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

[dlslhc]

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.

 

[Redbelly98]

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

 

[dlslhc]

2NiLBr sinθ = μmgr cosθ
mgsinθ = μmg cosθ

the answer is 2.45 if g = 9.8, i used 10 before lol, correct now?

 

[Redbelly98]

Looks good (I presume you are able to eliminate the sinθ & cosθ now).