The figure below shows a wooden cylinder with mass m = 0.125 kg and length L = 0

Question

The figure below shows a wooden cylinder with mass m = 0.125 kg and length L = 0.200 m, with N = 10.0 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 0.500 T, what is the least current i through the coil that keeps the cylinder from rolling down the plane?
A

Explanation / Answer

in this problem torque due to gravity is balanced by the magnetic torque
magnetic torque =Bsin => =NidLBsin

torque due to gravity =rFsin =>=dmgsin/2

NidLBsin =dmgsin/2

i =mg/2NLB

i =0.125*9.8/2*10*0.2*0.5

i=0.6125Amp

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