What is the least current to prevent the cylinder from rolling down the incline?

AI Thread Summary
To determine the least current required to prevent a wood cylinder from rolling down an incline, the problem involves a cylinder with a mass of 0.250 kg and a length of 0.100 m, wrapped with 10 turns of wire in a vertical magnetic field of 0.5T. The relevant equation is derived from the force equation, leading to the conclusion that the least current needed is 2.45A, assuming gravitational acceleration is correctly taken as 9.8 m/s². The discussion highlights the challenge of eliminating the sine and cosine components in the torque calculations. Clarifications on the calculations indicate that the values used for gravitational acceleration were corrected. The final answer confirms that 2.45A is indeed the correct current to prevent rolling.
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Homework Statement


A wood cylinder of mass m = 0.250 kg and length L = 0.100 m with N =10 turns of wire wrapped around it longitudinally, so that the plane of the wire coil contains the long axis of the cylinder. The cylinder is released on a plane inclined at an angle theta to the horizontal, with the plane of the coil parallel to the incline plane. If there is a vertical uniform magnetic field of magnitude 0.5T, what is the least current i through the coil that keeps the cylinder from rolling down the plane?

Homework Equations


force = NiL x B

The Attempt at a Solution


The answer is 2.45A, i try to find all the forces and solve when torque is zero, but i can't elimate the sin theta or cos theta thing. Please help, thankyou.
 
Last edited:
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Can you show what you worked out so far? It might be possible for somebody to spot your error (if any) that way.

p.s. if it helps, you can copy-and-paste this theta symbol: θ
 
2NiLBr sinθ = μmgr cosθ
mgsinθ = μmg cosθ

the answer is 2.45 if g = 9.8, i used 10 before lol, correct now?
 
Last edited:
Looks good (I presume you are able to eliminate the sinθ & cosθ now). :smile:
 

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