Current in relation to magnetic fields and velocity

AI Thread Summary
The discussion revolves around calculating the current flowing through an aluminum rod sliding down inclined conducting rails in a magnetic field. The problem specifies a vertical magnetic field of 0.047 T and an incline angle of 30 degrees. The user attempts to apply the formula F = ILBsin(θ) but struggles with incorporating the gravitational force correctly. They derive an equation for current, I, but suspect they are misapplying the gravitational component along the incline. Clarification on the correct approach to account for the incline's effect on gravitational force is sought.
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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
p21-35.gif

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?
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 F = ILB\sin{\theta}. My first approach was to use the relation F\,=\,MA and substitute that in so I had: MA\,=\,ILB\sin{\theta}

This turns into I\,=\, \frac{MA}{LB\sin{\theta}}

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

then I\,=\,49.5212766

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 \theta 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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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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