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Work done by magnetic field

  1. Jun 16, 2004 #1
    We all know that the work done by the magetic field is zero. But, how does the magnetic field lift iron objects?
     
  2. jcsd
  3. Jun 16, 2004 #2
    "We all know that the work done by the magetic field is zero."

    Only if the magnetic force is perpendicular to the motion, as with the cyclotron problem.
     
  4. Jun 16, 2004 #3

    Chi Meson

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    Work is only zero if the magnetic force (aka Lorenz force) is on a free, moving charged particle, as opposed to current in a wire or ferromagnetic atoms in a solid.

    THe magnetic force, by definition, can only act in a direction perpendicular to the field. If the object is bound within a solid (as in electrons moving through a wire) then this force can become constant in direction. WHereas, if the force acts on a single moving particle, the force will act as a centripetal force (perpendicular to direction of motion). Centripetal forces cannot change the speed of an object, so no work is done.

    Two magnets attract each other and one can lift the other since the charged particles within the magnets are not free to travel in the prescribed circles, so the net force on the lifted object will be constant in direction (well, more or less).
     
  5. Jun 18, 2004 #4

    Gza

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    It's funny, I asked my physics professor the same type of question reguarding why exactly a permanent magnet can move another magnet without violating the principal of the magnetic field not being able to do work, and he didn't have an answer.
     
  6. Jun 18, 2004 #5
    It's funny he couldn't answer this because the answer is pretty obvious. A magnet won't move another magnet unless someone first does the work of pushing them close enough together that their attraction draws them together. It's like a golfball; it won't fall down into the hole unless someone putts it over the edge of the hole.

    Once your magnets are stuck together, you won't get any more of this apparent work out of them unless you do the work of pulling them apart. In the end you are doing all the work, not the magnets.
     
  7. Jun 18, 2004 #6
    I'm not sure what you're saying here. The set up sounds like this to me: you have a wire carrying current and a permanent magnet sitting still near the wire. You are saying the magnetic field from the magnet is changing the kinetic energy of the electrons?
     
    Last edited: Jun 18, 2004
  8. Jun 18, 2004 #7

    ZapperZ

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    I think you missed the whole point of the question here. If you have two magnets floating in space, there is a force of attraction between the N-pole of one magnet onto the S-pole of another. This alone will cause the two to move towards each other, which by basic definition, is work done on one by the other. This shows that there's a force acting on the two that actually does work. So the original question then is if a magnetic field does no work, what's going on here?

    Now back to the original question. When a student, or anyone learning E&M, starts to ask question such as this, it means that he/she is beginning to understand and be aware of the subject matter, rather than blindly going through the material without even thinking. So this is a GOOD thing!

    As Chi Meson as alluded to, the "boundary" of the charges in the magnetic material may be playing a part in this. However, there is another aspect of this problem that may play a bigger role. If anyone has looked at the typical Stern-Gerlach experiment, where a beam of electrons is passed through a magnetic field, you'll notice that the field being used to separate out the spin orientation of the electrons is NOT UNIFORM. In fact, there is a deliberate magnetic field gradient transverse to the path of the electrons. It is this gradient in the magnetic field that exert a force onto the electrons of different spin orientation so that they separate out along the direction of the field!

    When you have a bar magnet, for instance, you'll remember from way back in elementary school by looking at the field lines using iron filings, that the magnetic field isn't uniform, i.e. they don't all come out in straight lines and with equal spacings between the lines over all space. If you put a magnetic dipole in such a field, there will always be a net force simply due to the gradient of the field. This is the origin of a substantial portion of the force you detect between two magnets.

    Zz.
     
    Last edited: Jun 18, 2004
  9. Jun 18, 2004 #8
    With the magnets in space, they will accelerate toward each other then stop dead when they finally contact. There is force and motion here, but not all force and motion equals work.

    Work
    Address:http://www.batesville.k12.in.us/physics/PhyNet/Mechanics/Energy/Work.html

    It is a tricky situation with the magnets, but it looks to me that, despite the force and motion, you end up with 0 change in kinetic energy for both magnets.

    It would be work if A. the magnets accelerated each other and then somehow kept going, or B. if they were already in motion and exerted force to stop each other.

    As it is, the force they exert is an accelerating force and the stopping is a matter of the resultant collision.

    I believe the case of two magnets in space most resembles the 4th example at that site, where a force is exerted on the wall, but the wall doesn't experience any net motion. In that case, no work is done.
     
  10. Jun 18, 2004 #9

    ZapperZ

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    Er...hello? All I need to show is that the FORCE causes a displacement of the object with a component along the same direction of the force! Acceleration isn't needed! Double check the definition of "Work done". It is the integral of the dot product between the displacement and the force! I have no idea why you are bringing up "acceleration".

    The reason why the magnetic component of the Lorentz force does NO work on the charge is because it is always perpendicular to the displacement of the charge. This isn't the case when you have the two magnets, as in the situation HERE!

    Zz.
     
  11. Jun 18, 2004 #10
    Don't give yourself an aneurism, ZapperZ.
     
  12. Jun 19, 2004 #11

    ZapperZ

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    I won't if you start paying attention to the stuff you were trying to "teach" me.

    Zz.
     
  13. Jun 20, 2004 #12

    Gza

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    You both brought up interesting points, but I think this argument stems from a disagreement with regards to semantics. To simplify the question, from my point of view: as a thought experiment, place a magnet in free space, and hold it fixed. Now place the N pole of another magnet aligned with the original magnet's N pole. This magnet obviously accelerates away, and my question is why?

    Since work is defined as:

    [tex]W = \int \vec{F} \cdot \vec{ds} [/tex]

    and [tex]\vec{v} = \frac {\vec{ds}}{dt}[/tex]

    obviously implying the velocity of an object to be in the same direction of ds; while v remains perpendicular to the force from the B field from the cross product in the relation:

    [tex] \vec{F_B} = q\vec{v} \times \vec{B}[/tex],

    and any beginning student in physics can tell you that any force perpendicular to the displacement of an object does no work. So the question is, what exactly is doing the work to change the kinetic energy of the magnet?
     
    Last edited: Jun 20, 2004
  14. Jun 20, 2004 #13

    ZapperZ

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    But that is EXACTLY what I was describing.

    (i) The SIMPLEST form of a "magnet" is a magnetic dipole. There are no monopoles.

    (ii) The B-field of a dipole isn't uniform, i.e. it doesn't produce the uniform field from an infinite plane of charge, for example.

    (iii) The non-uniformity of the field creates a gradient - this is the origin for the net force acting on the dipole.

    (iv) it does mean that if you place this dipole in a uniform field, such as at the very center of a very long solenoid, there will be no net force acting on it other than to cause the dipole moment to align with the field, i.e. there will be no net displacement.

    Zz.
     
  15. Jun 20, 2004 #14
    At some point a changing field induces an electric field. With an electric field, work can be done. The trick is figuring out where the field pops up - with permanent magnets it isn't obvious the way it is with a current-carrying wire.
     
  16. Jun 21, 2004 #15

    krab

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    A magnetic field does no work on a moving charge. But there are other forces in nature besides this Lorentz force. If there were magnetic monopoles, they would attract and repel each other with a law exactly analogous to Coulomb's law for electric charges. Since there are only dipoles, one can derive the force law between two magnets directly from this "Coulomb's Law" for magnetic "charges". It is a Different Force from the Lorentz force.
     
  17. Jun 21, 2004 #16

    Gza

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    But to me at least, it seems you must assume the existence of monopoles for this to work.
     
  18. Jun 21, 2004 #17

    krab

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    Why? A dipole is a combination of two opposite poles. The force between two dipoles is derivable from the force between each one of the 4 charges. It is immaterial whether or not the dipoles can be separated into individual charges. Secondly, a permanent magnet with its little individual dipoles all lined up is to a very good approximation a system where there are charges of type n at one end and charges of type s at the other. Think of two magnets: one of all n charges and the other of all s charges, occupying the same solid, but that the n charges are ever so slightly displaced w.r.t. the s charges. In between the N and S poles of the magnet, all the n's and s's cancel, leaving a surface at the S end with left over s charges, and the surface at the N end with left over n charges.
     
  19. Jun 21, 2004 #18
    The more I think about it, the less I think the free magnet will accelerate away.

    I think it will move away from the first magnet enough to do a 180 so the opposite pole faces the first magnet, whereupon it will be attracted.

    Edit: Nevermind. After experimenting with some magnets I think in space it probably would shoot far enough away fast enough to escape before the first magnet can turn it.
     
    Last edited: Jun 21, 2004
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