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Magnetic Braking

  1. Apr 21, 2007 #1
    When a rectangular loop of wire is drawn into a uniform magnetic field, there will be an induced current. What I don't understand is whether there are two components to the motion of the electrons, one of the current relative to the wire, and one of the whole wire relative to the field.

    If we just consider the constant current relative to the wire, then the top and bottom edge's magnetic force cancel out. However, after taking into account the movement of the whole loop, there should be a net downward force.

    However, my textbook states that there is only a force to the left. Maybe I'm looking at the wrong reference frame. Could someone explain this?

    Thanks.
     
  2. jcsd
  3. Apr 22, 2007 #2

    Andrew Mason

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    The motion of the electrons due to the motion of the wire does not contribute to an electromagnetic effect here and can be ignored.

    The induced emf in the loop is the time rate of change of the flux through the loop ie: the motion of electrons is caused by the magnetic flux inside the loop changing.

    If the loop is being drawn into a uniform magnetic field from a region where there is no field, so that part of the loop is outside and part is inside, there is a change in the magnetic flux ([itex]\phi = \int B\cdot dA[/itex]) through the loop and an emf in the loop: [itex]emf = d\phi/dt[/itex]. If it is simply moving in a uniform magnetic field, there is no change in the flux through the loop so there is no emf.

    [/quote]I might have to see a diagram here to understand where you are having a problem.

    I assume the rectangular loop is moving to the right into the magnetic field. The flux through the loop is increasing, so there is an induced emf in the wire. The emf in the loop creates a current in the loop. The current in the vertical section of wire on the right side of the loop creates a magnetic field around the wire that interacts with the applied magnetic field in a way that opposes its motion through the field - so it pushes to the left. Since the vertical section on the left is not in the applied field, there is no force on the left side of the loop. So the net force is to the left. The current in the horizontal sections of the loop interact with the applied magnetic field and push in opposite directions so they cancel out. The total net force is left, opposite to the direction of motion.

    AM
     
  4. Apr 22, 2007 #3
    Thanks for your response, however, I'm having trouble understanding this part:

    Isn't the net drift velocity of the electrons exceedingly slow? In that case, the motion of the wire might actually be faster than the motion of electrons relative to the wire.
     
  5. Apr 22, 2007 #4

    Andrew Mason

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    The motion of the wire moves the electons and protons at the same speed. The wire is electrically neutral. How are you suggesting that this motion creates an electromagnetic effect?

    AM
     
  6. Apr 22, 2007 #5
    the forces on the electrons will only be according to the velocity component which is perpendicular to the magnetic field, therefor it is only dependent on the velocity of the current in the wire. why? because the velocity of electrons can be made of two components, the velocity in reference to the wire, and the velocity of the wire itself - the velocity of the wire is in the same direction as the magnetic field therefor will not create force, and the other velocity is perpendicular to the magnetic field, therefor will create force.

    so the force will only be dependent on one component of the velocity, the one in relative to the wire.
     
  7. Apr 22, 2007 #6
    Oh, I think I see where I went wrong then. :cool: Thanks for your help!
     
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