The figure shows a wood cylinder of mass m = 0.193 kg and length L = 0.0989 m, w
ID: 1435194 • Letter: T
Question
The figure shows a wood cylinder of mass m = 0.193 kg and length L = 0.0989 m, with N = 30.0 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 ? to the horizontal, with the plane of the coil parallel to the incline plane. If there is a vertical uniform magnetic field of magnitude 0.650 T, what is the least current i through the coil that keeps the cylinder from rolling down the plane?
Explanation / Answer
Torque on coil = NIAB sin(theta)
Angular acc of coil = alpha
mgsin(theta) = ma
and alpha = a/r
so, alpha = gsin(theta)/r
Torquw = Igsin(theta)/r ( I is moment of inertia)
so NIAB sin(theta) = Igsin(theta)/r
NI*2*lB = m
I = 0.05 A
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