Magnetic flux through loop

Discussion in 'Physics' started by TsAmE, Oct 28, 2010.

  1. TsAmE

    Thread Starter Member

    Apr 19, 2010
    72
    0
    A rectangular loop of dimensions L and W moves with a constant velocity v away from a long wire which carries a current I in the plane of the loop. The total resistance of the loop is R. Derive an expression which gives the current in the loop at the instant the near side is a distance r from the wire.

    I dont understand why a clockwise current will be induced on the left side of the loop and an anti-clockwise current on the right side. Why isnt just one current induced?
     
  2. davebee

    Well-Known Member

    Oct 22, 2008
    539
    46
    What if you looked at this problem in two steps.

    Don't think of the motion as directly inducing current. Instead, think of the direct effect of the motion as generating EMFs (voltage) across the parts of the loop. You could work out the magnitude and direction of four separate EMFs, one for each section of the loop.

    Then, think of the total vector sum of the EMFs as what produces the overall resulting current flow.

    I think you'll find that the EMF of the far side of the loop is in the same direction as the near side, but because the magnitude of the nearside EMF is greater, the resulting current in the far side of the loop will be opposite to the direction you'd expect based only on the direction of the EMF induced in the farside section.
     
  3. steveb

    Senior Member

    Jul 3, 2008
    2,433
    469
    One current is induced.

    It's just a matter of perspective. If you go and stand on the other side of the loop (or equivalently spin the apparatus around), counter-clockwise becomes clockwise and vice versa.
     
    Last edited: Oct 28, 2010
  4. t_n_k

    AAC Fanatic!

    Mar 6, 2009
    5,448
    782
    Some thoughts on paper ...
     
    • loop.png
      loop.png
      File size:
      64.2 KB
      Views:
      115
  5. steveb

    Senior Member

    Jul 3, 2008
    2,433
    469
    Good answer!

    Actually, your answer made me realize that I misinterpreted the question. So, my answer should be ignored.

    So, if I can expand on your answer, you are saying that the current direction is the same in both legs of the loop (both going up on the right and left), but this translates into counter-clockwise and clockwise because of where the leg happens to be in the loop.
     
  6. TsAmE

    Thread Starter Member

    Apr 19, 2010
    72
    0
    Thanks. Why is it that there isnt any emf in the top and bottom parts of the loop? Also for the left and right sides of the wire the polarity I got was + at top and - at bottom by using the right-hand rule for magnetic forces: there is a magnetic force pointing upwards, which moves the protons to the top of the wire, and electrons to the bottom of the wire creating a clockwise current.
     
  7. Tesla23

    Active Member

    May 10, 2009
    318
    67
    Why not simply use Faraday's Law? If you sit on the loop you see the wire move away and you can work out \frac{\delta\Phi}{\delta t}
     
  8. t_n_k

    AAC Fanatic!

    Mar 6, 2009
    5,448
    782
    Yes I agree - my error on the diagram. I wasn't being particularly careful about the polarity.

    The ends of the loop won't develop any emf as they aren't cutting lines of flux as they move.
     
  9. TsAmE

    Thread Starter Member

    Apr 19, 2010
    72
    0
    Howcome? The horizontal parts of the loop are perpendicular to the magnetic field just like the vertical parts of the loop, so to me it looks like they still cut the flux lines. :confused:
     
  10. steveb

    Senior Member

    Jul 3, 2008
    2,433
    469
    You can use either Faraday's law or the Lorentz force. You get the same answer either way.

    Faraday's law doesn't clearly tell you where in the loop the EMF is being generated, while the Lorentz force law makes that clear.

    When Lorentz force is along the axis of the wire, current flows and voltage is developed across any resistance in the loop. The total loop resistance determines the loop current.

    When Lorentz force is perpendicular to the axial length of the conductor, you don't see a current generating EMF, however you do get a Hall effect, but that is hard to measure.
     
    Last edited: Oct 29, 2010
  11. steveb

    Senior Member

    Jul 3, 2008
    2,433
    469

    The Lorentz force is perpendicular to the velocity and the magnetic field. So, the horizontal legs have force that does not move charges in the direction of current flow. This side force is capable of creating a small voltage across the diameter of the wire, which is related to the Hall effect. The vertical legs have force along the axial length of the wire which is the direction of current flow in the loop.
     
  12. steveb

    Senior Member

    Jul 3, 2008
    2,433
    469
    Are you sure you were wrong? It looked correct to me when I looked at it.

    (I'll check it again later, but I'm tied up right now)

    EDIT: OK, I looked at it again, and agree with you. The force pushes electrons down, making the lower part negative.
     
    Last edited: Oct 29, 2010
  13. Tesla23

    Active Member

    May 10, 2009
    318
    67
    Interesting question - is it meaningful to say where the EMF is generated when it changes with the frame of reference?

    Also, if the wire is moving away I can't see how you get the equivalent of the Hall effect.

    They may give the same answer for the EMF, but they don't seem to be equivalent.
     
  14. steveb

    Senior Member

    Jul 3, 2008
    2,433
    469
    The attached paper goes into these issues better than I can.

    I haven't done a real analysis on this, so I could be wrong, but I'm thinking that the horizontal legs have current flowing in them and are in a magnetic field. Force will be exerted on these flowing electrons vertically (due to Lorentz force). Current can't flow vertically in the horizontal legs, but the force should build up negative charge on one side of the conductor. This is basically a Hall effect. This would be a small voltage, but Hall effects are generally small and sensors that use the Hall effect are made to be sensitive enough to measure these small voltages
     
    Last edited: Oct 30, 2010
Loading...