The figure shows a wood cylinder of mass m = 0.304 kg and length L = 0.0898 m, w

Question

The figure shows a wood cylinder of mass m = 0.304 kg and length L = 0.0898 m, with N = 25.0 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 0.568 T, what is the least current i through the coil that keeps the cylinder from rolling down the plane?

Explanation / Answer

Here ,

m = 0.304 Kg

L = 0.0898 m

N = 25

B = 0.568 T

for balancing the torque of the forces

2 * N * I * L * B * r * sin(theta) = u * m * g * r * cos(theta)

m * g * sin(theta) = u *m * g * cos(theta)

u * cos(theta) = sin(theta)

2 * 25 * I * 0.0898 = 0.304 * 9.8

solving for I

I = 0.664 A

the current in the cyclinder is 0.664 A

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