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Current in relation to magnetic fields and velocity

  1. Oct 28, 2007 #1
    Magnetic Field on an incline

    This is what looks to be a fairly simple magnetic field problem, but I don't have much experience in relating velocities to electrical currents, so I'm a bit stuck. The problem as described is
    So my variables to my knowledge are
    I = unknown
    m = 0.19 kg
    B = 0.047 T
    a = 9.8 m/s (gravity)
    L = 1.6 m

    The formula I know for this sort of thing is simply [tex]F = ILB\sin{\theta}[/tex]. My first approach was to use the relation [tex]F\,=\,MA[/tex] and substitute that in so I had: [tex]MA\,=\,ILB\sin{\theta}[/tex]

    This turns into [tex]I\,=\, \frac{MA}{LB\sin{\theta}}[/tex]

    So [tex]I\,=\,\frac{(.19 kg)(9.8 m/s)}{(1.6 m)(0.047 T)\sin{30}}[/tex]

    then [tex]I\,=\,49.5212766[/tex]

    However, that obviously isn't right. Did I miss an important detail? I found another problem which was identical except that it had a frictional coefficient they applied in the numerator and they had a [tex]\theta[/tex] of 90. Unless I misunderstood something it seems like this should be solvable the same way. Any help would be appreciated.
     
    Last edited: Oct 28, 2007
  2. jcsd
  3. Oct 28, 2007 #2
    I think the problem is that I'm using the gravity component in the y axis instead of for the plane, but I'm not sure how to do that correctly.
     
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