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Force on a Magnet in Lenz's Law

  1. May 9, 2015 #1
    1. The problem statement, all variables and given/known data
    As the bar magnet (see attached diagram) leaves the loop of wire, it experiences a force opposing its exit. I understand why, according to Lenz's law, a force would oppose the bar magnet's exit. I don't understand, however, what causes the force. In this case, the force is rightwards.

    In other words, what are the mechanics that cause the magnet to experience a rightwards force?

    2. Relevant equations

    See diagram.

    Lenz's Law: Current is induced in a surface to oppose the change in magnetic flux through it.
    3. The attempt at a solution

    Here, the current induced produces a magnetic field that points in the same direction as the magnet. Hence, as the magnet tries to leave, another magnetic field somehow opposes its motion...
     

    Attached Files:

  2. jcsd
  3. May 9, 2015 #2

    andrevdh

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    The moving magnet causes a change in magnetic flux through the coil. This induces current in the coil according to Faraday' s law of electromagnetic induction. The current in the coil now sets up a magnetic field of its own. Lenz's law helps us determine in which direction the induced current in the coil will be.
     
  4. May 9, 2015 #3

    Hesch

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    In short: Lenzes law says that the nature will counteract any changes in magnetic flux. So when pulling the magnet away from the loop, a current will be induced in the loop that substitutes the flux of the magnet by its own flux. All in all the amount (volume) of flux is increased.

    The density of energy in a magnetic field, Emagn = ½*B*H [ J/m3]. So when magnet and loop are separated, energy must be created, to fill out the "added volume" of magnetic energy density.

    F = ΔE / Δs = ( Emagn * ΔV ) / Δs = Emagn * A. ( s is distance, A is cross section area of field, V is volume of field ).
     
  5. May 9, 2015 #4

    andrevdh

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    All that Lenz's law enables us to say is that in this case the induced current will increase the flux through the coil and thereby opposing the removal of the magnet. It is just a qualitative law not quantitative. You seem to try and evaluate the opposing force that the magnet is experiencing from the induced magnetic field.
     
  6. May 9, 2015 #5

    Hesch

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    I'm finding the opposing force by means of energy in a magnetic field, and are not using Lenz's law to do this.
     
  7. May 9, 2015 #6

    andrevdh

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    Don't think I can help you there, but keep in mind that it is the change in magnetic flux that brings this about. That is the induced current causes an increase in flux, not necessarily such that the total flux is kept constant. In practice one would probably measure the required force to remove the magnet and thereby calculate the induced flux change responsible for it.
     
  8. May 9, 2015 #7

    Hesch

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    The flux is not increased, the induced current in the loop will try to keep flux steady, and could do so, were the loop a superconductor. But the volume of the fluxdensity is increased, thus the energy is increased.

    What is wrong with this:

    ? ?
     
    Last edited: May 9, 2015
  9. May 10, 2015 #8

    andrevdh

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    I would have thought that the induced current created more flux lines and thereby increased the overall energy density, but I am not at work now and can't reference my sources.
     
    Last edited: May 10, 2015
  10. May 11, 2015 #9

    andrevdh

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    The induced magnetic field from the current in the coil is in the same direction as that of the magnet that is removed
    from the coil. The coil thus acts as an additional magnet. It attracks the magnet as it is removed from the coil.
    The source of the force is thus the coil. This effect was also used in old car speedometers - Arago's rotation effect
    http://www.physics.montana.edu/demonstrations/video/5_electricityandmagnetism/demos/aragosdisc.html
    https://prezi.com/dyvhcumzt6k_/electromagnetism/
    The two objects become locked together and a force is required to separate them. Hope this is what you wanted to know.
    P5110034.JPG P5110035.JPG P5110036.JPG
     
    Last edited: May 11, 2015
  11. May 11, 2015 #10

    Hesch

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    This is wrong. The induced current in the loop will just try to keep the flux steady. It will never increase the energy density, but (here) the volume in which the steady energy density acts, thus increasing the total magnetic energy: ΔE = Emagn*Δvolume. The loop-current, creating some flux, will just substitute the disappearing flux from the magnet as it is removed. This leads to:
    I wanted to know if you agreed with that calculation?

    Say that the flux were increased through the loop, a back-emf would immediately be created in the loop since emf = dψ/dt.
     
    Last edited: May 11, 2015
  12. May 11, 2015 #11

    andrevdh

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    So your formula is suggesting the force is a result of a gradient created in the magnetic energy density?
    The term Emagn*ΔV (have for a long time) worries me.
    What does this term mean?
    What would the ΔV signify?
     
  13. May 11, 2015 #12

    Hesch

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    ΔV = change in volume.
    Emagn * ΔV = ( magnetic energy density ) * Volume. In units: [ J / m3 ] * [ m3 ] ) = [ J ].
     
  14. May 11, 2015 #13

    andrevdh

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    As you move a distance Δs along the field you go through a small volume ΔV where the field exists?
     
  15. May 11, 2015 #14

    Hesch

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    Yes: Δs * A = ΔV. ( A = cross section area ).
     
  16. May 11, 2015 #15

    andrevdh

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    I still would think that you need an energy gradient for a force to be present, that is the energy needs to change if
    you moved a small distance in the field, that is Emagn should change.
     
  17. May 11, 2015 #16

    Hesch

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    No, the total energy must change:

    ΔE / Δs is your gradient.
     
  18. May 11, 2015 #17

    andrevdh

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    According to some source I can get my hands on the force is proportional to - dH2/ds
    but I really don't know enought about the subject.
     
  19. May 11, 2015 #18

    Hesch

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    Emagn = ½*B*H = ½*μ0*H*H. Thus the force is proportional to - dH2/ds.
     
  20. May 11, 2015 #19

    andrevdh

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    Sorry, I am not happy with what you are saying.
    I would think that the force should be dependent on the change in the energy
    and not just its value, that is if the energy stays constant there would be no force.
     
  21. May 11, 2015 #20

    Hesch

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    Yes, that's what I'm saying: F = ΔEtotal / Δs.

    ΔEtotal = 0 → F = 0

    Etotal is not the same as Emagn
     
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