The figure shows a wood cylinder of mass m = 0.6 kg and length L = 0.12 m, with

Question

The figure shows a wood cylinder of mass m = 0.6 kg and length L = 0.12 m, with N = 12.0 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 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 0.54 T, what is the least current i through the coil that keeps the cylinder from rolling down the plane?

Explanation / Answer

from the figure

mgsintheta-fs= ma

net torque is zero

There are two sources of torque acting in opposite directions of rotation: the force of friction tries to make the cylinder to roll down while the currrent creates a torque that tries to make it go uphill

T = Tfriction-Tcurrent

Tcurrent = mue*B = i 2rLB sintheta* N

f =i 2LB sintheta* N

i = mg/2NLB = 0.6*9.8/2*12*0.12*0.54

i = 3.78 A

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