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Magnetic fields and loop of wire

  1. Aug 23, 2010 #1
    1. The problem statement, all variables and given/known data
    A rectangular loop of wire of size 5 cm x 15 cm is placed near a long straight wire with side CD at a distance of 5 cm from it, as shown in figure 6.29. What is the net force exetred on the loop (magnitude and direction)? How does your answer change if the current in the loop is reversed?


    2. Relevant equations
    F = BIl
    B = 4pi-7/2pi x I/r

    3. The attempt at a solution
    First I tried to figure out the net electric field at DC (same as AB). I added the magnetic field from the wire and the field from AB to get the net magnetic field at DC. But then I got confused, because what I have now is the net electric field. What lenght from the equation F=BIl should be used, is that 15 cm as that is the length of the loop? Or should I figure out the magnetic field from the wire and AB indivdually, calculate the force of each and then add them?
     

    Attached Files:

    • 6.29.bmp
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  3. Aug 23, 2010 #2

    kuruman

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    There is no electric field. Current carrying wires and loops are electrically neutral. The attraction is magnetic. Calculate the magnetic force on each loop segment and then find the net force. Note that the total force on the segment of the loop perpendicular to the wire is equal and opposite to the segment parallel to it on the other sides.
     
  4. Aug 23, 2010 #3
    I meant to write magnetic field, not electric field. But I need to calculate all for sides? The parallel are easy, but when it comes to the perpendicular, how do I add the force of these to the rest since they are perpendicular and not parallel?
     
  5. Aug 23, 2010 #4
    Nevermind my comment about the perpendicular sides, because they cancel out, right?
     
  6. Aug 23, 2010 #5

    kuruman

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    Right. At any particular distance from the wire there are equal and opposite forces acting on the perpendicular segments so the net force contributed by the perpendicular segments is zero.
     
  7. Aug 24, 2010 #6
    Okay, thank you!
     
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