Torque on a Frictionless Incline? I'm trying to think of some conceptual aspects of a problem similar to this one.
http://physicstasks.eu/550/a-cylinder-on-an-inclined-plane
The problem basically asks us to consider a cylinder with wire loops wrapped around the cylinder such that the wire loops are parallel to the incline. Then the question asks us to determine what current flow keeps the cylinder from rolling down the incline in the presence of a magnetic field of known strength.
I basically set the magnetic torque equal to the torque caused by gravity and solved for the current. This worked well and I got the right answer, but my question is, would the magnetic torque still oppose the downward force of gravity on a frictionless surface? All the free body diagrams I drew for the problem assumed a surface with friction, but I don't think this would work on a frictionless surface.
My thought is that it wouldn't and, if the cylinder was on a frictionless incline, the cylinder would still rotate because of the magnetic torque, but would slide down the incline because of the force of gravity. 
edit.
Picture of basic set up.

 A: If you make a free body diagram for the system you will realise that without friction it is impossible for the system not to move.

You don't even need to consider the torque. If you apply Newton's 2nd law you can see it.
$$\sum_i \overrightarrow{F_i} = \overrightarrow{F_{B, 1}}+\overrightarrow{F_{B, 2}}+\overrightarrow{W}+\overrightarrow{N}=m\overrightarrow{a}$$
The two forces due to the magnetic field cancel out, and you are left with:
$$\overrightarrow{W}+\overrightarrow{N}=m\overrightarrow{a}$$
But $\overrightarrow{W}$ and $\overrightarrow{N}$ are not colinear so there is no way that they cancel out. The cylinder has to slide down. The net force is,
$$mg~\sin{(\alpha)}\hat{u}_T=m\overrightarrow{a}$$
where $\{\hat{u}_N, \hat{u}_T\}$ are the unit normal and tangent vectors to the plane.
As you can see, the motion of the center of mass is exactly the same as the case for a "normal" frictionless incline.
What happens with the torque? If we assume that both the magnetic field $\overrightarrow{B}$ and the current $\overrightarrow{I}$ are constant, then the forces $\overrightarrow{F_{B, i}}$ are constant. This will result in a rotation that is very similar to a pendulum (picture the location where $\overrightarrow{F_B}$ acts as the pendulum mass, there is no gravity but the constant force produces the same motion).
So the cylinder slides down the incline while at the same time it oscillates "rotationally". 
A: Your intuition is exactly right. In particular, the torque "from gravity" is actually coming from the friction at the point of contact. Because we're worried about rotation about the center of mass of the object (as it rolls), gravity itself cannot provide a torque (because the lever arm is zero). So gravity only acts on the center of mass and generates only linear motion. 
Similarly, the forces on the wire from the magnetic field cancel so that there is no net force on the cylinder's center of mass, so the linear motion is unaffected. The torques, however, add up and you get rotation. In this way, if the ramp is frictionless, the linear motion of the cylinder down the ramp and the rotational motion from the magnetic field completely decouple and you can treat them as two completely separate problems. 
In particular, the rotational motion is interesting because the cylinder will rotate back and forth, oscillating about the orientation aligned with the magnetic field while it slides down the ramp with the usual $g \sin(\theta)$ acceleration. 
