The rotor in a certain electric motor is a flat, rectangular coil with 79 turns
ID: 1446884 • Letter: T
Question
The rotor in a certain electric motor is a flat, rectangular coil with 79 turns of wire and dimensions 2.69 cm by 3.68 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 3.57 Times 10^3 rev/min. Find the maximum torque acting on the rotor. Find the peak power output of the motor. Determine the amount of work performed by the magnetic field on the rotor in every full revolution. Your response differs from the correct answer by more than 10%. Double check your calculations. J What is the average power of the motor?Explanation / Answer
part (C) for one full revolution
the amount of work done by the magnetic field W = 4NIAB
Where N = number of loops I = Current A = Area of Loop B = Magnetic Field.
so that W = 4*79*0.0109A*0.0269m*0.0368m*0.800T
W = 2.72*10-3 T
but you got 1.3638*10-3 T that is the ans for half revolution.
formula clarification
It comes from the potential energy formula for the magnetic moment, which is the dot product of the moment and the B field:
U= - mu B cos()
where (mu=NIA)
and is the angle between the moment (perpendicular to the loop) and the B field.
Now look at half of a revolution.
At the beginning, the moment is opposite the B field. What is the potential energy? After one-half of a revolution, the moment is in the same direction as the B field; what is the potential energy then? The work done is the (negative of the) change in the potential energy, so that gives the answer for half of a revolution. Doubling that gives the coefficient of 4 for the full revolution.
Do you get that answer?
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