- #1
Potatochip911
- 318
- 3
Homework Statement
A wood cylinder of mass ##m## and length ##L## with ##N## turns of wire wrapped around it longitudinally, so that the plane of the wire coil contains the long central 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 ##B##, what is the least current ##i## through the coil that keeps the cylinder from rolling down the plane?
Homework Equations
##\vec{\tau}=\vec{\mu}\times \vec{B}##
##\vec{\tau}=\vec{r}\times \vec{F}##
The Attempt at a Solution
Since the cylinder isn't rolling down the plane we have that ##\tau_{net}=0##, the torque from coil on the cylinder is ##\tau_{coil}=\vec{\mu}\times\vec{B}=NiAB\sin\theta## and the torque from the cylinder's weight is ##\tau_{w}=\vec{r}\times\vec{F}=rmg\sin\theta##, now setting these equal to each other solves the problem but I'm confused about the direction of these forces from the right hand rule/cross product. I'll draw a picture to show you:
Which makes no sense and I also can't get the torque from the cylinders weight to go in the correct direction, if someone could show/tell me what's wrong with my vector diagram that would be great.