You wind a square coil 0.125 m on a side from a 5.00-m long piece of wire that h

Question

You wind a square coil 0.125 m on a side from a 5.00-m long piece of wire that has a mass of 0.240 kg. You then suspend the coil from one side (with a hinge arrangement) in a 0.0400 T magnetic field directed vertically upward by running a current of 2.00 A through the coil as shown.

(a) When the coil is in rotational equilibrium, determine the angle that the plane of the coil makes with the vertical.
° with respect to the vertical

(b) Determine the magnitude of the torque acting on the coil due to the magnetic force.
N · m

Explanation / Answer

a) Number of turns, n = 5m/(0.125*4) =10

Torque about the hinge at equilibrium = 0

mg sin beta *0.125/2 - niLB*0.125 cos beta = 0

0.240*9.8*0.125/2   tan beta - 10*2*0.125*0.04*0.125 = 0

0.147 tan beta = 0.0125

tan beta = 0.0125/0.147 = 0.085

beta = 4.86 degree with vertical

b) Magnitude of torque due to magnetic forces = niLB*0.125 cos beta

                                              = 10*2*0.125*0.04*0.125 cos 4.86 degree

                                             = 0.012455 Nm

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