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Electromagnetic Induction Related Circuit

  1. Dec 22, 2008 #1
    1. The problem statement, all variables and given/known data

    [​IMG]

    Examine the circuit shown in figure 12. When the switch is closed, assume the magnetic field produced by the solenoid is very strong.

    a) Explain what will happen to the copper ring (which is free to move) when the switch is closed
    b)Explain what will happen to the copper ring when there is a steady current in the circuit.
    c)Explain what will happen to the copper ring when the switch is opened.
    d)How would your answers change if the terminals on the power supply are reversed? Explain your answer.

    2. Relevant equations

    Law of electromagnetic induction
    An electric current is induced in a conductor whenever the magnetic field int he region of the conductor changes with time.

    Lenz's Law
    When a current is induced in a coil by a changing magnetic field, the electric current is in such a direction that its own magnetic field opposes the change that produced it.

    3. The attempt at a solution

    a) When the switch is closed current will flow from the positive to the negative. Based on the right hand rule the bottom of the coil will be "N" while the top of the coil will be "S". But I don't know how the copper ring will react. It'll probably be attracted towards the coil but I don't know why.

    b) When there is a steady current in the circuit the copper ring will no longer be attracted to the current since the current becomes constant? Again I have no idea why.

    c) When the switch is open the current will stop flowing and the copper ring will no longer be affected?

    d) I have no idea.

    Could anyone explain some of the concepts I need to know to solve this problem? I'm having difficulty understanding what the copper ring and the coil, current, etc do.

    Any help will be greatly appreciated.

    Thanks
     
  2. jcsd
  3. Dec 22, 2008 #2

    Defennder

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    Well then why did you think it will be attracted? Think in terms of what it means for a initial changing magnetic flux through the ring when the switch is first turned on.

    I can't remember if copper is a magnetic material, but knowing that should tell you the answer.

    Again use Faraday's law and think of applying Lenz law and the direction of the counter-induced magnetic field.

    Hint. The copper ring isn't a magnet.
     
  4. Dec 22, 2008 #3
    Ok so copper is not a ferromagnetic material so it shouldn't be affected by the magnetic current generated by the solenoid?

    a) When the switch is closed a current can be seen in the solenoid (curl) part of the circuit. This generates a magnetic field but the copper will not move because it is not a ferromagnetic material?

    b) Steady current means that the solenoid curl part will have no current.

    c) When the switch is opened there will be a brief current in the solenoid curl part, one that is opposite the current measured in a)

    I don't understand what the significance of the copper is when it is not a ferromagnetic material.
     
  5. Dec 22, 2008 #4
    Copper won't move so why does the diagram hint that it is free to move?

    What will happen to copper is that it will have a current in it for a) and c) as a counter induced magnetic field, whereas for b) nothing will happen
     
  6. Dec 23, 2008 #5

    Defennder

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    Though it is not a ferrogmagnetic material, it is a conductor. And that means that you should expect there to be a magnetic field due to induced current.

    Not quite. I assume gravity plays no role here because otherwise the ring would fall. What is the nature of the magnetic field due to the solenoid? How does this affect the B-flux through the ring and what does this in turn cause?

    Don't know what you mean by curl but I interpret it to mean the shape of the solenoid. And no that's not true. You will have a current when there is a closed circuit with a voltage source.

    Incomplete. The question asks what happens to the ring.

    Suppose the copper ring were replaced by an iron ring. Then what can you say about how you expect the iron ring to move under steady current in the neighbouring circuit? And with this in mind, how would you deduce what would happen when the B flux through the iron ring is altered?

    Why won't it move?

    Think of the induced magnetic field due to the induced current caused by the changing magnetic flux.
     
  7. Dec 23, 2008 #6
    I still don't really understand all of this induced current business but here's an educated guess.

    a) The copper ring will move towards the solenoid. This is because the solenoid generates a magnetic field where the solenoid's top is S, and the bottom is N. Due to this magnetic field an induced current in the copper ring will form where the current flows to the left, thus the copper ring generates a magnetic field of its own that is attracted towards the south end of the solenoid.

    b) no change in current therefore the copper will not move since there is no induced current.

    c) the copper will be repelled upwards. this is because a voltage opposite to that of a) develops in the solenoid due to a change in current. I really don't understand why though?

    Am I on the right track?

    I still don't think I understand the fundamental principles involved in this scenario, which is why I have no idea what the heck is going on. This is frustrating, I'm good at kinematics but electromagnetism is confusing for me.
     
  8. Dec 23, 2008 #7

    Defennder

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    Not quite. Bear in mind that the induced magnetic field in caused by the copper ring's induced current always opposes the change in magnetic flux producing it. It's kind of confusing here, but you can understand it easily by remembering that the magnetic field induced by a changing magnetic flux always opposes the B-field which causes the change.

    Correct, since the copper isn't ferromagnetic.

    This part is very similar to that of a).

    Well the part on Lenz law is admittedly confusing. For me, I found mechanics more difficult than electromagnetism.
     
  9. Dec 23, 2008 #8
    Wait, so for both A) and C):

    The copper ring will move away from the solenoid?

    This is because the copper ring has an induced current that produces a magnetic field that opposes the field created from the solenoid?

    Am I correct though that the copper ring will have current that goes to the left? As such, the top of the copper ring will be N, while the bottom S which will lead to a repulsion between the copper ring and the solenoid (since the top of the solenoid is S).
     
  10. Dec 23, 2008 #9

    Defennder

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    Half correct. I did say they were similar, but not exactly the same. For c), when the solenoid's magnetic field weakens the induced current in the ring will generate a magnetic field to reinforce that decaying field. What is the direction and polarity of that magnetic field?

    It depends on whether you are opening or closing the switch.
     
  11. Dec 26, 2008 #10
    Hey Defennder,

    Thanks for sticking with me on this difficult concept. I hope you had a great Christmas.

    So to summarize this:

    [​IMG]

    First of all when the switch is turned on the bottom of the solenoid is N, while the top is S.

    Therefore for

    A) The copper ring will have an induced current that goes to the left. This in turn generates an induced magnetic field, where the bottom of the copper ring is S, while the top is north. This is because the copper ring is trying to resist the sudden field of the solenoid. Anyways due to this interaction the copper ring will be repelled and will move away from the solenoid.

    C) In this case since the solenoid's magnetic field is deteriorating, the copper ring will try to strengthen the field of the solenoid. As such, the copper ring's induced current will go to the opposite direction of right. This in turn generates an induced magnetic field, where the bottom of the copper ring is N, while the top is S. As such the copper ring will move towards the solenoid because it is attracted to the deteriorating field of the solenoid.
     
  12. Dec 27, 2008 #11
    Thats true.
     
  13. Dec 27, 2008 #12

    Defennder

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    Yeah, that appears correct, except perhaps for the direction of induced current. The direction of the induced current should be consistent with the magnetic field generated to oppose the change.
     
  14. Dec 5, 2009 #13
    what would be the answer to this?
     
  15. Nov 13, 2010 #14
    Hi Everyone, I'm new here and I was hoping to get some help with this question. I understand parts a, b, and c, but I'm having trouble with d.

    Since the terminals are reversed, the current would flow in the opposite direction. This would reverse the poles on the solenoid so that the top would be the North pole and the bottom would be the South pole. But, all my answers in a, b, and c would remain unchanged, would they not?

    For (a) the copper ring would still induce a magnetic field opposing the solenoid, the difference being that the poles of the copper ring's magnetic field would be reversed. But, the copper ring would still move up.
    For (b) the copper ring would remain in its place of equilibrium since the solenoid is not inducing a magnetic field when there is a steady current.
    For (c) the copper ring would have a magnetic field to strengthen the solenoid and would be pulled down towards the solenoid. The difference again being that the poles on the copper ring would be reversed in relation to the original question.

    From this I would assume that there is no change in the answers. But, I think I must be wrong because it does not make sense that the text would ask the question if there were no changes.

    Thank you in advance :)
     
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