Why Does the Calculation of Work in an Electric Motor Involve Multiplying by 4?

[Ithryndil]

Problem:

The rotor in a certain electric motor is a flat, rectangular coil with 90 turns of wire and dimensions 2.50 cm by 4.00 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, it carries a current of 9.1 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 3600 rev/min.

(a) Find the maximum torque (max) acting on the rotor.
Nm
(b) Find the peak power output (max) of the motor.
W
(c) Determine the amount of work (W) performed by the magnetic field on the rotor in every full revolution.

(d) What is the average power (avg) of the motor?

"Then part C is given by 4(9.1 x 10-3A)(90)(0.025mx0.04m) = 0.00262J"
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The above was asked by another user. I also had this problem and had a question on part C. The user above correctly finds C, I am wondering why the answer is what it is. The rest of the problem makes sense to me. Here's what I have below, I am not sure where the 4 is coming from.


4NIAB

Where
N = number of loops
I = Current
A = Area of Loop
B = Magnetic Field.

Thanks.

 

[alphysicist]

Hi Ithryndil,

Problem:

The rotor in a certain electric motor is a flat, rectangular coil with 90 turns of wire and dimensions 2.50 cm by 4.00 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, it carries a current of 9.1 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 3600 rev/min.

(a) Find the maximum torque (max) acting on the rotor.
Nm
(b) Find the peak power output (max) of the motor.
W
(c) Determine the amount of work (W) performed by the magnetic field on the rotor in every full revolution.

(d) What is the average power (avg) of the motor?

"Then part C is given by 4(9.1 x 10-3A)(90)(0.025mx0.04m) = 0.00262J"
------------------------------------------------------------------------

The above was asked by another user. I also had this problem and had a question on part C. The user above correctly finds C, I am wondering why the answer is what it is. The rest of the problem makes sense to me. Here's what I have below, I am not sure where the 4 is coming from.


4NIAB

Where
N = number of loops
I = Current
A = Area of Loop
B = Magnetic Field.

Thanks.

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(θ)

where (mu=NIA) and θ 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?

 

[baba89]

Hello,

Thank you for your question. It seems like there may be some confusion regarding the units in the solution provided for part C. The answer should be in Joules (J), not Newton meters (Nm). Additionally, the 4 in the solution represents the number of times the rotor rotates in one full revolution (since the rotor turns through one-half revolution in each step and the process is repeated twice), not a constant value. Therefore, the correct solution for part C should be:

W = 4(9.1 x 10^-3 A)(90 turns)(0.025 m)(0.04 m) = 0.00262 J

I hope this clarifies the solution for you. Let me know if you have any other questions or concerns.