1. PF Contest - Win "Conquering the Physics GRE" book! Click Here to Enter
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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.
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook