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?