The figure below shows a wooden cylinder with mass m = 0.225 kg and length L = 0

Question

The figure below shows a wooden cylinder with mass m = 0.225 kg and length L = 0.800 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 0 to the horizontal, with the plane of the coil parallel to the incline plane. If there is a vertical uniform magnetic field of magnitude 0.400 T, what is the least current i through the coil that keeps the cylinder from rolling down the plane?

Explanation / Answer

In this 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 =0.225*9.8/2*30*0.8*0.4

i=0.115 A

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