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Magneitic fields can do no work

  1. Sep 7, 2009 #1
    Hello,

    I know this topic has been posted and discussed much, but I am still not satisfied and would like some thoughts.

    The common phrase is 'magnetic fields can do no work' and the source of this phrase is attributed to [tex] \mathbf{F} = \mathbf{v} \times \mathbf{B} [/tex]. I think it is clear that in the case of a current the magnetic field induces an electric field which then does the work. I always assumed this was the case for permanent magnets as well, but now have some doubts. (An often touted answer is the person moving the magnets is doing the work, but that is obviously unsatisfactory since that is the case in any conservative field.)

    Now Griffith's says: "Magnetic forces do no work" and he goes on to say that "...it can be a very subtle matter to figure out what agency does [do the work]"

    Physics Forums member Gokul43201 often links this page in such discussions: https://www.physicsforums.com/showpost.php?p=997210&postcount=14 and he says that the magnetic field will do work on a magnetic moment and it is this force that attracts magnets to each other, or iron filings to a magnet.

    So now I see contradiction between Griffiths and
    [tex]\mathbf{F} = \mu \left( \frac{\partial \mathbf{B}}{\partial z} \right) [/tex].
    Is it not correct to say that this inhomogeneous magnetic field creates an electric field which does the work? And if so, then Griffith's is plainly wrong, no?

    Thx for your insight!
     
  2. jcsd
  3. Sep 7, 2009 #2
    Hi Academic-
    Griffiths is correct in stating that a magnetic field (with or without a gradient) can do no work on a moving charged particle, because the force is always perpendicular to the direction of motion. But a magnetic field with a gradient can attract a magnetic dipole, just like an electric field with a gradient can attract an electric dipole.
    As for magnetic fields doing work on charged particles, a magnetic field changing in time (dB/dt) can accelerate charged particles.
    Bob S
     
  4. Sep 7, 2009 #3
    Consider a permanent magnet (fixed in the lab frame) and a charged particle moving in its vicinity. The force on the particle is [itex]\vec F = v \times \vec B + \vec E q[/itex]

    The magnet does no work on the particle [itex](\vec F \cdot \vec v = 0)[/itex], the particle is deflected and its kinetic energy remains constant.

    But suppose we move the magnet. Then we create an electric field (in the lab frame) at the particle location and this does work on the particle.

    I should also consider how things appear in the rest frame of the particle when the magnet is stationary in the lab frame.
     
  5. Sep 7, 2009 #4
    Ahhh, but he doesn't stipulate on a moving charged particle! He says never, as in never ever. At least that is how I read it. But now I am not not believing that, I think the magnetic field can do work - it can do work on a magnetic dipole.

    He talks about the difficulty in figuring out what agency does do the work, which to me suggests that the magnetic field is not the agency doing the work on a magnetic moment.

    Is he being sloppy with his language? Am I misinterpreting what he says? What is happening when a magnetic field does work on a dipole moment?
     
  6. Sep 7, 2009 #5
    Hi Academic-
    This is a very good question. It is obvious that energy is required to accelerate a magnetic dipole, and this energy has to somehow come from the field.
    I will explain this using the electric field analogy, because it is easier to visualize.

    Consider a large vacuum capacitor (no dielectric) with a capacitance C, a 2-cm gap, and a battery with voltage V on it. The total charge on each plate is Q = CV. Suppose a fast electron with charge -e enters from the left side in the center of the gap. An image charge, +e/2, appears on the two plates. This is the same image charge that was following the electron, and will disappear when the electron exits the capacitor on the right side. No net work is done on either the electron or the image charge, as long as the electron exits in the center of the gap. But the electron is attracted to, and moves toward the positively charged plate. If the elecron moves halfway to the positive plate, then the image charge on the positive plate is +3e/4, and +1e/4 remains on the negative plate. The +1e/4 image charge has to move through the external wires and the battery to get to the positive plate from the negative plate. This represents +eV/4 of work done by the battery on the image current. A similar thing would occur with an electric dipole in an electric field with a gradient.
    Energy (work) is also required to deflect a magnetic dipole in a magnetic field with a gradient.
    In the case of a magnetic dipole, windings on the electromagnet supply the energy for moving the dipole in the magnetic field.
    Bob S
     
  7. Sep 7, 2009 #6
    hmmm... In your example I would say the battery does work on the field and then the field does work on the charge. The battery cant do work directly on the charge, it must work on the field first as an intermediary. Is that not correct?

    So taking that analogy to the B field, the supply does work on the field, then the field does work on the moment. Hence the B field can do work.

    Begging to what supplied the energy initially is sidestepping the issue I think. Im gonna think about it some more though, thx for your input.
     
  8. Sep 7, 2009 #7
    That's right. Electric fields can do work: F dx = qE dx


    Right again.


    OK
    Bob S
     
    Last edited: Sep 7, 2009
  9. Sep 8, 2009 #8
    Last edited: Sep 8, 2009
  10. Sep 8, 2009 #9
    This is not correct. A changing magnetic field does not accelerate charged particles! The correct mechanism is that a changing current produces BOTH a changing B field and the E field that accelerates the particles. Hence while the B field is related to the causal source of the acceleration , it is not the CAUSE of it. The E field is producing the acceleration. Proof. Charges are accelerated even in regions where B = 0.
     
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