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Understanding motional emf

  1. Apr 30, 2014 #1
    I was reading this: http://web.mit.edu/8.02t/www/materials/StudyGuide/guide10.pdf
    (specifically page 10-6), and came across the formula;
    ## \mathcal{E} = -Blv##
    which is used for motional emf.

    Next, I was trying to solve a problem on page 10-33 (bottom problem)
    with a spinning bar. They broke up the bar into segments (dr) and used that formula and integrated to find the induced emf.

    I don't get why you can use that formula though in this problem's case, because isn't it only for a bar moving across a circuit?

    I guess i'm lacking in the theory; can someone explain why we can use it?
     
  2. jcsd
  3. Apr 30, 2014 #2
    It is for a bar moving in a magnetic field. No need for a circuit to have an emf.
     
  4. Apr 30, 2014 #3

    BiGyElLoWhAt

    User Avatar
    Gold Member

    Emf is the rate of change of flux with respect to time. When you have a bar (or a circuit) in your hand, and you give it a velocity v directed into a magnetic field what you have is

    ##\frac{\Delta \phi}{\Delta t} = \frac{\Delta (B\cdot A)}{\Delta t}## and if B is constant, then the only Chang in flux is due to the change in area, and the change in area is due to the velocity granting the equation you have.

    B is constant and delta a is l times delta width/delta t and delta width is the velocity
     
  5. May 1, 2014 #4
    So does #\mathca{E} = -Blv# give the emf produced by the rod moving?
     
  6. May 1, 2014 #5
    By the rod moving in a magnetic field when the velocity is perpendicular to the field.
    To be more precise.
     
  7. May 1, 2014 #6
    Two things:

    1] Physically, does this mean that if we attach wires to the ends of the bar and to a light bulb while it is moving, the light bulb will turn on?
    2] Is the correct version ##-Blv\sin\theta## where ##\theta## is the angle between the field and velocity?
     
  8. May 1, 2014 #7
    1. It depends on the geometry of the connectors. The emf is only well-defined on closed loops, and it is proportional to the time derivative of the flux through the loop. So how you connect the wires will affect how the flux changes.

    2. Again it depends on the geometry, but your basic trig will give you the net flux if the surface normal vector enclosed by the wire loop is not parallel the magnetic field. So in short, yes.
     
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