Nuru
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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
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.
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![]()
The two conducting rails in the drawing are tilted upwards so they make an angle of 30.0° with respect to the ground. The vertical magnetic field has a magnitude of 0.047 T. The 0.19 kg aluminum rod (length = 1.6 m) slides without friction down the rails at a constant velocity. How much current flows through the bar?
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.
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