1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

How does a permanent magnet lift iron filings

  1. Jun 16, 2004 #1
    Evryone knows that the work done by a magnetic field is zero. Then how does a permanent magnet lift iron filings.
  2. jcsd
  3. Jun 16, 2004 #2


    User Avatar
    Science Advisor

    We do not all know that: many of know that any force field does no work on an object moving perpendicular to its lines of force. Magnetic fields certainly can do work.

    (And please don't double post.)
  4. Jul 4, 2004 #3
    I'm afraid I'll have to disagree

    Magnetic forces do not work since

    F= q (v x B) is a vector perpendicular to the infinitesimal displacement dr, so W= F dr=0

    Back to the original question, magnetic forces change the direction of another force. When lifting something, the one doing the work is most likely the generator or motor

  5. Jul 5, 2004 #4


    User Avatar
    Science Advisor

    No. That's the force of a magnetic field on a charged particle. There are other forces as well. Hold two magnets apart; now they have potential energy. Let them go; now the field does work on the magnets and the result is that they accelerate towards one another.
  6. Jul 5, 2004 #5
    To bring the magnets together requires changing the magnetic field, and a changing B induces an E. Against this E is the work done. At the beginning and at the end there is no E, but in between there is definitely one.
  7. Jul 5, 2004 #6


    User Avatar
    Science Advisor

    You need not change any magnetic field. If big magnets are confusing you, think of 2 elementary particles with no charge, but with magnetic moment. There are magnetic forces between them. These forces are not perpendicular to velocities and so they do work.

    I will repeat. A magnetic field does no work on an electric charge. But a magnetic field does work on a magnetic charge. It is true that there are no free magnetic charges, but that is a detail.

    Read this thread: https://www.physicsforums.com/showthread.php?t=32547
    You did not contribute to that thread, so maybe you missed it. I will not repeat again.
  8. Jul 5, 2004 #7
    So you're talking about fiticious magnetic monopoles?

    Because a real magnet is made up of tiny magnetic dipoles which are nothing but tiny currents, which are composed by charges with a certain velocity. If we're talking about forces on magnetic dipoles then said force is F=grad(m B) but this formula comes from the Lorentz force law.

    As far as I know, magnetic field exert forces only on moving charges.
  9. Jul 5, 2004 #8


    User Avatar
    Science Advisor

    didn't read the other thread, did you?

    Not so. So you have an isolated neutron. It has a dipole moment. So what are the circulating charges that result in the tiny current loop? Other elementary particles?
  10. Jul 6, 2004 #9
    The neutron isn't an elementary particle; it's made up of quarks. Wouldn't they be the source of the magnetic dipole moment?
  11. Jul 6, 2004 #10
    A magnetic field cannot do work on a charged particle. This was stated quite precisley in Griffith even stated this explicitly in his text "Introduction to
    Electrodynamics - 1st Ed.," on page 179
    The magnetic field cannot do work on a current element either, and for the same reason. But for a finite current element such as a current loop I believe the answer to that is yes but I'm not 100% sure.

    It only seems like the magnetic field does work. As a worked example of how it appears to do work and the correct explaination please see the AJP article I posted at


    re - "When lifting something, the one doing the work is most likely the generator or motor"

    To be precise its the conductors doing work on the charges. A magnet can be modeled as a collection of dioples which in turn can be modeled as a collection of conducting loops of wire. The magnetic field facilitates the work. I.e. Work done by B-field = 0, Work done by wire = Function of B.

    re - "fiticious magnetic monopoles" - Undiscovered is a more accurate term. Its not quite correct to call them fictitious if you don't know if they exist or not. That'd be like Newton saying that particles which make up atoms are fictitious because he's never observed them.

    Last edited: Jul 6, 2004
  12. Jul 6, 2004 #11


    User Avatar
    Science Advisor

    A very good question! Before the quark model, it was known that a free neutron decayed to an electron and a proton. It was tried to model neutrons as rotating combinations of these particles, with rotation rate set at whatever gives the correct magnetic dipole moment. But these models went nowhere, and in particular did not help in deriving the currently-accepted quark model. In fact, even earlier with the advent of QM, it was learned that the only bound electron-proton systems are: hydrogen atoms.

    Another example is the electron by itself. It has never exhibited any behaviour suggesting it is a composite particle. We can investigate a model where its magnetic moment arises from an extended distribution of charge, spinning at the appropriate rate. But we find that we end up with no coherent picture.

    Further questions relating to elementary particles should be posted on the Nuclei and Particles forum.
  13. Jul 6, 2004 #12
    Except that "undiscovered" means you expect to find them, rather than having them be something excluded by Maxwell's equations.
  14. Jul 6, 2004 #13
    But do the motions of the quarks account for the neutron magnetic dipole moment, and if so, isn't that an answer to you question about the presence of circulating charges?
  15. Jul 7, 2004 #14
    Maxwell's equations does not exclude monopoles. People just tend to set the monopole density to zero since applications to date don't use them. If they are discovered then the monopole density will be non-zero. But keep that in mind. That they are zero is an assumption and to effect that assumption in Maxwell's equations, the divergence of the magnetic field is set to zero. But it need not. Assuming monopoles exist then you set that divergence to the monopole density in the region that you're interested in.
  16. Jul 7, 2004 #15
    Im not an expert on this subject, but a magnetic field can't do work on a charged particle, it can only changed the particle's direction, which is how a particles in particle accelerators are made to go in a loop. But the thing is that like it was said in the biginning, a permanent magnet can pick up iron fillings or another magnet Picking it up takes force, and the force is applied through a distance....soooo, it seems like work is done.
  17. Jul 7, 2004 #16


    User Avatar
    Science Advisor

    Magnetic Moments; References

    The iron filings have a magnetic dipole moment -- among other things because the motions of the charges are highly constrained -- primarily due to the magnetic moments of the atoms. The energy term that results is

    -mB where m is the magnetic moment, B is the magnetic field.Magnetic fields also generate torques, which is a key idea behind the galvanometer.

    Resnick and Halliday give a good basic presentation of forces, and energetics in a magnetic field. More advanced texts, Panofsky and Phillips, Symthe, Landau and Lihschitz (Electrodynmics of Continuous Media) will tell you more than you ever cared to know about the subject.

    Regards, Reilly Atkinson
  18. Jul 14, 2004 #17
    I don't want to get into baryons and the internal structure of neutrons because that's off topic.

    All I'm saying is this: the force on a magnetc dipole is derived from the Loretz force law, which means that the magnetic force does not work.

    The point I'm trying to make is that Lorentz force law implies no work done by the magnetic force. Reading your posts I get the impression that you're saying that it's only valid for electric charges (ie. Lorentz force law), so I would like you to show me a formula for a force between an electric field and something other than an electric current. I'm honestly asking you, I'm not being a smart-ass.

    Best regards, Javier
  19. Jul 14, 2004 #18


    User Avatar
    Science Advisor

    All I'm saying is that's just wrong. If you have say a uniform field [itex]\vec{B}[/itex] and a magnet with dipole moment [itex]\vec{m}[/itex], you have to do work to pull it out of alignment with B. The torque to do this is


    There is a potential energy U:


    Rotating the dipole out of alignment increases the potential energy. Let it go and it will rotate on its own as the potential energy decreases and the rotational kinetic energy increases. This is known as doing work. The magnetic field is doing work on the dipole, just as gravity does work on a mass that is falling. These formulas are directly from Jackson. Lorentz force is not relevant here. You are mis-applying something else, namely that a magnetic field does no work on a charged particle.

    Did you mean magnetic field?

    Anyway, the force law between two magnetic poles p_1 and p_2 in free space is this

    [tex]F={p_1p_2\over r^2}[/tex]

    in the direction of the vector r from p_1 to p_2. This can be described as p_2 in the magnetic field


    of p_1. Equally well, describe it by a potential


    So you complain that monopoles don't exist. Fine. They only come in pairs. The potential from an isolated dipole is found be adding the two together:


    where [itex]\delta[/itex] is the separation of the two opposite poles [itex]\vec{\delta}=\vec{r_2}-\vec{r_1}[/itex], and [itex]\theta[/itex] is the angle it makes with the observation point. We call [itex]p_1\delta[/itex] the dipole moment m. So


    This is the same as the formula given by Jackson.

    If you have a cylindrical permanent magnet, with magnetization [itex]\vec{M}[/itex], then it acts to a very good approximation as if it has magnetic monopoles on its North surface with a charge density [itex]\sigma_M=|\vec{M}|[/itex], with a surface charge density of exactly opposite sign at the South pole (this is directly from Jackson...). Get two such magnets together and they attract or repel eachother with forces given by applying the formula for F that I gave above to these surfaces and integrating. These forces have potentials and everything, just as in gravity and electrostatics.

    Here's a quote from my first post "Hold two magnets apart; now they have potential energy. Let them go; now the field does work on the magnets and the result is that they accelerate towards one another." Sheesh. I'm repeating myself...
    Last edited: Jul 14, 2004
  20. Jul 14, 2004 #19
    Thank you for your input krab, I'm still not convinced, but I'll go through your argument with more detail later nonetheless.
  21. Jul 17, 2004 #20
    to anybody
    So when a magnet picks up iron filings off a table -- we know work is being done -- but by what?
    Note that the end field is different from the start field as more filings will not be picked up. Any explanations Ray.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook