How Does the Angle of Incline Affect Current in a Sliding Rod Experiment?

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SUMMARY

The discussion centers on calculating the current flowing through a sliding aluminum rod in a magnetic field during an experiment involving inclined rails. The setup includes a vertical magnetic field of 0.048 T and an angle of incline of 30.0°. The participant initially misapplied the sine function in the force calculation but later corrected it to use sin(60°) for the angle, leading to the correct current calculation. The final current determined was approximately 14.036 A.

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  • Understanding of Newton's Second Law (F=ma)
  • Knowledge of electromagnetic induction principles (F=ILB sin(θ))
  • Familiarity with trigonometric functions and their application in physics
  • Basic concepts of magnetic fields and forces on conductors
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  • Study the principles of electromagnetic induction in detail
  • Learn about the effects of angles on forces in inclined planes
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Students and educators in physics, particularly those focusing on electromagnetism and mechanics, as well as anyone conducting experiments involving magnetic fields and inclined planes.

kpangrace
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:rolleyes: 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 0.048 T. The 0.22 kg aluminum rod (length = 1.6 m) slides without friction down the rails at a constant velocity. How much current flows through the bar?
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ok so I'm really completely lost here, but let me give you guys what I'm thinking, see if you could lead me in the right direction

I'm expecting that you don't use that 30 degrees as part of F= ILB sin(theta), but rather you use it to find force with F=ma right? if that's the case, how would you find the acceleration? would it just be 9.8 for gravity, or would it be something else because you're not going straight down... but then i have a problem with velocity being constant, which means there would be no acceleration... but there has to be gravity...

(i'm sorry if i seem all over the place :confused: just confused)

so here's what I've done and it didn't work

F=ma

F= .22 * 9.8sin30
F= 1.078
F= I L B sin(theta)
1.078= I 1.6*.048 sin 90
I =14.036

thats not the right answer, but to me that seems the only logical way to go about answering this problem... any info would help thank you!
 
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Nevermind i figured it out, its sin(60), not sine of 30 or 90... you can kill this thread if you wish!
 

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