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Lenz Law copper tube problem modification

  1. Nov 30, 2012 #1
    1. The problem statement, all variables and given/known data

    If we drop a bar magnet down a copper tube, it will travel down very slowly. If we now cut vertical slits in the copper tube, will the magnet travel down faster or slower as compared to the original problem?

    2. Relevant equations

    [tex] \int \vec E \cdot d \vec l = -\frac{d \Phi}{dt} [/tex]
    [tex] V = Q/C [/tex]


    3. The attempt at a solution

    As a model, I imagine horizontal strips of the copper tube located above and below the location of the magnet. If we look at a strip below the magnet, we can view the strip as a circuit with several capacitors. Given the same EMF, a circuit with capacitors will have a lower current as compared to a similar circuit without capacitors. The force on the magnet is proportional to the current in the strips, so there will be less force if we cut vertical slits in the tube. Thus the magnet will drop more quickly.

    Is my model/logic correct here? Any comments would be great!

    Thanks,

    Eric
     
  2. jcsd
  3. Dec 1, 2012 #2
    The magnet falls slowly in the "full" tube because the interaction force between the magnet and the induced magnetic dipole (induced current in the tube) points upwards.
    If you cut vertical slits the induced current drops a lot and so does the interaction force. Thus the magnet falls in a shorter time.
    The slits you cut on the tube's surface look like capacitors but they're not.
     
  4. Dec 1, 2012 #3

    rude man

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    Think of the copper tube a a series of coils around the perimeter of the tube. Each coil sees a changing magnetic field as the magnet drops above and then below it. Each coil thus develops a current by i = emf/R where emf = -d(phi)/dt, phi = coil area x B, and R = resistance of the coil. By Lenz's law, what is the direction of the force on the falling magnet produced by each coil's magnetic field? If you then cut a vertical slit as described, what does that do to R of each coil, and consequently to i and B?

    Why is it irrelevant that the coils are "shorted" to each other in the vertical direction?
     
  5. Dec 1, 2012 #4
    Why not? There is a common problem where you view a slit in a fat wire as a capacitor. Why can't we use the same idea here? It leads me to the same conclusion that the magnet drops in less time if slits are added...
     
  6. Dec 1, 2012 #5
    I dont know. That is why I put together a model that views the coil as a circuit with capacitors. This led me to the conclusion that there will be less current in each coil as compared to a coil without slits, and thus less force. Is this a valid way to see the problem?

    What exactly do you mean by the coils being "shorted" to eachother??

    Thanks!
     
  7. Dec 1, 2012 #6
    Usually, the slits aren't cut along the full length of the tube. Thus you don't have the "plates" you need in a capacitor. Certainly, the slits will perform in a complicated way at microwave frequencies, but this isn't the case. Rude man gave a good, alternative explanation.
     
  8. Dec 1, 2012 #7

    rude man

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    No. As others have pointed out, capacitance does not figure in this situation.

    I don't think I drew the picture adequately. The coils are SINGLE-TURN ones running down the tube, stacked one on top of the other, like rings on a ring-tailed cat. You can for the moment assume they're insulated from each other vertically by tiny strips of non-conducting tube. So each 1-turn coil sees a dB/dt as the magnet drops into its opening from the top, and a current is induced.

    Your tube is actually a continuous piece of metal so the 1-turn coils are actually connected to each other vertically. Hint: the induced currents run circularly along the perimeter of the tube and not vertically.

    It should be obvious what happens to that (and all the other) single-turn coil currents if each coil is cut by the vertical slit. R = ∞, i = 0 and therefore B = 0 and force = 0 also.
     
  9. Dec 2, 2012 #8
    This doesn't seem right to me.. If we are viewing the tube as single-turn coils piled on top of eachother, then when we cut slits in the coils, we have exactly the situation of the fat wire problem (cut a slit in a fat wire and it acts as a capacitor). The falling magnet certainly produces an EMF in the coils, so there should be a current (as there is in the fat wire problem if an EMF is introduced)!
     
  10. Dec 2, 2012 #9
    I am hearing a lot of different conclusions here. Rude man says the magnet will fall as if there is no copper tube at all. Gordianus says the magnet will drop quicker when the slits are introduced but still be slowed a bit by the tube. What am I to make of this?

    Both of you seem to think the model with the capacitors has no merit but I am not quite convinced of this either...
     
  11. Dec 2, 2012 #10
    Actually I just found this video showing the magnet fall slowly in the tube with slits but not as slowly as compared to a tube without slits

     
    Last edited by a moderator: Sep 25, 2014
  12. Dec 2, 2012 #11

    rude man

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    I checked that out & it appears the slits do not go all the way up & down the entire tube.

    If the slit is part-way only, then circular currents will still be induced in the section without the slit. If the slit were to run all the way up & down the tube there would be no braking action at all on the falling magnet.

    BTW it only takes one slit. Adding extra slits parallel to the first one changes nothing.

    Eric, you will have to explain the details of how capacitance across (?) the slit affects the falling magnet for us.

    EDIT - on second viewing I'm not persuaded the magnet fell more slowly with the slit tube than it would in air. The video I saw showed only two tubes, one without slits & one with, and that the magnet fell faster with the slits. I did not see aa comparison between the slit tube and just in air.
     
    Last edited by a moderator: Sep 25, 2014
  13. Dec 2, 2012 #12
    Look again... there is no horizontal strip on the bar unaffected by a slit. When he turns the bar you see there are extra slits to make sure of this. And the footage is sped up. The magnet in the tube with no slits takes 20 seconds to fall. The other magnet follows shortly after... surely much more slowly than a free-fall.

    In the original post I have provided an explanation of the capacitor model. Looking at a horizontal strip of copper tube, we see something like a fat wire with gaps in it. The EMF produced due to the falling magnet pushes electrons towards the edges of the gaps, which i believe should make the gaps act as capacitors (as they do if we were to cut slits in a fat wire... which I think is roughly analogous)
     
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