magnets, but can't do any work on charged objects, just because they are charged
  9. Mar 14, 2010 #8
    Interesting point. Glad you brought that up! Thanks.
  10. Mar 14, 2010 #9
    Magnetic fields could do work on magnetic charges (monopoles), but no one has ever found any. But magnetic fields can do work (torque) on magnetic dipoles, like permanent magnets. Watch a compass needle oscillate about the direction of magnetic North.

    Also, if the total energy in a magnetic field is

    W = (1/2μ0)∫B2 dVvolume,

    then there is a force in the x direction given by

    Fx = ∂W/∂x.

    and this force can do work.

    Bob S
  11. Mar 14, 2010 #10


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    I'm not sure about the last, Bob S. You identify the field energy as work instead and state that the field has an intrinsic force contained within it. There is no force without an object to work against, however, as in my example.
  12. Mar 14, 2010 #11
    If you have two large neodymium magnets 1 cm apart, with a large attractive force between them, you can harvest this magnetic energy by letting the gap decrease to, say, 5 mm. So the stored magnetic field energy can do work.

    Bob S
  13. Mar 14, 2010 #12


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    Technically though, with the absence of magnetic monopoles, the magnetic dipoles are microscopic loop currents. Saying that magnetic fields do work on magnetized objects is just another shortcut that is taken as the magnetization can be modeled as macroscopic bound currents.
  14. Mar 14, 2010 #13


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    In your example, the field from one magnet is doing work on the other, and vice versa. That is how a field can do work. The field from a single magnet in the absence of the second generates no force, and I think your earlier wording was prone to this interpretation.

    True, but such problems are usually solved by referring to the magnetization M or dipole moment m. Calculating the force between a field and the equivalent magnetic surface currents on the solid, or actual microscopic currents within, is an academic exercise only.
  15. Mar 15, 2010 #14
    Yes, that's where the mistake comes from and why there is a false assumption that magnetic force can do no work. But, if there was no separate displacement due to interaction of magnetic fields we would have never known about this force.

    The most important thing here is that fields alone do not make any work in any case, for that you need TWO, you need at least two magnetic fields or there will be no magnetic FORCE, and without the force of course there will be no work done, which is very different from the actual ability to perform the work.

    Momentum, you mean direction or velocity? Is that not what is force supposed to do and how you measure the work done? Fields on their own can do nothing, you need the force, which means at least TWO of these fields. What is "only" about actually causing a charge to displace?

    So, electrons move in one direction in a wire and the displacement is caused by the electric force because there are electric fields interacting along the wire. But you can not expect the wire to be moving sideways because you only have magnetic field around the wire and to have a force, displacement and work done, we therefore need at least one more wire or permanent magnet, something to interact with this magnetic field of a moving charges in our wire.

    Sure enough, two parallel current carrying wires attract and if the current goes in opposite directions they will repel due to Lorentz force (Ampere's force), we can measure this displacement and hence the work done by the magnetic force.
  16. Mar 15, 2010 #15
    That is not true, it is yet another popular misconception.

    In complex situations you have to decompose displacements, ALL OF THEM, in vectors and find out forces and split them in vector components too, then see what displacements match with what force to find out work done on particular object by a particular force.

    When pushing against a wall work is done but the displacement is not obvious as it is in a form of muscle tension, bone friction, blood displacement, hart-rate and all the rest of biological and chemical reactions that go along. Some things are not obvious, but that that does not mean they do not exist.
  17. Mar 15, 2010 #16


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    The magnetic field isn't actually directly doing any work here. According to The Lorentz Force Law (from which the equation you've posted is derived), magnetic fields never (barring the existence of magnetic monopoles) do any work. The work, in this case , and all cases where it appears that a magnetic field is doing work, is caused by an electric field which is induced by a changing magnetic field. The electric field is what actually directly does the work.
  18. Mar 15, 2010 #17


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    Ok, I'll post this once more, from Griffith's textbook:


    There is no stipulation that you need two magnetic fields to create a force. A single magnetic field will result in a force when any charged particle passes through the field.

    Force describes the change in momentum. Work is the change in momentum along the object's trajectory. The magnetic field can only change the charge's direction, but not its speed. And again, you do not need two fields, only one field.

    In the case of the two current carrying wires, as I explained earlier, the work is done by the electric fields, not the magnetic fields. This is because in the frame of the charges, the magnetic fields transform to electric fields which provide a force that allows for work to be done. This is an example shown in many textbooks such as Griffiths or Grant & Phillips.
  19. Mar 15, 2010 #18
    I think we can safely say that the "magnetic fields do no work" is an oversimplification of classical electrodynamics despite born2bwire's text. What's on the following page born2bwire?

    Rather, instead, the zeroth derivative of the magnetic field does no work on electric charge according to the Lorentz force equation.

    More, the Lorentz force is defined without regard to any specific inertial or non-inertial frame of the observer, where E and B are local fields to the charges in question.

    Yet, one simple thought experiment seems to say the Lorentz force is wrong or incomplete (it isn't):

    A charge neutral, ring of current is subject to a force in a time invariant magnetic field. To obtain the magnetic field we can use another axially aligned ring of current.

    Who can explain the work done by one fixed ring of current, upon on the other, using the Lorentz force equation?
    Last edited: Mar 15, 2010
  20. Mar 15, 2010 #19
    The operative word here is model.
  21. Mar 15, 2010 #20


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    Well we could use the original microscopic atomic loop currents as well, it's no different. In fact I recollect that Griffiths does a treatment of this in his text. We just take the shortcut of superposition to bring it up to macroscopic bound currents. This is the relationship between magnetization and the current densities in Maxwell's equations.
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