The figure shows a wood cylinder of mass m = 0.327 kg and length L = 0.141 m, wi

Question

The figure shows a wood cylinder of mass m = 0.327 kg and length L = 0.141 m, with N = 13.0 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 0.619 T, what is the least current i through the coil that keeps the cylinder from rolling down the plane? the tolerance is +/-2%

Explanation / Answer

Torque T = N i * A * B * sin theta

Torque created by weight T = r * F = m g R sin theta

m g R = N i A B

m g = 2 N i L B

0.327 * 9.8 = 2 * 13 * 0.141 * 0.619 * i

i = (0.327 * 9.8) / (2 * 13 * 0.141 * 0.619)

i = 1.412 A

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