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Magnetic Field Induced Current Problem

  1. Mar 9, 2012 #1
    1. The problem statement, all variables and given/known data

    http://tinypic.com/r/rlidg7/5

    The picture above shows four different cases of a conducting square loop (in orange), traveling at speed v and passing through a rectangular region. In the left half of the rectangular region, the magnetic field is uniform and directed into the screen; in the right half, the magnetic field is uniform and directed out of the screen. The field has the same magnitude in each half of the rectangular region, and there is no magnetic field outside the rectangular region.

    Rank the four cases based on the magnitude of the total magnetic flux passing through the loop, from largest to smallest, at the instant shown in the picture.

    Rank the four cases based on the magnitude of their induced current, from largest to smallest, at the instant shown in the picture.

    What is the direction of the induced current for each case: counter-, clockwise, or none.

    2. Relevant equations

    I don't think I will be needing any equations for this problem.

    3. The attempt at a solution

    I think I got part a down. Cases A and C have zero total magnetic flux due to opposing directions of magnetic field. And Case B has the highest magnitude due to most field lines.

    I am really confused for the other two parts. How can there be an induced current at an instant? If it is moving for an instant afterwards, how can I tell what the magnitude of the induced current is?

    Additionally for the last part, the direction, I think Case C would have no direction. Case A seems to be getting more out of the page direction, so the opposing field of the induced current should be into the page. Using the right hand rule, would the direction of the induced current be clockwise? Using the same method, I feel that B would also be clockwise and D would be counterclockwise.

    Thanks for taking your time to read this and helping me out!
     
  2. jcsd
  3. Mar 9, 2012 #2

    NascentOxygen

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    Staff: Mentor

    http://img96.imageshack.us/img96/5725/red5e5etimes5e5e45e5e25.gif [Broken]
    I agree.
    At the moment the photo is taken, what is the current doing.
    I agree.
    I would say current is CCW, but your explanation is too vague, seems like a guess. :uhh:
     
    Last edited by a moderator: May 5, 2017
  4. Mar 11, 2012 #3
    Thanks NascentOxygen for the feedback and welcome.

    For the 2nd part, you asked what the current was doing at the moment the picture is taken.

    I first thought that if there was no movement, there would be no induced current, so all would be equal to zero. However, since there is a velocity at the instant, I thought that would be false.

    For case A, the loop is moving into the out of the page magnetic field. For case B, it is moving into the out of the page field. For case C, the the loop does not change direction of fields or strength. For case D, it is moving out of the out of the page field.

    Using the above, case A's induced current is opposing what will happen, so it will be CW. B will also be CW, C will have no direction, and D will be opposing the flux to becoming zero and be CCW.

    I feel that I am doing all of this incorrectly. And I have no idea on how to figure out the magnitude ranking of the induced current. Is it looking at how many field lines are through the loop in each case?

    The "instant that the picture is taken" is really throwing me off! Any further clarification would be appreciated.
     
  5. Mar 11, 2012 #4

    NascentOxygen

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    Staff: Mentor

    I think this question is tricky. Think of each loop as comprising 4 wires. Each wire is moving through a magnetic field. Some of the wires will have no induced voltage, such as where that wire is not cutting any field lines. Only if the sum of the 4 voltages comes to a non-zero value can we say there is a voltage induced in the loop.

    You need to apply one of the right/left hand rules. For example, this right hand generator rule: http://www.diracdelta.co.uk/science/source/f/l/flemings right hand rule/source.html

    Remember the rule applies to a straight piece of wire *
    and we have 4 straight pieces that make up each loop.
     
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