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Do magnetic fields do work?

  1. Aug 22, 2007 #1
    I started asking this question after one of my students had trouble with a problem asking for the net force on a current loop in a homogeneous magnetic field (on my first day of TAing, no less!). A current loop in a magnetic field will experience a torque, but not a net force. I always assumed that this was because magnetic fields do no work, until I noticed that a magnetic dipole in a non-homogeneous magnetic field will accelerate. Bar magnets are ab example of this.

    So here's my question. David Griffiths says, in his textbook, that magnetic fields never do any work. My E&M professor, however, told me today that magnetic fields are able to do work. Which is it?
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  3. Aug 22, 2007 #2
    well, magnetic fields induce CENTRIPETAL forces on charges. The force due to the magnetic field causes the charge to take on a circular path, but it does NOTHING to increase the speed of the charge. If the speed is not changing then the kinetic energy is not changing, and therefore no work has been done on the charge. Remember, if net work is done on a particle then the particle's KE must be changing. No change in speed means no change in KE, which in turn means no work is being done. Just because you have net forces doesn't mean you are doing work.

    This has always been my understanding. I'm not sure if this is correct.

    However, as you say, if a magnetic field causes a force couple in a current loop resulting in a torque there is a net force on the loop because the forces act on different points of the loop causing an acceleration (angular acceleration to be exact) and therefore work is done on the loop. Similarly, two bar magnets attract or repel because the B-field from one bar magnet is not PERPENDICULAR to the current 'loop' formed by the eddy currents in the second bar magnet, but forms some angle resulting in an attractive force or a repulsive force.

    So, can magnetic fields do work? I guess sometimes they can. IT seems that magnetic fields can do work on moving charges that are 'stuck' in a conductor and are forced to move in a certain direction, but magnetic fields can't do work on free charges that are free to move in any direction. Perhaps Griffiths was just referring to magnetic fields acting on FREE charges.
    Last edited: Aug 22, 2007
  4. Aug 22, 2007 #3
    HOWEVER, as the loop rotates due to the magnetic field there is a changing magnetic flux through the loop resulting in an induced EMF. This EMF reduces the current through the loop. So, really, what is causing the angular acceleration of the loop is the voltage source causing the current to flow. This voltage source overcomes the induced, or back, EMF in the loop. This is a complicated issue for me also.
    Last edited: Aug 23, 2007
  5. Aug 22, 2007 #4
    This issue has troubled me greatly. I've reached the conclusion that only static magnetic fields do now work, but changing magnetic fields do work. I know that's contrary to what it says in most textbooks, but I see no other way to explain the following situation:
    Consider two infinitely long straight line parallel currents going from left to right. The bottom current will experience an upward force due to the magnetic field of the top one. Similarly, the top current will experience a downward force due to the magnetic field of the bottom one. Thus, they will attract one another. Since in each case, the force to the magnetic field is in the same direction as the motion of the wire, the magnetic field DOES do work in this case. My only explanation is that changing magnetic fields CAN do work, because the magnetic fields change as the wires move closer together.
  6. Aug 22, 2007 #5
    What are the exact words that Griffith says?
  7. Aug 22, 2007 #6


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    Last edited: Aug 22, 2007
  8. Aug 22, 2007 #7

    He says "Magnetic forces do no work." He says it in bold font, and he draws a box around it for emphasis. :smile:

    In the following paragraphs he admits this fact frequently appears false at first glance, noting that "it can be a very subtle matter to figure out what agency does [perform work]", he then goes through an explicit example, and gives a simpler analogy (a mechanics problem of raising a cart, where the normal force from an inert inclined plane might at first appear to do work).

    Although (from previous threads) the rule might not apply at the quantum scale (e.g. intrinsic spin).
    Last edited: Aug 22, 2007
  9. Aug 23, 2007 #8


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    Ok, explain how to model this scenario:

    Take an unmagetized iron bar, on a pivot, pointing east-west. Hold it fixed in the E-W direction and magnetize it (e.g. with another magnet). Remove what ever device you magetized it with.

    Now let the bar go so it starts to turn to point N-S like a compass needle. What is doing the work that creates the kinetic energy in the moving bar, if it isn't "the magnetic forces"?

    Sure the energy originally came from some place else (when the bar was magnetized) but that's not the point. When the bar turns from E-W to N-S, something does mechanical work on the bar. If it isn't the magnetic forces, what is it?
  10. Aug 23, 2007 #9
    The possible caveat I mentioned was the quantum scale: I suspect ferrous materials might derive their properties from the alignment of atomic quantum spin axes, so I can't break that explanation down further myself. Doesn't mean Griffiths, author of the best undergraduate texts to read (on QM, electrodynamics, or particle physics), has neglected scenarios of that general nature:

    ".. The fact that magnetic forces do no work is an elementary and direct consequence of the Lorentz force law, but there are many situations in which it appears so manifestly false that one's confidence is bound to waver. When a magnetic crane lifts the carcass of a junked car [emphasis added], for instance, something is obviously doing work, and it seems perverse to deny that the magnetic force is responsible. Well, perverse or not, deny it we must, and it can be a very subtle matter to figure out what agency does deserve the credit in such circumstances. .."
  11. Aug 23, 2007 #10


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    I suggest that you reconsider the statement that I put in boldface. Start with a straight wire parallel to the x-axis, with a constant current in the +x direction. If we ignore random variations from collisions, the electrons have a constant velocity in the -x direction. Put the wire in a uniform magnetic field that points in the +y direction. It experiences a constant magnetic force in the +z direction, and acquires a constant acceleration in that direction. After the wire has started moving, are the electrons still moving in a straight line in the -x direction? Are they still moving at constant speed? Is the magnetic force on the electrons in the same direction as the motion of the wire as a whole (the +z direction)?

    For simplicity, I assumed a uniform magnetic field, rather than the non-uniform field from another wire, but I think this changes only the details rather than the essential features of the situation.
    Last edited: Aug 23, 2007
  12. Aug 23, 2007 #11
    well if the charge is a moving one then by biot savart law it becomes amust that it experiences no force .but whta about magnets and ferromagnetic materials they are governed in some sense by couloumb's law.but that does not neccisate that the work done be 0?
  13. Aug 23, 2007 #12


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    Well, I deliberately didn't choose that example, because I expected somebody would make an argument that it was the electrical power supply doing work against the back EMF generated in the coil when the car moves and changes the magnetic field pattern, or something similar. That may or may not be the case, but it does't explain my permanent magnet question.
  14. Aug 23, 2007 #13
    Griffiths also writes briefly about electrons in atomic orbitals (in the part on paramagnetism, etc), basically saying that their average orbital speed changes if a magnetic field is applied (with a footnote linking this change to the electric field associated with varying the magnetic field).

    The argument seems to be that, just as an inclined plane can convert kinetic energy to gravitational potential, so to a permanent magnet lifting iron filings merely converts sub-atomic rotational kinetic energy to gravitational potential, with neither the plane nor magnetic field performing work itself.
  15. Aug 23, 2007 #14
    But you see, if a force acts on an object, and the velocity of the object has a nonzero component parallel to the force, the force definitely does work. It doesn't matter whether or not the force in question "caused" the motion in that direction or not. So even if the magnetic force wasn't the "agency which deserves credit" for the motion, as long as the magnetic force is not perpendicular to the velocity, it undoubtedly does nonzero work.
  16. Aug 23, 2007 #15
    Thanks Cesiumfrog for quoting Griffiths. I'll find the page number when I go to my office today. As Cesiumfrog alluded, Griffiths likens the magnetic force to the normal force that acts on blocks on inclines. If force is applied to move a block up a incline, the normal force never does work on the system, but rather it redirects the applied force.

    I was talking to some of the other grad student about this at the bar yesterday (yes, I talk about physics at the bar...). Apparently there's a diversity of opinions on this. I don't suppose there are any peer reviewed papers on this apparently non-trivial isue?
  17. Aug 23, 2007 #16
    Yes, but the normal force is always perpendicular to the motion, at least in the circumstance of pushing a block up an incline.
  18. Aug 23, 2007 #17
    Yes, but I think Griffiths' point is that at the microscopic level, the magnetic force is always perpendicular to the motion. And since the Lorentz force is the only way that magnetic fields can ever interact with matter (let me know if this is wrong in the classical limit), magnetic fields can thus do no work.
  19. Aug 23, 2007 #18
    Maybe its just the wording:

    they never do work (by themselves); and, they can do work (if associated with something else--like a rotor)
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