Current in relation to magnetic fields and velocity

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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, an incline angle of 30.0°, and a rod mass of 0.19 kg. The participant attempts to apply the formula F = ILBsin(θ) but struggles with incorporating the gravitational force correctly. The correct approach involves resolving the gravitational force along the incline to accurately determine the current using the formula I = (mg sin(θ)) / (LB). The participant's initial calculation of I = 49.5212766 A is incorrect due to improper handling of the gravitational component.

PREREQUISITES
  • Understanding of magnetic fields and forces (Lorentz force)
  • Knowledge of inclined plane physics
  • Familiarity with Ohm's Law and current calculation
  • Basic trigonometry for resolving forces
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  • Study the Lorentz force law and its applications in magnetic fields
  • Learn how to resolve forces on inclined planes in physics
  • Explore the relationship between current, magnetic fields, and motion
  • Review examples of similar problems involving friction and inclined planes
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Physics students, educators, and anyone interested in electromagnetism and mechanics, particularly those dealing with problems involving magnetic fields and inclined planes.

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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.
 
Last edited:
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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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