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 2.49 cm by 4.12 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 10.9 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 .93 x 103 rev/min. (a) Find the maximum torque acting on the rotor. 7.69E-4 (b) Find the peak power output of the motor. 0.3164 (c) Determine the amount of work performed by the magnetic field on the rotor in every full revolution 0.002808 X Your response is within 10% of the correct value. This may be due to roundoff error, or you could have a mistake in your calculation. Carry out all intermediate results to at least four-digit accuracy to minimize roundoff error.J (d) What is the average power of the motor?

Explanation / Answer

In one half revolution the work is

W = Umax - Umin = -mu*B*cos180 - mu*B*cos0

W = 2*mu*B

mu = NIA

t = NIAB

t = 7.69 x 10^-4 N-m

W = 2 * 7.69 * 10^-4 = 1.538 x 10^-3

for 1 full revolution = W = 2 * 1.538 x 10^-3 = 3.076 x 10^-3 J

part d )

time for 1 revolution = dt = 60s/3930rev

Pavg = W/dt = 0.201 W

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