The figure shows a cylinder of mass 3.1 kg, radius 3.9 cm and length 8.9 cm with

Question

The figure shows a cylinder of mass 3.1 kg, radius 3.9 cm and length 8.9 cm with 110 turns of wire wrapped around it lengthwise, so that the plane of the wire loop is parallel to the incline and contains the axis of the cylinder. What is the least current which while flowing through the loop will prevent the cylinder from rolling or sliding down the inclined plane in the presence of a vertical magnetic field of B = 1.12 T? The angle of inclination of theta = 43 degrees. The plane of the windings is parallel to the inclined plane. You should assume that the wires are wound much tighter than the figure implies (ie, assume that the wire loop has the same dimensions as the cylinder).

Explanation / Answer

here,

m = 3.1 kg

r = 3.9 cm = 0.039 m
L = 8.9 cm = 0.089 m
Area = 2rh+2r^2
A = 2 *3.14 * 0.039 * 0.089 + 2 * 3.14 * 0.039^
A = 0.031 m^2

n = 110 turns
B = 1.12 T
Theta = A = 43 degrees

as we know Torque = N*I*A X B

Writing Components of force acting on cyclinder,

Fx = 0 = F - mg sinA
F = mgSinA
F = 3.1 * 9.8 * Sin 43
F = 20.719 N

around Centre of mass :
Torque = 0
tb + tf = 0
NIA X B + 20.719 * r = 0
NI*2*r*L*B*SinA 20.719 * r = 0

I = mg/(2*n*l*B*SinA)

I = (3.1 *9.8)/(2*110*0.089*1.12*Sin43)

I = 30.38/(220*0.09968*0.68199)

I = 2.0313 A

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