The rotor in a certain electric motor is a flat, rectangular coil with 86 turns

Question

The rotor in a certain electric motor is a flat, rectangular coil with 86 turns of wire and dimensions 2.29 cm by 3.84 cm. The rotor rotates in a uniform magnetic field of 0.800 T. When the plane of the rotor is perpendicular to the direction of the magnetic field, the rotor carries a current of 11.1 mA. In this orientation, the magnetic moment of the rotor is directed opposite the magnetic field. The rotor then turns through one-half revolution. This process is repeated to cause the rotor to turn steadily at an angular speed of 3.27 × 103 rev/min. (a) Find the maximum torque acting on the rotor. 0.00067 N·m (b) Find the peak power output of the motor. Your response differs from the correct answer by more than 10%. Double check your calculations. w (c) Determine the amount of work performed by the magnetic field on the rotor in every full revolution. (d) What is the average power of the motor?

Explanation / Answer

torque = N*I*A*B*sintheta

torque = 86*11.1*10^-3*0.0229*0.0384*0.8*sin90

maximum torque = 0.00067 Nm

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power = torque*angular speeed

angular speed = 3.27*10^3 rev/min = 3.27*2pi/60 = 0.342 rad/s

P = 0.00067*0.342

P = 0.000229 W

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C)

work = N*I*A*B*(costheta1-costheta2)

theta2 = 180

theta1 = 0

work = 86*11.1*10^-3*0.0229*0.384*(cos0-cos180)

work = 0.0168 J

===============

d)

average power = peak power/2 = 0.000229 /2 = 0.0001145 W

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